Khan Academy Link: Limits and continuity

All topics covered in Khan Academy:
Green underlined topics are topics at least partially covered on my website and red topics are topics not yet covered on my website. Note that even green topics might not be covered in full detail on my page.

1. Defining limits and using limit notation
2. Estimating limit values from graphs
3. Estimating limit values from tables
4. Determining limits using algebraic properties of limits: limit properties
5. Determining limits using algebraic properties of limits: direct substitution
6. Determining limits using algebraic manipulation
7. Selecting procedures for determining limits
8. Determining limits using the squeeze theorem
9. Exploring types of discontinuities
10. Defining continuity at a point
11. Confirming continuity over an interval
12. Removing discontinuities
13. Connecting infinite limits and vertical asymptotes
14. Connecting limits at infinity and horizontal asymptotes
15. Working with the intermediate value theorem
16. (Optional) Formal definition of limits

Calculus is all about change! In the world of calculus, we often deal with very small changes. And I mean very, very small changes, even changes that are infinitely small. But what does that even mean? How do we define “infinitely small”?

Let’s start our journey with one of the most important problems in calculus, the tangent line problem. But before we get into it, let’s see an example of this problem coming up in real life. It’s always good to relate the math concepts you learn to the real world!

Mathematics is a way for us to quantitatively describe our world. It might feel like knowing math is just knowing how to manipulate numbers and symbols to solve abstract problems, but math is used all the time around us to describe real-world situations and solve all sorts of problems related to them. And often, new math concepts are introduced to solve new types of problems.

For example, let’s say that you were on a road trip one day and you drove 120 miles between 4 and 6 PM. How fast have you been driving on average during those 2 hours? We can represent this situation with a simple equation, using the fact that speed is distance divided by time:

Simple arithmetic is enough to describe this situation. But what if you wanted to know exactly how fast you were driving at a specific time, say, 5:02 PM? Not some sort of average over a time period, but your exact speed at that instant. In this hypothetical scenario, you were driving on a freeway during this time, so you suspect your speed was greater than 60 miles per hour.

One idea is to use a very small time period to estimate your speed at 5:02 PM. Because speed is equal to distance divided by time, you could take the distance that you traveled between 5:02 and 5:03 PM and divide that by the time elapsed, 1 minute (1/60 of an hour). Let’s say you traveled 1.1 miles within that 1 minute. Using this information, we can once again calculate an average speed.

But this doesn’t tell you exactly how fast you were going at 5:02, that just tells you your average speed between 5:02 and 5:03. How could we do better?

We could use an even smaller time period, like 1 second, and now we would be taking the distance traveled between 5:02:00 and 5:02:01 PM and dividing by 1 second (1/3,600 hour). This will get us closer to our speed at 5:02, but it still won’t be exact.

Here’s what it might hypothetically look like if we calculated our speed over smaller and smaller time intervals:

As our time interval gets smaller and smaller, we get closer and closer to the true answer of how fast we were moving at exactly 5:02 PM. (Think about it, our average speed in the last second should be closer to our current speed than our average speed in the last minute.) Looking at the table, it seems like our true speed at 5:02 is 65 mph, but how could we know for sure?

To solve this problem, we need a way to repeat this process to infinity. In other words, we need a way to measure our speed over an infinitely small time interval.

We might try an interval of 0 seconds, but then we run into a problem: no matter how fast we travel, over a period of 0 seconds, we travel a distance of 0 miles, because anything multiplied by 0 is 0. Let’s try finding the speed with our distance over time formula:

As you can see, using a time interval of zero doesn’t work out. So we need a way to measure speed over a time interval that isn’t zero but is also as small as possible. This is where normal arithmetic breaks down. We need a new branch of mathematics to help us. That branch is calculus!

Let’s go back to the tangent line problem. In our speed problem, we were trying to find our speed at a single instant of time. The tangent line problem boils down to the same question, but in a more abstract context:

How do we find the slope of a function at a single point?

Knowing the exact slope (or rate of change) of a function at a single point has many applications. The slope at a point tells us how fast the function is increasing or decreasing at that point — exactly how fast it’s changing at that particular instant! If we had this knowledge, we could model how functions change with better accuracy and precision than ever.

How can we find the slope of the red function at the blue point? In other words, how fast is the red function increasing at the blue point?

Finding the slope of a line between two points is simple: you simply calculate the change in \(y\) between the two points divided by the change in \(x\), or “rise over run”.

But what if we want to find the slope of a function at just one point? The rise over run formula won’t work in this case — it gives the average rate of change between two points, but not the exact slope at a single point. We need a way to define what we mean by “slope at a single point”.

Here’s an idea: to find the slope at a single point, we take a second point on the function and move it closer and closer to the first point. Then, we determine what the slope between those two points approaches.

We are trying to find the slope of the function at the red point, and to do that, we are moving the blue point closer and closer to the red point. As we do that, the slope of the green line approaches a certain value. If we could find this value, we could say that this value is the slope at the red point!

How do we mathematically describe this situation? The blue point is moving closer to the red point, but how close do the two points have to be before the slope of the green line actually reaches the slope at the red point?

It turns out that no matter how close the points are to each other, the green line will never reach the actual slope we are trying to find (as it changes a little each time we move the second point). So is all hope lost?

No: there is a way to make sense of this. Notice that the distance between the points is getting smaller and smaller; in other words, it is approaching zero. We need a way to describe what happens to the slope as the distance between the points approaches zero, and this introduces us to a fundamental tool of calculus - the limit!

I will go into more detail about limits in the next section, Intro to Limits.

In our attempt to find the slope at one point, we took a second point and moved that second point closer and closer to the first point. We are trying to find what the slope between the two points approaches. But how do we define “approach”? We use something called a limit.

This is the graph of \(f(x) = 2x\). What happens to the \(y\)-value as \(x\) approaches 2? We can use a limit to answer this question.

Use the slider below to explore what happens to \(f(x) = 2x\) as \(x\) approaches 2:

\(f(x)\) approaches 4 as \(x\) approaches 2. However, we only tested what happens if \(x\) starts at a value smaller than 2 and increases towards 2. Let’s see what happens if \(x\) starts above 2:

\(f(x)\) still approaches 4. Let’s summarize our results in the language of mathematics now!

A limit in mathematics is the value an expression or function approaches as a variable approaches a certain value. For example, the “limit of \(\class{red}{2x}\) as \(\class{blue}{x}\) approaches 2” is 4, because as \(\class{blue}{x}\) gets closer to 2, the value of \(\class{red}{2x}\) gets closer to 4. This is written in mathematical notation as:

Importantly, we can make \(2x\) as close to 4 as we want (“arbitrarily close” to 4) as we move \(x\) closer to 2. Also, it doesn’t matter if \(x\) is increasing or decreasing towards 2, \(2x\) will approach 4 either way. These two conditions are required for a limit to exist.

\(\displaystyle\lim_{x\to 2} 2x = 4\) because as shown by the tables, we can make \(2x\) as close to 4 as we want by moving \(x\) closer and closer to 2. It doesn’t matter if \(x\) is increasing (1.9 → 1.99 → 1.999 → ...) or decreasing (2.1 → 2.01 → 2.001 → ...), \(2x\) will approach 4 either way.

It is super important that both of these tables approach the same value (4)! If these tables approached different values, the limit would not exist. In general, a limit must approach the same value from both sides in order to be considered a valid limit.

You might have noticed that using a limit was unnecessary here because we could have just plugged in \(x = 2\) into the expression \(2x\) to get the same answer 4. However, limits become much more powerful when we can’t just plug in the \(x\)-value into the expression.

One example of a function that can be problematic is \(f(x) = \frac{x^2}{x}\). If we plug \(x = 0\) into this function, we get \(\frac{0}{0}\), which is undefined! However, limits can come to the rescue here. Use these sliders to see what happens to \(f(x) = \frac{x^2}{x}\) as \(x\) approaches 0.

This is the graph of \(f(x) = \frac{x^2}{x}.\) This function is undefined at \(x = 0\), but we can still describe what happens near \(x = 0\) by using a limit! Looking at the graph, what happens to \(f(x)\) as \(x\) approaches 0 from either side? (The open circle at \((0, 0)\) means that \(f(x)\) is undefined at \(x = 0\).)

Limits are especially useful in situations where we would otherwise be dividing by 0. In this situation, we found that as \(x\) approaches 0, \(\frac{x^2}{x}\) also approaches 0 from both sides. This means that \(\displaystyle\lim_{x \to 0}{\frac{x^2}{x}} = 0\).

Notice how we were still able to get an answer for the limit as \(x\) approaches 0 even though the function is undefined at \(x = 0\). That’s because, importantly, the value of \(f(0)\) is not relevant when taking the limit as \(x\) approaches 0. The limit describes what happens when we approach 0, not what happens when we actually get to 0.

That’s a very important thing to keep in mind: for any value \(c\), the value of \(f(c)\) or whether it is defined or not will never have an impact on the limit as \(x\) approaches \(c\). This next example will make this very clear.

Here is a mysterious function \(f(x)\), designed to test your knowledge of limits. Use these two sliders to figure out the limit of \(f(x)\) as \(x\) approaches 2.

Here is a table of the function’s value at a few points. \(f(x)\) is equal to 100 if \(x = 2\) and is equal to \(2x\) otherwise.

Don’t be fooled! Even though the value of \(f(x)\) at \(x = 2\) is 100, that has no impact on the limit as \(x\) approaches 2. Remember, limits describe what happens as you approach a value, not what happens when you reach it! \(\displaystyle\lim_{x\to 2} f(x)\) is equal to 4 in this case.

I said previously that if a limit approached different values from both sides, the limit would not exist. Are there any other situations where limits don’t exist? Keep reading to find out in the next section!

There’s one very important thing to keep in mind about limits. In some cases when trying to find a limit, a limit simply won’t exist (there is no value that could reasonably be considered the limit). There are three situations where a limit will not exist.

For each of these three situations, I will have an example function that you can explore using some sliders. Try to figure out why the limit doesn’t exist in all three of these examples!

To summarize, here are the 3 causes for limits not existing:

1. The limit from the left side is different from the limit from the right side. In other words, \(f(x)\) approaches a different value if \(x\) increases towards \(c\) than if \(x\) decreases towards \(c\).
2. \(f(x)\) increases or decreases without bound as \(x\) approaches \(c\). In other words, \(f(x)\) increases or decreases forever as \(x\) gets closer to \(c\) and never approaches a finite value. This is known as approaching positive or negative infinity.
3. \(f(x)\) oscillates between different values as \(x\) approaches \(c\). In other words, as \(x\) gets closer to \(c\), \(f(x)\) bounces up and down between different values but never approaches one specific value.

One of the reasons a limit might not exist is if the limit is different from both sides. However, in these cases, we can still describe what happens as \(x\) approaches some value from the left side and right side separately. We just have to use one-sided limits.

Consider the piecewise function \(f(x) = \begin{cases} x + 2 & \text{if } x > 0 \\ x - 1 & \text{if } x < 0 \\ 1 & \text{if } x = 0 \\ \end{cases}\)

What is \(\displaystyle\lim_{x \to 0}{f(x)}\)? Let’s break it down. As \(x\) approaches 0 from the left side (meaning that it starts below 0 and increases towards 0), \(f(x)\) is equal to \(x - 1\) (since \(x\) is always less than 0). This means the limit as \(x\) approaches 0 is -1 from the left side.

As \(x\) approaches 0 from the right side (starting above 0 and decreasing towards 0), \(f(x)\) is equal to \(x + 2\) (since \(x\) is always greater than 0). This means the limit as \(x\) approaches 0 from the right side is 2.

Important reminder: the value of \(f(x)\) at \(x = 0\) is not relevant to either of these limits.

To find these one-sided limits, we looked at the piecewise function and only paid attention to the parts that were relevant for each limit. For example, to find the right-side limit, we only looked at what \(f(x)\) is equal to when \(x\) is greater than 0 (since \(x\) will be greater than 0 when approaching 0 from the right).

In this case, \(\displaystyle\lim_{x \to 0}{f(x)}\) does not exist because the limit from the left side is different from the limit from the right side. However, we can still describe the two conflicting limits separately with one-sided limits.

In this case, the left-side limit is \(\displaystyle\lim_{x \to \class{red}{0^-}}{f(x)} = -1\) and the right-side limit is \(\displaystyle\lim_{x \to \class{blue}{0^+}}{f(x)} = 2\). Notice the “-” and “+” next to the 0 in each limit expression that shows whether it is a left-side or right-side limit.

Look at the two sets of functions below. What is the main difference between them?

The first set of functions don’t have any gaps in them, unlike the second set of functions. You could draw each of the first set of functions on a piece of paper without lifting your pencil, which is something you can’t say for the second set of functions.

The mathematical term for this is continuity. The first graph shows continuous functions and the second graph shows discontinuous functions.

There are three main types of discontinuities, or ways that a function can become discontinuous. All three ways are shown in the second graph.

• Point discontinuities / removable discontinuities: a single hole in the graph, as shown by the blue function in the second graph above. They are “removable” because you can remove the discontinuity just by adding a single point (at the hole) to the function.
• Jump discontinuities: when the graph suddenly jumps from one \(y\)-value to another, as shown by the green function in the second graph above.
• Asymptotic discontinuities / infinite discontinuities: a vertical asymptote; when the graph goes up to infinity or down to negative infinity, as shown by the red function in the second graph above.

An example of each of the three main types of discontinuities.

The function \(f(x) = \frac{1}{x}\) is continuous over the open interval \((0, 10)\). Can you figure out why?

A function is continuous over an interval if there are no discontinuities in that interval. For example, the function \(f(x) = \frac{1}{x}\) is continuous over the interval \((0, 10)\) because there are no discontinuities between \(x = 0\) and \(x = 10\). (Because \((0, 10)\) is an open interval [meaning it doesn’t include the start and end values, as shown by the parentheses around \((0, 10)\)], the function’s behavior at \(x = 0\) and \(x = 10\) is not relevant. This means the discontinuity at \(x = 0\) doesn’t affect the continuity over this interval.)

More formally, a function \(f(x)\) is continuous at the point \(x = c\) if \(\displaystyle\lim_{x \to c}{f(x)} = f(c)\). In other words, \(f(x)\) is continuous at \(x = c\) if the limit as \(x\) approaches \(c\) from both sides is equal to the function’s actual value at \(x = c\).

Try finding the limit as \(x\) approaches 0 for each of these discontinuous functions - they won’t equal the function’s value at \(x = 0\). (In fact, all three functions are undefined at \(x = 0\).)

The value of \(f(x) = x^2\) at \(x = 2\) is 4, and the limit as \(x\) approaches 2 of \(f(x)\) is also 4, so \(f(x)\) is continuous at \(x = 2\).

Here are some examples of functions that are continuous across their entire domain:

• Power functions with positive exponents: \(x\), \(x^2\), \(3x^3\), etc.
• Polynomials: \(x^2 + 3x\), \(x^4 - 30x + 2\), etc. (Remember, polynomials cannot contain negative exponents.)
• Exponential functions: \(e^x\), \(2^x\), etc.
• Logarithmic functions: \(\ln(x)\), \(\log_{10}(x)\), etc.
• Sine and cosine functions: \(\sin(x)\), \(\cos(2x)\), etc.

This is a graph of the continuous function \(f(x) = x^3 + 1\). We will only focus on the interval \([-1, 1]\) (the blue region in the image above). Notice how as the function progresses through this interval, the function takes on every single \(y\)-value between 0 and 2 at some point (every \(y\)-value in the green region). Why 0 and 2? Because \(f(-1) = 0\) and \(f(1) = 2\). Those are the \(y\)-values of \(f(x)\) at the start and end of this interval.

This is an example of the intermediate value theorem (IVT). It states that if a function \(f(x)\) is continuous over an interval \([a, b]\), it will take on every value between \(f(a)\) and \(f(b)\) somewhere in that interval. In our example above, our interval is \([-1, 1]\), so \(a = -1\), \(b = 1\), \(f(a) = 0\), and \(f(b) = 2\). \(a\) and \(b\) are the bounds of the blue region, and \(f(a)\) and \(f(b)\) are the bounds of the green region in the graph above. The intermediate value theorem only applies if \(f(x)\) is continuous over the interval \([a, b]\)!

The intermediate value theorem makes sense intuitively, since if a function is continuous over the interval \([a, b]\), there’s no way for it to skip over any \(y\)-value in between \(f(a)\) and \(f(b)\). (In order for a function to skip over a \(y\)-value, you need to pick up your pencil when drawing its graph, creating a discontinuity!)

You should try it yourself: try drawing a function in the above graph that is continuous and passes through \((-1, 0)\) and \((1, 2)\) (the bottom left and top right corners of the square). You will find that the function will take on every \(y\)-value between 0 and 2 at some point within the square!

More formally, the intermediate value theorem states that if \(f(x)\) is continuous over \([a, b]\), then for every value \(L\) between \(f(a)\) and \(f(b)\), you can find at least one value \(c\) in the interval \([a, b]\) where \(f(c) = L\). In the above example where \(f(x) = x^3 + 1\), for every \(y\)-value \(L\) between 0 and 2, there is at least one corresponding \(x\)-value \(c\) between -1 and 1 where \(f(c) = L\).

For example, if \(L = 1\), then there is a corresponding value \(c = 0\) where \(f(\class{blue}{c}) = \class{green}{L}\) since \(f(\class{blue}{0}) = \class{green}{1}\). If \(L = 1.5\), then \(c \approx 0.794\), since \(f(\class{blue}{0.794}) \approx \class{green}{1.5}\). Because of the intermediate value theorem, I will be able to find a corresponding \(c\)-value between -1 and 1 for every value of \(L\) between 0 and 2. Remember, \(L\) is a \(y\)-value of the function and \(c\) is an \(x\)-value of the function where \(f(\class{blue}{c}) = \class{green}{L}\).

Looking at this continuous function \(\class{red}{f(x)}\), for every \(L\)-value in the interval \(\class{green}{[0, 2]}\), there are one or more corresponding \(c\)-values in the interval \(\class{blue}{[-1, 1]}\) where \(f(\class{blue}{c}) = \class{green}{L}\).

Try selecting a value of \(L\) with the slider below. For every value of \(L\) between 0 and 2, there is a value of \(c\) between -1 and 1. This is due to the intermediate value theorem!

\(L =\)

\(c \approx\)

\(f(\)\() \approx\)

---

Here’s an example problem where we can use the intermediate value theorem. We have a function \(f(x) = x^2 - 2x\) and we want to know if \(f(x) = 0\) at some point between \(x = -1\) and \(x = 1\).

The first thing we need to check before using IVT is the function’s continuity. \(f(x)\) is continuous everywhere, so we can use IVT. The intermediate value theorem tells us that from \(x = -1\) to \(x = 1\), \(f(x)\) will take on every \(y\)-value from \(f(-1)\) to \(f(1)\).

\(f(-1) = \class{red}{3}\) and \(f(1) = \class{blue}{-1}\), so we know \(f(x)\) will take on every value between -1 and 3 within the interval \([-1, 1]\). This means that there must be some value \(c\) between -1 and 1 where \(f(c) = 0\), because 0 is in between -1 and 3. To answer our original question, there is in fact a point between \(x = -1\) and \(x = 1\) where \(f(x) = 0\). (That point turns out to be \(x = 0\).)

Now that you know what a limit is, it’s time to start solving for some limits! Sometimes it’s very easy to find a limit: all you have to do is plug in an \(x\)-value into the function. This is known as direct substitution.

This technique will only work if our \(x\)-value approaches a point \(x = c\) that is continuous in our function. This is because if \(f(x)\) is continuous at \(x = c\), then \(\displaystyle \lim_{x \to c}{f(x)} = f(c).\) Remember, this is the formal definition of continuity!

For example, for the function \(f(x) = 2x\), the value of \(\displaystyle\lim_{x \to \class{red}{2}}f(x)\) is simply \(f(\class{red}{2}) = 4\), since \(f(x) = 2x\) is continuous at \(x = \class{red}{2}\).

The function \(f(x) = 2x\) is continuous at \(x = 2\), so to find the limit of \(f(x)\) as \(x\) approaches 2, we just need to plug \(x = 2\) into \(f(x)\)! You can tell by the graph that \(\displaystyle \lim_{x \to 2}{f(x)} = f(2)\).

Here are some more examples where you can use direct substitution:

Note that direct substitution won’t always work! Consider this example:

We get \(\frac{0}{0}\) which is undefined! However, this doesn’t mean the limit doesn’t exist, it just means that direct substitution doesn’t work in this case. There are other ways to solve for limits that you will learn later, and in the next section, you will find out how to algebraically show that this limit is actually equal to 8.

The graph of \(f(x) = \frac{x^2 - 16}{x - 4}\) is undefined at \(x = 4\), which is why direct substitution doesn’t work. You can still tell graphically that the limit as \(x\) approaches 4 is is 8.

When you use direct substitution and get a result of \(\frac{0}{0}\), that means it’s an indeterminate form. In this case, you need to do more work to find out what the limit is (if it even exists). If you get a nonzero numerator divided by 0, that means that the limit doesn’t exist, usually because of a vertical asymptote in the function’s graph.

Let’s go back to our previous example, \(\displaystyle\lim_{x \to 4}\frac{x^2 - 16}{x - 4}\). We can’t use direct substitution here because that results in a division by 0, so let’s try rewriting the expression \(\frac{x^2 - 16}{x - 4}\) in an equivalent form that does let us use direct substitution. We will use factoring:

It is important to include \(x \ne 4\) in the final expression because the original expression \(\frac{x^2 - 16}{x - 4}\) is undefined when \(x = 4\). However, when we are taking the limit as \(x\) approaches 4, the value at \(x = 4\) does not matter! So we can ignore the domain restriction \(x \ne 4\) when we solve for the limit.

For \(x \ne 4\), \(\frac{x^2-16}{x-4}\) is equal to \(x + 4\). The value of that expression at \(x = 4\) is irrelevant when we take the limit as \(x\) approaches 4, so we can use direct substitution here.

All that’s left now is to substitute \(x = 4\) into the expression \(x + 4\) to get our limit of 8. This result agrees with the graph in the previous section!

We can also use a similar technique to solve limits involving square roots. Here’s an example:

Using direct substitution gives \(\frac{0}{0}\), so we need to rewrite this in an equivalent form. To do that, we will multiply both the numerator and denominator by the conjugate of the numerator in order to get rid of the square root.

Now we can use direct substitution because it won’t result in a division by 0.

We can also use trigonometric identities to create equivalent expressions. In the example below, we use the identity \(\sin(2x) = 2 \sin(x) \cos(x)\). List of trigonometric identities

The key is to use direct substitution whenever possible and only use algebraic manipulation if needed. If direct substitution gives you \(\frac{0}{0}\), simplify the expression further using algebraic manipulation (if possible) and use direct substitution again. Repeat this process until direct substitution gives you either a real number or a fraction in the form \(\frac{b}{0}\) where \(b \ne 0 \). If you get a real number, you found the limit! If you get \(\frac{b}{0}\), that means the limit doesn’t exist.

Let’s say we are evaluating a limit that involves two functions, like \(\displaystyle\lim_{x \to c} [f(x) + g(x)]\). How do we do that? We use the following limit properties:

The last property only works if \(\displaystyle\lim_{x \to c}{g(x)}\) exists and \(f(x)\) is continuous at the point \(x = \displaystyle\lim_{x \to c}{g(x)}\).

These properties work exactly the way you would expect them to, so there’s not much to memorize! Here’s an example:

We can use the multiplication of two functions property to split the limit into two limits, then multiply them:

Here’s another example. Let’s say that for some function, we know that \(\displaystyle\lim_{x \to c}{f(x)} = 7\). In that case, what is \(\displaystyle\lim_{x \to c}[5 \cdot f(x)]\)?

We can use the multiplication by constant rule to rewrite the limit as \(5 \cdot \displaystyle\lim_{x \to c}{f(x)}\). We know that \(\displaystyle\lim_{x \to c}f(x) = 7\), so we can find that the limit is \(5 \cdot 7 = 35\).

What is the limit of \(\class{red}{f(x)}\) as \(x\) approaches 0? Based on the graph, we can estimate the limit to be 0, but graphs aren’t precise enough to tell you the exact limit for sure. I’ll give you a few more pieces of information just so you can be sure:

1. For all values of \(x\), \(\class{green}{g(x)} \ge \class{red}{f(x)} \ge \class{blue}{h(x)}\). In other words, the value of \(\class{red}{f(x)}\) is always in between the values of \(\class{blue}{h(x)}\) and \(\class{green}{g(x)}\).
2. \(\displaystyle\lim_{x \to 0}{\class{green}{g(x)}} = 0\) and \(\displaystyle\lim_{x \to 0}{\class{blue}{h(x)}} = 0\).

Now, can you be sure that \(\displaystyle\lim_{x \to 0}{\class{red}{f(x)}} = 0\)? Yes, because of the squeeze theorem! Notice how \(\class{red}{f(x)}\) is being “squeezed” in between \(\class{green}{g(x)}\) and \(\class{blue}{h(x)}\). Because we know the limits of \(\class{green}{g(x)}\) and \(\class{blue}{h(x)}\) are 0 as \(x\) approaches 0, we can conclude that the limit of \(\class{red}{f(x)}\) as \(x\) approaches 0 is also 0 because the value of \(\class{red}{f(x)}\) has to be in between the other two functions as \(x\) approaches 0. This means the only possible value for \(\displaystyle\lim_{x \to 0}{\class{red}{f(x)}}\) is 0.

More formally, the squeeze theorem states that for any functions \(\class{red}{f(x)}\), \(\class{green}{g(x)}\), and \(\class{blue}{h(x)}\), if \(\class{green}{g(x)} \ge \class{red}{f(x)} \ge \class{blue}{h(x)}\) is always true over some interval including \(c\) (the three functions do not necessarily have to be defined at \(x = c\)), and \(\displaystyle\lim_{x \to c}{\class{green}{g(x)}} = \lim_{x \to c}{\class{blue}{h(x)}} = L\), then \(\displaystyle\lim_{x \to c}{\class{red}{f(x)}} = L\).

Let’s use the squeeze theorem to find a famous limit: \(\displaystyle\lim_{x \to 0}{\frac{\sin(x)}{x}}\). In order to find this limit, we’re first going to have to do some geometry. Let’s draw a unit circle with a positive angle \(x\) in radians between 0 and \(\frac{\pi}{2}\) (between 0 and 90 degrees).

Looking at \(\triangle ABE\), \(\displaystyle\sin(x) = \frac{\text{opposite}}{\text{hypotenuse}} = \; \)\(\displaystyle \frac{BE}{1} = BE\). Looking at \(\triangle ACD\), \(\displaystyle\tan(x) = \frac{\text{opposite}}{\text{adjacent}} = \;\)\(\displaystyle \frac{CD}{1} = CD\). Here is the same image with both of these lengths labeled:

The image shows that \(\class{green}{\text{area of }\triangle ABD} \le \; \)\(\class{blue}{\text{area of sector }ABD} \le \; \)\(\class{purple}{\text{area of }\triangle ACD}\). This will be true for any \(x\) between 0 and \(\frac{\pi}{2}\) radians (0 to 90 degrees). The area of \(\class{green}{\triangle ABD}\) is \(\frac{1}{2}bh = \frac{1}{2} \cdot 1 \cdot \sin(x) = \frac{1}{2} \sin(x).\) Similarly, the area of \(\class{purple}{\triangle ACD}\) is \(\frac{1}{2}bh = \frac{1}{2} \cdot 1 \cdot \tan(x) = \frac{1}{2} \tan(x).\)

Now for the area of sector ABD. Because there are \(2\pi\) radians in a circle, the fraction of the circle that sector ABD takes up is \(\frac{x}{2\pi}\). The area of a unit circle is \(\pi r^2 = \pi(1)^2 = \pi\), so the area of sector ABD is \(\pi \cdot \frac{x}{2\pi} = \frac{1}{2}x\).

Our inequality can thus be rewritten as \(\class{green}{\frac{1}{2}\sin(x)} \le \class{blue}{\frac{1}{2}x} \le \class{purple}{\frac{1}{2}\tan(x)}\). Multiplying the inequality by 2 gives us \(\sin(x) \le x \le \tan(x)\). Using the identity \(\tan(x) = \frac{\sin(x)}{\cos(x)}\) gives us \(\sin(x) \le x \le \frac{\sin(x)}{\cos(x)}\).

We will then take the reciprocal of all three terms in the inequality. When we do this, we need to flip the signs of the inequality because all three terms have the same sign. (If \(x\) is positive and between 0 and \(\frac{\pi}{2}\) radians, then \(\sin(x)\) and \(\frac{\sin(x)}{\cos(x)}\) will also be positive.) To explain why we need to flip the signs of the inequality, consider the inequality \(1 \le 2 \le 3\). If we take reciprocals, we need to switch the signs to keep the inequality true (\(1 \ge \frac{1}{2} \ge \frac{1}{3}\)). Taking the reciprocal of \(\sin(x) \le x \le \frac{\sin(x)}{\cos(x)}\) gives us \(\frac{1}{\sin(x)} \ge \frac{1}{x} \ge \frac{\cos(x)}{\sin(x)}\).

Multiplying the inequality by \(\sin(x)\), we arrive at \(1 \ge \frac{\sin(x)}{x} \ge \cos(x)\). We just showed that this inequality is true for \(0 < x < \frac{\pi}{2}\). This inequality is also true for some negative values of \(x\): because \(\frac{\sin(-x)}{-x} = \frac{\sin(x)}{x}\) and \(\cos(-x) = \cos(x)\), it’s also true for \(-\frac{\pi}{2} < x < 0\), meaning that the inequality is true for any \(x\) between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), excluding \(x = 0\).

Now let’s get back to the limit \(\displaystyle\lim_{x \to 0}{\frac{\sin(x)}{x}}\). Because we are trying to find the limit of \(\class{red}{\frac{\sin(x)}{x}}\) as \(x\) approaches 0, and we know the inequality \(\class{green}{1} \ge \class{red}{\frac{\sin(x)}{x}} \ge \class{blue}{\cos(x)}\) is always true for values of \(x\) near 0, we can finally use the squeeze theorem! We know that \(\displaystyle\lim_{x \to 0}\class{green}{1}\) and \(\displaystyle\lim_{x \to 0}{\class{blue}{\cos(x)}} = \cos(0)\) are both equal to 1. This means that \(\displaystyle\lim_{x \to 0}\class{red}{\frac{\sin(x)}{x}}\) must equal 1! Here’s a diagram showing \(\frac{\sin(x)}{x}\) being “squeezed”:

And here are tables that demonstrate the limit:

This limit will show up again as you learn calculus, so stay tuned! For example, this limit means that for small values of \(x\), \(\sin(x) \approx x\) (because for small values of \(x\), \(\frac{\sin(x)}{x} \approx 1\)), an important fact in calculus.

Using this limit, another important limit in calculus can be found, \(\displaystyle\lim_{x \to 0}{\frac{1 - \cos(x)}{x}}\).

Remember that point discontinuities or “holes” in a graph are also called removable discontinuities. In this section, you’ll see why.

Let’s say that we have a function with a removable discontinuity, like \(f(x) = \frac{x^2-7x+12}{x-4}\). Here’s a graph of that function:

This function has a removable discontinuity at \(x = 4\).

The name “removable” suggests that we can remove this discontinuity somehow. And we could by just adding a single point to the function. So the question is, what value should \(f(4)\) take on in order to make \(f(x)\) continuous everywhere?

Remember that by definition, a function is continuous at a point if the limit as \(x\) approaches that point equals the function’s actual value at that point. In this case, in order for \(f(x)\) to be continuous at \(x = 4\), \(\displaystyle\lim_{x \to 4}{f(x)}\) must equal \(f(4)\). So the question is essentially just asking for the limit of \(f(x)\) as \(x\) approaches 4. We can use factoring to solve that.

Looking at the graph above, to make the function continuous, we just need to fill in the point at \((4, 1)\). This means a value of \(f(4) = 1\) will make the function continuous. Notice that the \(y\)-value of this point is just \(\displaystyle\lim_{x \to 4}{f(x)}\). Here’s our new continuous function if we make this change:

What if we have a piecewise function that we want to make continuous? Let’s say we have this function:

Notice how \(f(x)\) isn’t defined at \(x = 2\). What value should \(f(x)\) take at this value to make it continuous?

How can we make this function continuous? We can do the same thing we did before: find the limit of \(f(x)\) as \(x\) approaches \(\class{red}{2}\). Using direct substitution, the limit from the left is \(\class{red}{2} + 2 = 4\) and the limit from the right is \(-\class{red}{2} + 6 = 4\). So the overall limit is 4 and that’s the value we need to set \(f(2)\) to in order to make the function continuous.

Here is the function \(f(x) = \frac{1}{x^2}\). What is \(\displaystyle\lim_{x \to 0}{f(x)}\)?

Since \(f(x)\) goes up forever as \(x\) approaches 0, we would normally say that the limit doesn’t exist, but that doesn’t give us too much information. Saying that “the limit does not exist” doesn’t tell us why the limit doesn’t exist or exactly what happens to the function as \(x\) approaches that value. We can use infinite limits to give us a little bit more information.

Notice how as \(x\) approaches 0 from either side, \(f(x)\) increases forever without bound. This is known as “approaching positive infinity”. We write this as \(\displaystyle\lim_{x \to 0}{f(x)} = \infty\). If a function \(f(x)\) keeps decreasing without bound as \(x\) approaches \(c\), we write that as \(\displaystyle\lim_{x \to c}{f(x)} = -\infty\), since it approaches negative infinity. Infinite limits can appear whenever we have a vertical asymptote. In this case, \(f(x) = \frac{1}{x^2}\) has a vertical asymptote at \(x = 0\), which explains the infinite limit as \(x\) approaches 0.

Note that in both these cases, the limit still doesn’t exist by our definition of a limit! That’s because in both cases, the function isn’t approaching any specific number (infinity isn’t considered to be a number). Saying that a limit equals \(\infty\) or \(-\infty\) is just a notation that we use to describe the idea of increasing or decreasing without bound.

Here’s another function with a vertical asymptote: \(f(x) = \frac{1}{x}\). What is \(\displaystyle\lim_{x \to 0}{f(x)}\)?

The function approaches negative infinity from the left, but it approaches positive infinity from the right. In this case, we can’t even use our new infinite limit notation because there are two different limits from both sides. So all we can say is that the limit does not exist.

We can still use one-sided limits to describe the function’s behavior from either side: \(\displaystyle\lim_{x \to 0^-}{f(x)} = -\infty \) and \(\displaystyle\lim_{x \to 0^+}{f(x)} = \infty \).

Let’s get some practice finding infinite limits. What are the one-sided limits of \(f(x) = \frac{1}{x - 3}\) as \(x\) approaches 3?

Limit from the left: as \(x\) approaches 3 from the left, the denominator \(x - 3\) approaches 0 from the left side (meaning it stays negative but gets closer and closer to 0). This means that the fraction \(\frac{1}{x-3}\) is going to get bigger and bigger in absolute value, approaching negative infinity.

Limit from the right: as \(x\) approaches 3 from the right, \(x - 3\) approaches 0 from the right (staying positive), meaning that \(\frac{1}{x-3}\) gets larger and larger, approaching positive infinity.

Let’s continue analyzing our classic function \(f(x) = \frac{1}{x}\). What happens to \(f(x)\) as \(x\) gets larger and larger?

As \(x\) gets larger and larger, it gets closer and closer to 0. With another extension of our limit notation, we can describe this using a limit: \(\displaystyle\lim_{x \to \infty}{f(x)} = 0\). In this case, \(x \to \infty\) simply means “as \(x\) gets larger and larger without bound”, described more concisely as “approaching positive infinity”. Similarly, \(\displaystyle\lim_{x \to -\infty}{f(x)} = 0\), because as \(x\) keeps decreasing, approaching negative infinity, \(f(x)\) also approaches 0.

Just like how infinite limits are associated with vertical asymptotes, limits at infinity are associated with horizontal asymptotes. In this case, \(f(x) = \frac{1}{x}\) has a horizontal asymptote at \(y = 0\), which explains why both limits at infinity are equal to 0.

Here is a technique for evaluating limits at infinity without using a graph or table. Consider this example:

This might look scary at first, but all you have to pay attention to are the most dominant terms. Let’s look at the numerator \(5x^5 + 4x^3 + 20x^2\) first.

As \(x\) gets larger and larger, the \(5x^5\) term will become much larger than the \(4x^3\) and \(20x^2\) terms. For example, here’s what these terms equal at \(x = 10\):

• \(5x^5 = \text{500,000}\)
• \(4x^3 = \text{4,000}\)
• \(20x^2 = \text{2,000}\)

You can clearly see that \(5x^5\) is much larger than the other terms. As \(x\) heads towards infinity, this difference will only get larger and larger, to the point where \(5x^5\) is the only term that actually matters. So when we take the limit at infinity, we can ignore the less dominant terms \(4x^3\) and \(20x^2\).

When we deal with polynomials, the most dominant term is the term with the highest exponent. Coefficients are not relevant to determining the dominant term.

We can do the same with the denominator. The most dominant term in the denominator is \(x^5\), since as \(x\) goes to infinity, \(x^5\) will very quickly become much larger than 1000 (the other term in the denominator).

This means that for large values of \(x\), \(\large{\frac{5x^5 + 4x^3 + 20x^2}{x^5 + 1000}}\) is approximately equal to \(\large{\frac{5x^5}{x^5}}\), which can be simplified to just 5. That means the limit of that expression as \(x\) approaches infinity is simply 5.

Here’s another example:

We can ignore the \(-5x\) and \(7\) in the denominator since they are less dominant than the \(3x^3\) term. That gives us \(\large{\frac{2x^2}{3x^3}}\) as our approximation for large values of \(x\). (I’m using “large” in this context to mean negative numbers with a large absolute value, like -1,000 or -10,000. I’m not referring to positive values of \(x\), since we are taking the limit as \(x\) approaches \(-\infty\).) The approximation \(\large{\frac{2x^2}{3x^3}}\) can be further simplified to \(\large{\frac{2}{3x}}\). As \(x\) approaches \(-\infty\), the denominator \(3x\) approaches \(-\infty\), meaning that we’re dividing by negative numbers with a larger and larger magnitude. As the denominator’s absolute value gets larger and larger, \(\frac{2}{3x}\) approaches 0, meaning our limit is 0.

Here’s one final example:

Ignoring the less dominant terms, this expression is approximately equal to \(\large{\frac{x^4}{6x^3}}\) for large values of \(x\), which can be simplified to \(\large{\frac{x}{6}}\). However, as \(x\) approaches \(-\infty\), \(\large{\frac{x}{6}}\) will not approach any finite value, instead approaching negative infinity. This means that the limit does not exist, or more precisely, the limit is \(-\infty\).

I’ve said that the limit of a function \(f(x)\) at some point \(x = c\) is the value that \(f(x)\) approaches as \(x\) approaches \(c\). But what exactly do we mean by “approach”? How can we mathematically describe this idea of a limit?

This is our current definition of a limit:

“The value that \(f(x)\) approaches as \(x\) approaches \(c\)”

We can improve this a little bit by rewording it like this:

“The value that \(f(x)\) is close to when \(x\) is close to \(c\)”

Why does this help us? Because we can write down mathematically what it means for two numbers to be close to each other. The distance between two numbers is the absolute value of the difference between them. For example, the distance between 3.8 and 4 is 0.2 because \(|3.8 - 4| = 0.2\).

The smaller the distance between two numbers, the closer they are. So two numbers \(a\) and \(b\) are close if \(|a-b|\) is small.

Now that we have this definition of closeness, we can make our definition more mathematical. If we say that \(L\) is the limit of \(f(x)\) as \(x\) approaches \(c\), then our definition of a limit becomes:

“The value \(L\) such that \(|f(x)-L|\) is small when \(|x-c|\) is small”

But we still need to be more specific. How small exactly do we want these two values to be?

One important part of a limit \(L\) is that the function \(f(x)\) can be arbitrarily close to \(L\): in other words, we can make the distance between \(f(x)\) and \(L\) as small as we want. So let’s say we want \(f(x)\) to be within \(\epsilon\) (the Greek letter epsilon) units of \(L\). Let’s add that into our definition:

“The value \(L\) such that \(|f(x)-L| \lt \epsilon\) when \(|x-c|\) is small”

The value \(\epsilon\) represents how close we want \(f(x)\) to be to the limit \(L\). For example, if we want \(f(x)\) to be within 0.1 of \(L\), we should choose \(\epsilon = 0.1\), and if we want \(f(x)\) to be within 0.01 of \(L\), we should use \(\epsilon = 0.01\).

Because we want \(f(x)\) to be able to get arbitrarily close to \(L\), this condition should be satisfied no matter how small of an \(\epsilon\) we choose. If \(L\) is truly the limit of \(f(x)\) as \(x\) approaches \(c\), for any \(\epsilon\) greater than 0, our condition should be satisfied. We will add this to our definition:

“The value \(L\) such that \(|f(x)-L| \lt \epsilon\) for every \(\epsilon \gt 0\) when \(|x-c|\) is small”

Finally, how small do we want \(|x-c|\) to be? We want \(f(x)\) to be close to \(L\) (i.e. \(|f(x)-L| \lt \epsilon\)) as long as \(c\) is within some distance of \(x\). We will call this distance \(\delta\) (the Greek letter delta).

We want to choose our \(\delta\) such that if \(x\) is within \(\delta\) units of \(c\), it is guaranteed that \(f(x)\) will be within \(\epsilon\) units of \(L\).

However, because the value of \(f(c)\) is irrelevant to the limit of \(f(x)\) as \(x\) approaches \(c\), we don’t want \(\delta\) to equal 0 (because that would imply that \(x = c\)). To avoid this, we will simply add on the requirement that \(\delta\) must be greater than 0.

This gives us our final definition of a limit. If \(\displaystyle\lim_{x\to c}f(x) = L\), then:

For every \(\epsilon \gt 0\) (no matter how small), we can find a \(\delta \gt 0\) such that if \(0 \lt |x-c| \lt \delta\), it is guaranteed that \(|f(x)-L| \lt \epsilon\).

This is sometimes known as the epsilon-delta definition of a limit.

Here’s a simple example of how to use this definition to formally prove a limit.

Problem: Consider the function \(f(x) = x\). Prove that \(\displaystyle\lim_{x \to 1}f(x) = 1\) using the formal definition of a limit.

This limit might seem very obvious, but I’m just trying to get you familiar with this definition of a limit.

Our goal is to prove that for any positive number \(\epsilon\), we can find a positive number \(\delta\) that meets our requirement. The easiest way to do this is to come up with a formula that gives a valid \(\delta\) in terms of \(\epsilon\). This formula works as a sort of “\(\delta\)-generator” that we can use to generate a \(\delta\) for any value of \(\epsilon\) that we give it.

To create this \(\delta\)-generator, let’s start off with what we want to achieve. We want \(|f(x) - L| \lt \epsilon\), so let’s plug in the values of \(f(x)\) and \(L\). In this case, our function \(f(x)\) is equal to \(x\), and \(L\) is equal to 1 (since we are trying to prove that the limit is 1).

The symbol \(\implies\) means “implies” (i.e. if the preceding statement is true, then the following statement must be true). You’ll see that symbol a lot in this section!

Remember that we want this to be true when \(0 \lt |x - c| \lt \delta\). Because we are trying to find the limit as \(x\) approaches 1, \(c\) is equal to 1.

This is interesting because we want \(|x-1| \lt \epsilon\) when \(0 \lt |x-1| \lt \delta\). This means that if we simply set \(\delta = \epsilon\), \(0 \lt |x-1| \lt \delta\) would automatically imply that \(|x-1| \lt \epsilon\). This means that our \(\delta\)-generator is simply \(\delta = \epsilon\).

To finish our proof, we just need to show that \(0 \lt |x-c| \lt \delta\) implies that \(|f(x) - L| \lt \epsilon\). Every epsilon-delta proof ends in this way.

Just like that, we have proven that \(\displaystyle\lim_{x\to 1}x = 1\)! It’s quite a bit of work for such an obvious limit, but that’s just what you need to do if you want to be rigorous.

Now let’s try a harder limit to prove:

Problem: Consider the function \(f(x) = x^2\). Prove that \(\displaystyle\lim_{x \to 3}f(x) = 9\) using the formal definition of a limit.

Just like the last problem, we will start with \(|f(x) - L| \lt \epsilon\) and try to find a \(\delta\)-generator.

We want to manipulate this expression into the form \(|x-c| \lt \text{[something]}\) so that we can find a value of \(\delta\) in terms of any \(\epsilon\). Remember, we want \(0 \lt |x-c| \lt \delta\) to imply that \(|f(x) - L| \lt \epsilon\).

Here, it might seem that \(\delta = \frac{\epsilon}{|x+3|}\) could work, but we want our formula for \(\delta\) to only be in terms of \(\epsilon\) (and not \(x\)). How can we eliminate the \(x\) in this formula?

Here, we are finding the limit as \(x\) approaches 3, so we can assume that \(x\) will be somewhere near 3. Let’s assume that \(x\) is within 1 unit of 3 (i.e. \(\delta \lt 1\)). This means that \(2 \lt x \lt 4\), which implies that \(5 \lt |x+3| \lt 7\) (keep this in mind). Let’s keep going to find bounds for \(\frac{\epsilon}{|x+3|}\):

When we take the reciprocals of all three terms, we must reverse the equality.

Doing this gives us an upper bound for \(\frac{\epsilon}{|x+3|}\): we can be sure it’s less than \(\frac{\epsilon}{7}\). If we set \(\delta\) equal to this upper bound \(\frac{\epsilon}{7}\), we can complete our proof! I will demonstrate by starting with \(0 \lt |x-c| \lt \delta\) and getting to \(|f(x) - L| \lt \epsilon\):

Note that this won’t work if \(\epsilon\) is greater than 7, because that would cause \(\delta\) to be greater than 1, which violates the assumption we made. In that case, we could just set \(\delta = 1\). Note that \(\delta = 1\) still implies that \(5 \lt |x+3| \lt 7\). Here’s the proof for the case where \(\delta = 1\):

In conclusion, given any \(\epsilon\), we should set \(\delta\) equal to 1 or \(\frac{\epsilon}{7}\), whichever is smaller. Doing so will guarantee that \(0 \lt |x-3| \lt \delta\) implies that \(|x^2-9| \lt \epsilon\), so \(\displaystyle\lim_{x\to 3}x^2 = 9\).

The formal definition of a limit is important because it allows us to rigorously prove all of the limit properties that will make our lives a lot easier as we learn more calculus!

• Limits describe what happens to a function \(f(x)\) as \(x\) approaches a specific value \(c\).
• The limit of \(f(x)\) as \(x\) approaches \(c\) doesn’t exist if \(f(x)\) doesn’t approach a finite value or approaches two different values from the left and right side of \(c\).
• One-sided limits describe what happens to \(f(x)\) as \(x\) approaches \(c\) either from the left or from the right.
• \[ \text{Limit from the left:} \lim_{x \to c^{-}}f(x) \] \[ \text{Limit from the right:} \lim_{x \to c^{+}}f(x) \]
• A function is continuous if its graph is a single line without any gaps or holes (i.e. you can draw the graph without lifting your pencil).
• A function \(f(x)\) is continuous at a point \(x = c\) if the limit as \(f(x)\) approaches \(c\) equals \(f(c)\).
• There are three main types of discontinuities (things that make a function not continuous): point discontinuities, jump discontinuities, and infinite discontinuities.

• The intermediate value theorem states that if a function \(f(x)\) is continuous over an interval \([a, b]\), it will pass through every \(y\)-value in between \(f(a)\) and \(f(b)\).
• To find limits, you can use direct substitution (directly plugging in the \(x\)-value into the limit expression), algebraic manipulation, limit properties, and/or the squeeze theorem.
• The squeeze theorem states that if you have three functions \(f(x)\), \(g(x)\), and \(h(x)\), and \(g(x) \ge f(x) \ge h(x)\) for all points near a point \(x = c\), then \(\displaystyle\lim_{x \to c}{f(x)} = \lim_{x \to c}{g(x)} = \lim_{x \to c}{h(x)}\).
  • Important limits:
  • \[ \lim_{x \to 0}\frac{\sin(x)}{x} = 1 \] \[ \lim_{x \to 0}\frac{1 - \cos(x)}{x} = 0 \]
• If a function \(f(x)\) has a removable discontinuity at \(x = c\), you can remove it by redefining \(f(c)\) to be \(\displaystyle\lim_{x \to c}{f(x)}\).
• Infinite limits describe what happens when a function \(f(x)\) increases or decreases without bound as \(x\) approaches a value \(c\). If \(\displaystyle\lim_{x \to c}{f(x)} = \infty \), then \(f(x)\) increases forever as \(x\) approaches \(c\), and if \(\displaystyle\lim_{x \to c}{f(x)} = -\infty \), then \(f(x)\) decreases forever as \(x\) approaches \(c\).
• Limits at infinity describe what happens to a function \(f(x)\) as \(x\) increases or decreases without bound. \(\displaystyle\lim_{x \to \infty}{f(x)}\) is the value \(f(x)\) approaches as \(x\) increases forever, and \(\displaystyle\lim_{x \to -\infty}{f(x)}\) is the value \(f(x)\) approaches as \(x\) decreases forever.

Khan Academy Link: Differentiation: definition and basic derivative rules

All topics covered in Khan Academy:
Green underlined topics are topics at least partially covered on my website and red topics are topics not yet covered on my website. Note that even green topics might not be covered in full detail on my page.

1. Defining average and instantaneous rates of change at a point
2. Defining the derivative of a function and using derivative notation
3. Estimating derivatives of a function at a point
4. Connecting differentiability and continuity: determining when derivatives do and do not exist
5. Applying the power rule
6. Derivative rules: constant, sum, difference, and constant multiple: introduction
7. Derivative rules: constant, sum, difference, and constant multiple: connecting with the power rule
8. Derivatives of \(\cos(x)\), \(\sin(x)\), \(e^x\), and \(\ln(x)\)
9. The product rule
10. The quotient rule
11. Finding the derivatives of tangent, cotangent, secant, and/or cosecant functions

In the Intro to Limits section, I gave a preview into the tangent line problem. In this unit, we’re going to actually solve it!

I mentioned a way to find the slope of a function at a single point: You take a second point and move it closer and closer to the first point, then calculate what the slope between the two points approaches.

In this animation, the green line is known as a secant line: a line that passes through two points on a function. A secant line tells you the average rate of change between two points: how fast the function is increasing between those two points on average.

The slope of a secant line can be calculated as the change in \(y\) divided by the change in \(x\). Therefore, a secant line that passes through two points on a function’s curve \((a, f(a))\) and \((b, f(b))\) has a slope of:

But the green line in the animation is approaching what is known as a tangent line: the line that just barely touches the function at a certain point.

The blue line is a tangent line because it is tangent to the function!

A tangent line tells you the exact slope of a function at a given point — exactly how fast the function is increasing at that instant! Because of this, the slope’s value is known as the function’s instantaneous rate of change at that point.

For example, if the red function in the image above represented your position over time, the slope of a line tangent to a point on the function would tell you your velocity (speed and direction) at that time, because velocity describes how fast your position is changing at a given instant (and in what direction).

Specifically, velocity is the instantaneous rate of change of position. This means that your velocity at any point in time describes how fast and in what direction your position is changing in that instant.

This circle is moving across the screen. What is the relationship between its position and velocity?

Position: pixels

Velocity: pixels/second

This circle’s velocity is the instantaneous rate of change of its position.

Graph of circle’s position:

The slope of the tangent line tells you the circle’s velocity at that point in time.

To find the slope of a secant line, we can use basic arithmetic. However, finding the slopes of tangent lines is more complicated and will require us to use calculus.

This is the main idea behind derivatives: they tell you how to find the slopes of tangent lines. The process of finding a derivative is called differentiation. To differentiate a function is to find its derivative. In the next few sections, you’ll find out exactly what derivatives are and how to find them!

Derivatives are essentially like speedometers for functions, because they describe how fast functions change and in what directions. As you learn more calculus, you’ll see derivatives pop up in many places, from solving optimization problems (that require you to find the maximum or minimum value of a function) to evaluating integrals (another huge calculus concept that is closely related to differentiation).

Differentiation is extremely useful to describe how a function behaves, and it’s used in physics to describe the motion of objects (since motion is really just about change!)

Remember, derivatives and differentiation are all about slope and rate of change!

(Note: In this section, I will use the word “slope” to refer to instantaneous rate of change.)

What is the slope of the tangent line of \(\class{red}{f(x) = x^2}\) at \(\class{blue}{x = 2}\)?

Consider the function \(f(x) = x^2\). What is the slope at \(x = 2\)?

To find this slope, first, we’ll choose two points on the function with \(x\)-values close to 2. We’ll choose the points with \(x\)-values of 2 and 2.1, which are \((2, 4)\) and \((2.1, 4.41)\). (Remember, the \(y\)-values of these points are equal to \(x^2\), because that’s our function!)

The purple line is the secant line between the points \(\class{red}{(2, 4)}\) and \(\class{blue}{(2.1, 4.41)}\).

Let’s find the slope of the secant line between these two points. Remember, slope is defined as “rise over run”, or change in \(y\) divided by change in \(x\).

Now let’s move the second point closer to the first point. We’ll change the \(x\)-value of the second point to 2.01, giving us the point \((2.01, 4.0401)\). Let’s calculate the slope between this point and our other point \((2, 4)\).

If we do this slope calculation again with the even closer points \((2, 4)\) and \((2.001, 4.004001)\), we get a slope of 4.001.

Here is a table summarizing our results, as well as testing what happens when the second point is to the left of the first point:

Using this slider, explore how the position of the second point changes the slope of the secant line. What happens to the slope as the change in \(x\) approaches 0 from either side?

The table and slider show that as the points get closer and closer, the slope gets closer and closer to 4. In fact, we can make the slope as close to 4 as we want (arbitrarily close to 4) by moving the points closer and closer to each other. Because of this, 4 is defined as the slope of the function at \(x = 2\). That is the limit of the slope of the secant line as the distance between the two points approaches 0.

Importantly, just like with limits, the slope of the secant line must approach the same value whether the second point is approaching from the left or from the right. If the slope approaches a different value from the left than from the right, the slope at that point does not exist.

The slope of \(f(x) = x^2\) at \(x = 2\) is 4. This is shown with the green tangent line, which has a slope of 4.

Let’s try finding the slope of this function at some other points. Use the buttons to control the \(x\)-coordinate of the first point (i.e. the point we are trying to find the slope at), and use the slider to control the \(x\)-coordinate of the second point.

Here is a table showing the slopes of \(f(x) = x^2\) at different \(x\)-values:

Take a moment to think about what this table is telling us. For example, the slope is greater at \(x = 3\) than at \(x = 2\), which makes sense because if you look at the graph of \(f(x) = x^2\), it is increasing faster at \(x = 3\) than at \(x = 2\).

At negative values of \(x\), the graph of \(f(x) = x^2\) is decreasing, which explains why the slopes at negative values of \(x\) are also negative. At \(x = 0\), the parabola turns around, meaning it is neither increasing nor decreasing at that point, hence the slope of 0.

Did you notice a pattern in the table? It turns out that for any \(x\)-value on the function \(f(x) = x^2\), the slope at \(x\) is always equal to \(2x\). (In a future section, you will learn how to prove this!)

This means that the derivative of \(f(x) = x^2\) is \(f'(x) = 2x\). (I will explain in more detail what \(f'(x)\) means in the next section. For now, just know that it represents the derivative of \(f(x)\).) A derivative of a function is an expression that tells you the slope of that function at any particular \(x\)-value.

For example, if a function has a derivative of \(3x\), at \(x = 10\), that function has a slope of \(3 \cdot 10 = 30\). At \(x = 100\), that function has a slope of \(3 \cdot 100 = 300\).

Another way to think of this derivative of \(2x\) is that if I changed \(x\) by a tiny amount (let’s call it \(\Delta x\) for “change in \(x\)”, pronounced “delta x”), then the value of \(x^2\) would change by approximately \(\Delta x\) times our derivative \(2x\), or \(2x\Delta x\). For most functions, this approximation involving the derivative will get more accurate as \(\Delta x\) gets smaller.

For example, for the function \(f(x) = x^2\), at \(\class{red}{x = 10}\), the derivative is \(2\class{red}{x} = 2 \cdot \class{red}{10} = 20\). If I increased \(\class{red}{x}\) by a tiny amount \(\class{blue}{\Delta x}\), let’s say 0.1, then the value of \(f(x)\) would increase by about \(2\class{red}{x}\class{blue}{\Delta x} = 2\cdot \class{red}{10} \cdot \class{blue}{0.1} = 2\). This is the derivative at \(\class{red}{x = 10}\) (which is equal to 20) multiplied by \(\class{blue}{\Delta x}\).

In fact, the value of \(f(10 + 0.1)\) is 2.01 greater than \(f(10)\), so our approximation was really good! To summarize, the derivative tells us how much a function’s output value will change in response to small changes to its input value.

Changing \(x\) by \(\Delta x\) increases \(f(x)\) by about \(\class{blue}{20} \cdot \Delta x\), since 20 is the derivative of \(f(x)\) at \(x = \class{red}{10}\). The derivative gives us the number we need to multiply \(\Delta x\) by to calculate the change in \(f(x)\).

The change in \(f(x)\) (written as \(\Delta y\)) divided by the change in \(x\) is approximately equal to \(f'(\class{red}{10})\), the derivative of \(f(x)\) at \(x = \class{red}{10}\). The deriviative is the limit of \(\frac{\Delta y}{\Delta x}\) as \(\Delta x\) approaches 0.

I said that the derivative of \(x^2\) is \(2x\), but to be more precise, I should have said that the derivative of \(x^2\) with respect to \(x\) is \(2x\). What this means is that as I change \(x\) by some amount, the value of \(x^2\) also changes, and the rate of change of \(x^2\) as I change \(x\) is the derivative \(2x\).

However, in the world of calculus, we won’t always be differentiating with respect to \(x\)! For example, if I referred to the derivative of some hypothetical function with respect to \(y\), I would be describing how that function responds to changes in the variable \(y\).

Often in real-world problems, you’ll be differentiating with respect to the variable \(t\), which represents time. The derivative of a function with respect to \(t\) tells you how that function changes as time passes. For example, if \(f(t)\) gives you the altitude of a plane at time \(t\), the derivative of \(f(t)\) represents how fast the plane’s altitude changes as \(t\) increases (i.e. as time goes on).

Wait, what does \(f'(x)\) even mean? \(f'(x)\) means the derivative of \(f(x)\), and is pronounced “f-prime of x”. But that’s not the only way of notating a derivative. This \(f'(x)\) notation is known as Lagrange’s notation, after Joseph-Louis Lagrange. An expression like \(f'(2)\) means the derivative of \(f(x)\) at \(x = 2\), or the slope of the function at \(x = 2\).

Another way of denoting a derivative is Leibniz’s notation, named after Gottfried Leibniz. Here’s the logic behind it.

The first step to finding the derivative at a point is to take two points on the function and then find the slope of the secant line through those two points. The slope is defined as \(\frac{\Delta y}{\Delta x}\), where \(\Delta y\) means change in \(y\) (the difference in \(y\) between the two points) and \(\Delta x\) means change in \(x\) (the difference in \(x\) between the two points). To find the slope at a single point, we take the limit of the slope as the second point approaches the first. As we move the second point towards the first, \(\Delta x\) will approach 0. The limit can thus be written like this:

This notation can be shortened to:

Therefore, in this notation, the derivative of a variable \(y\) with respect to \(x\) is written as \(\dv{y}{x}\). For example, if \(y = x^2\), then \(\dv{y}{x} = 2x\).

You can think of the \(d\) as meaning “differential”, “derivative”, or even “delta”. You can think of \(\dv{y}{x}\) of the ratio of the change in \(y\) to change in \(x\) when we change \(x\) by an infinitesimal (infinitely small) amount \(\dd{x}\).

\(\dv{y}{x}\) may look like a fraction, but it’s not really a regular fraction since \(\dd{y}\) and \(\dd{x}\) don’t represent specific numbers. There are some places in calculus where you can essentially treat \(\dv{y}{x}\) as a fraction, but you have to be very careful and only do this when it’s allowed.

The derivative operator, which takes in a function and outputs its derivative, is written as \(\dv{}{x}\). So the derivative of a function \(f(x)\) is written as \(\dv{}{x}f(x)\) or \(\dv{[f(x)]}{x}\). As an example, \(\dv{}{x}(\class{red}{x^2}) = \class{blue}{2x}\) because \(\class{blue}{2x}\) is the derivative of \(\class{red}{x^2}\) with respect to \(x\).

The derivative operator can also be used with variables other than \(x\). For example, \(\dv{\class{red}{t}}f(\class{red}{t})\) represents the derivative of \(f(\class{red}{t})\) with respect to \(\class{red}{t}\). The variable after the \(d\) in the “denominator” of the derivative operator is the variable we are differentiating with respect to.

Likewise, the \(\dv{y}{x}\) notation can be used with any two variables, not just \(y\) and \(x\). For example, \(\dv{a}{b}\) represents the derivative of \(a\) with respect to \(b\), or how the variable \(a\) responds to changes in \(b\).

Throughout this page, I will be switching between these two notations depending on which is most convenient, so it’s important to understand both of them. There are more ways to denote derivatives, but these are the most common.

This isn’t a lesson on its own, but rather an interactive demo I’ve created to help you understand a concept better.

Here, you can experiment with how different values of \(x\) affect the function \(f(x) = x^2\) and its derivative \(f'(x) = 2x\). Here are some important questions to consider:

• What happens to \(f'(x)\) as \(x\) takes on different values?
• When is \(f'(x)\) negative?
• What does it mean when \(f'(x)\) is negative?
• When is the value of \(f'(x)\) the largest?
• What does it mean when \(f'(x)\) is large?
• What happens to \(f(x)\) and \(f'(x)\) when \(x = 0\)? What does this mean?

Try figuring these out on your own first, then use the “Give me some hints...” button if you get stuck.

Imagine if we had to go through the process detailed in Derivatives: An Example every time we wanted to find a derivative! That’s simply just too much work. We need a more mathematical way of describing derivatives that we can use to find derivatives quickly.

Let’s start with the first step of our process: if we want to find the derivative of a function \(f(x)\) at a certain \(x\)-value, we choose two points on the function: one point on the function at that exact \(x\)-value and another point that is nearby on the function. Now, our two points are separated by some amount, so let’s call the horizontal distance between these two points \(h\).

This means the first point has an \(x\)-value of \(\class{red}{x}\) and the second point has an \(x\)-value of \(\class{blue}{x+h}\). To get the \(y\)-values of these points, we plug in their \(x\)-values into the function \(f(x)\). By doing this, we can figure out that the coordinates of the two points are \(\class{red}{(x, f(x))}\) and \(\class{blue}{(x+h, f(x+h))}\).

Now let’s find the slope of the secant line between these points. The “rise”, or change in \(y\), between the points is \(f(x+h) - f(x)\), and the “run”, or change in \(x\), is simply \(h\) by our definition. Let’s plug these values into the slope formula:

To actually find the slope (instantaneous rate of change) at the first point, we need to move the second point closer and closer to the first point. Let’s try an example where we do just that!

As the points get closer to each other, \(h\) will get smaller and smaller, approaching 0. So to get an expression for instantaneous rate of change, we need to take the limit of the secant line’s slope as \(h\) approaches 0:

Now we have an expression that we can use to find the slope of any function at any point! Let’s try using it with \(f(x) = x^2\) and \(x = 5\):

We can’t use direct substitution here to solve for this limit because that would result in a division by 0. Let’s try simplifying this further:

Now we can use direct substitution, giving us a final slope of \(\displaystyle\lim_{h \to \class{red}{0}}({10 + h}) = 10 + \class{red}{0} = 10\). What this means is that the slope (instantaneous rate of change) of \(f(x) = x^2\) at \(x = 5\) is 10. It also means that the slope of the line tangent to \(f(x) = x^2\) at \(x = 5\) is 10.

Being able to find the slope at one point mathematically is cool, but what’s much more powerful is being able to find the slope at any \(x\)-value. We can do that with our slope equation simply by not substituting any specific value for \(x\). Let’s try this with \(f(x) = x^2\):

What this means is that for any \(x\)-value on the function, the slope is equal to \(2x\). This is the derivative of \(f(x) = x^2\)! Using this limit definition of a derivative, we can find the derivative of many different functions.

There is an alternative definition of a derivative that also uses a limit. Let’s take the diagram from above and make some small changes:

We are now trying to find the slope at \(x = c\) (the red point). Now instead of defining \(h\) as the horizontal distance between the points, we give the first point an \(x\)-value of \(c\) and the second point an \(x\)-value of \(x\), making the horizontal distance between them \(x - c\). Let’s find the slope of the secant line using the slope formula.

We are still moving the second point towards the first point, so now \(x\) is changing instead of \(h\) as we move the second point closer to the first. As we do this, the value of \(x\) approaches \(c\). So to get the slope of the tangent line, we need to take the limit as \(x\) approaches \(c\):

This is an expression that gives us the slope of \(f(x)\) at \(x = c\). This alternative definition of the derivative is sometimes easier to work with, although the previous definition is used more often when finding derivatives.

Let’s say you were tasked to determine the current speed of a car. You are given three pieces of information: the average speed of the car over the last 10 minutes, over the last 2 minutes, and over the last 2 seconds. What is the best guess for the current speed of the car?

The answer is the average speed of the car over the last 2 seconds. Why do you think this is the best answer?

Well, the average speed of the car over the last 10 minutes (or even the last 2 minutes) likely won’t tell us much about the car’s current speed. These average speeds might be close to the current speed if the car has maintained a constant speed for a long time, but if the car was driving through a city, its speed would be constantly changing. This means that its current speed is probably nowhere near its average speed over the last few minutes.

However, the average speed of the car over the last 2 seconds is likely to be very close to its current speed, since the car is not very likely to have sped up or slowed down in the last 2 seconds. If we knew the car’s average speed over the last 0.2 seconds, that would be an even better estimate of the car’s current speed.

The idea is that the smaller the interval we look at, the better we can estimate how fast something is changing (the instantaneous rate of change). We can use this concept to estimate derivatives of a function at a point when we have limited information.

Let’s say we have a function \(f(x)\), and we only know its values at a few points. (Maybe this data came from a scientific experiment, and values were only measured every once in a while.)

These points are the only information we have of \(f(x)\).

Our task is to estimate this function’s derivative at \(x = 1.5\) (i.e. estimate the value of \(f'(1.5)\)). How can we come up with the best possible estimate?

Since we only know the function’s value at a few points, we’re going to have to use average rates of change to come up with our estimate. We could find the average rate of change from \(x = 0\) to \(x = 4\), but we could do better.

My analogy with the car demonstrates that looking at a smaller interval gives a more accurate estimate. So to get the best estimate, we need to look at the smallest interval possible.

We are trying to estimate \(f'(1.5)\), so within the points we do know the values of, we need to find the smallest interval that includes \(x = 1.5\).

That interval is between \(x = 1\) and \(x = 2\). We can’t choose an interval that is smaller than this because we don’t know the values of any points on the function between \(x = 1\) and \(x = 2\).

The best estimate for the derivative at \(x = 1.5\) is the slope of the secant line, or average rate of change, between \(x = 1\) and \(x = 2\). We can calculate that with the slope formula:

To be clear, we don’t actually know what the derivative is at \(x = 1.5\). Instead, we’re making an educated guess based on the average rate of change around that point. Most likely, the actual slope of the function at \(x = 1.5\) is somewhere near this average rate of change.

We don’t know what the slope of the function is at \(\class{blue}{x = 1.5}\), but we can make an educated guess based on the slope of the secant line between \(x = 1\) and \(x = 2\). Unless the function behaves erratically, \(f'(1.5)\) should be near the slope of the secant line.

Just like when finding limits, when trying to finding derivatives, sometimes these derivatives just don’t exist. Differentiability is whether or not a function has a derivative (a slope/instantaneous rate of change) at a certain point.

When will derivatives not exist? There are three main conditions that will cause a derivative to not exist at a point.

The first is if a function has a vertical tangent line at a point. A vertical tangent line doesn’t have a defined slope, so a function is not differentiable at a point if it has a vertical tangent at that point. Here’s an example:

What is the derivative of \(f(x) = \sqrt[3]{x}\) at \(x = 0\)?

We’ll try to find the derivative at \(x = 0\) by finding what slope the secant line approaches as a second point approaches \(x = 0\).

If the second point approaches from the right, the slope of the secant line seems to keep increasing without bound.

The same happens if the second point approaches from the left.

It turns out that the tangent line to \(x = 0\) is a vertical line. That means that its slope is undefined at \(x = 0\), so the function is not differentiable there.

The second way a function can be not differentiable is if there is a sharp turn in its graph. Here’s an example:

What is the derivative of \(f(x) = |x|\) at \(x = 0\)?

Let’s use the same technique to find the slope of this function at \(x = 0\).

From the right side, the slope of the secant line approaches 1. (In fact, it stays at 1 the entire time.)

And from the left side, the slope of the secant line approaches -1.

Because the slope approaches a different value from the left than from the right, the function does not have a slope at \(x = 0\), making it not differentiable at that point. In fact, you can draw an infinite number of “tangent” lines that just barely touch the function at \(x = 0\). But a function can’t have infinitely many slopes at a point, so its slope is undefined at \(x = 0\).

The third and final thing that can cause a function to not be differentiable is if it is not continuous at that point. I will go into more detail in the next section!

We’ve explored what affects differentiability and when derivatives won’t exist. Now let’s explore how continuity affects differentiability.

Let’s say a function is undefined at a certain \(x\)-value. That means that it is discontinuous at that point. That also means there is no way to find its derivative at that point. (It doesn’t make sense for a function to have a slope where it’s undefined!) So in this case, the function is discontinuous and also not differentiable.

But just because a function is discontinuous at a point doesn’t necessarily mean it’s undefined at that point. Let’s see what happens to a derivative if a function is discontinuous at a point yet still defined at that point.

Remember, there are three main ways a function can be discontinuous. (You can review them in the Intro to Continuity section.) The first is a point discontinuity - a hole in a function’s graph.

This function has a point discontinuity at \(x = 1\).

Let’s try to find the derivative at \(x = 1\) by taking a second point and moving it closer and closer to the first point.

With the second point coming from the right side, the slope of the secant line approaches negative infinity. Now let’s try from the left side...

Now the slope approaches positive infinity! Because we can’t find a finite slope at the point \(x = 1\), this function is not differentiable at \(x = 1\). So a point discontinuity makes a function not differentiable at that point.

Now let’s see how a jump discontinuity affects differentiability.

This function has a jump discontinuity at \(x = 0\).

Let’s bring in our second point and try to find the slope at \(x = 0\).

From the right side, the slope of the secant line approaches positive infinity.

From the left side, the slope of the secant line is 0.

Because we have two different limits for the slope (the secant line approaches a different slope from the left than from the right), the function does not have a derivative at \(x = 0\). This means that jump discontinuities also make a function not differentiable.

For our last example, let’s see what happens with a vertical asymptote.

This function has a vertical asymptote at \(x = 0\), but is still defined at \(x = 0\).

From the right, the slope approaches positive infinity.

And from the left, the slope also approaches positive infinity.

The slope does not approach any finite value, so the function is not differentiable at \(x = 0\). So a vertical asymptote also results in a loss of differentiability.

What we’ve seen is that all three types of discontinuity lead to the derivative not existing at the discontinuity. This means that if a function is not continuous at a point, it is also not differentiable. This also means that if a function is differentiable at a point, it must also be continuous at that point. In other words, differentiability implies continuity.

However, this does not mean that if a function is continuous, then it must be differentiable! We’ve seen an example in the previous section with the absolute value function \(f(x) = |x|\). That function is continuous everywhere but not differentiable at \(x = 0\).

So differentiability implies continuity, but continuity does not imply differentiability! In other words, a function must be continuous in order to have a chance at being differentiable, but not all continuous functions are differentiable.

Now that we know how to find derivatives using the limit definition, let’s try out a few more derivative examples. We’ve already figured out that the derivative of \(x^2\) is \(2x\). But what is the derivative of \(f(x) = x^3\)? Let’s try figuring it out.

The binomial theorem comes in handy for expanding \((x+h)^3\).

We find that the derivative of \(x^3\) is \(3x^2\). Let’s try finding the derivative of \(x^4\) now. Maybe some pattern will appear with these derivatives, who knows?

Here are the derivatives we’ve found so far:

Do you see a pattern? Each time we increase the exponent of \(f(x)\) by 1, both the coefficient and exponent of \(f'(x)\) increase by 1.

Can you think of a way to generalize the derivative of \(x^n\) for any \(n\)? In other words, can you write the derivative of \(x^n\) in terms of \(x\) and \(n\)? Give it a try!

With Lagrange’s notation:

\[ \large{f(x) = x^\class{red}{n}} \] \[ \large{f'(x) = \class{red}{n}x^{\class{red}{n}-1}} \]

With Leibniz’s notation:

\[ \large{\dv{}{x}(x^\class{red}{n}) = \class{red}{n}x^{\class{red}{n}-1}} \]

This result is known as the power rule, and it is an extremely important tool we can use to find derivatives! This is the first derivative rule you’ve learned, and you will continue to learn more rules that make your life easier. (Imagine if you had to use the limit definition every time you wanted to find a derivative! That would surely be annoying.)

The power rule works for any real number exponent, including fractions and irrational numbers. Here are some examples:

\[ \dv{}{x}(x^\class{red}{10}) = \class{red}{10}x^{\class{red}{10}-1} = \class{red}{10}x^9 \] \[ \dv{}{x}\left(\frac{1}{x}\right) = \dv{}{x}(x^\class{red}{-1}) = \class{red}{-1}x^{\class{red}{-1}-1} = \class{red}{-1}x^{-2} = \class{red}{-}\frac{\class{red}{1}}{x^2} \] \[ \dv{}{x}\left(\sqrt[3]{x^2}\right) = \dv{}{x}(x^\class{red}{2/3}) = \class{red}{\frac{2}{3}}x^{\class{red}{2/3}-1} = \class{red}{\frac{2}{3}}x^{-1/3} \] \[ \dv{}{x}(x^\class{red}{\pi}) = \class{red}{\pi}x^{\class{red}{\pi}-1} \] \[ \dv{}{x}(x) = \dv{}{x}(x^\class{red}{1}) = \class{red}{1}x^{\class{red}{1}-1} = \class{red}{1}x^0 = \class{red}{1} \]

The last example should make intuitive sense. \(f(x) = x\) is simply a line that always has a slope of 1, so it makes sense that its derivative is 1.

The graph of \(f(x) = x\) always has a slope of 1, so its derivative is 1.

Show me a proof of the power rule.

Here is a proof of the power rule for positive integer values of \(n\).

In this proof, we will be using the binomial theorem, which looks like this:

\[ (a + b)^n = a^n + {n \choose 1}a^{n-1}b + {n \choose 2}a^{n-2}b^2 + \cdots + {n \choose {n-1}}a^{1}b^{n-1} + b^n\]

The notation \(n \choose k\) is pronounced “n choose k”, and it tells you how many unique ways there are to choose \(k\) objects out of a group of \(n\) objects. The formula for this is:

\[ {n \choose k} = \frac{n!}{k!(n-k)!} \]

\(n!\) is pronounced “n factorial”, and equals the product of all positive integers less than or equal to \(n\): \(n! = n(n-1)(n-2) \cdots (2)(1)\).

Importantly, if both \(n\) and \(k\) are positive integers and \(k\) is less than or equal to \(n\), \(n \choose k\) will equal a positive integer. If \(k = 1\), \(n \choose k\) equals:

\[ \begin{flalign} {n \choose 1} &= \frac{n!}{1!(n-1)!}\\ &= \frac{n(n-1)(n-2)\cdots(2)(1)}{(n-1)(n-2)\cdots(2)(1)}\\ &= n \end{flalign} \]

Now we’re ready to start our proof. We start by using the limit definition of the derivative.

\[ \dv{}{x}f(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} \] \[ \dv{}{x}x^n = \lim_{h \to 0}\frac{(x+h)^n - x^n}{h} \]

Here, we can use the binomial theorem to expand \((x+h)^n\).

\[ \begin{flalign} \dv{}{x}x^n &= \lim_{h \to 0}\frac{(x+h)^n - x^n}{h}\\ &= \lim_{h \to 0}\frac{(x^n + {n \choose 1}x^{n-1}h + {n \choose 2}x^{n-2}h^2 + \cdots + {n \choose {n-1}}x^{1}h^{n-1} + h^n) - x^n}{h}\\ &= \lim_{h \to 0}\frac{{n \choose 1}x^{n-1}h + {n \choose 2}x^{n-2}h^2 + \cdots + {n \choose {n-1}}x^{1}h^{n-1} + h^n}{h}\\ \end{flalign} \]

Here, we can factor out an \(h\) from the numerator:

\[ \begin{flalign} \dv{}{x}x^n &= \lim_{h \to 0}\frac{{n \choose 1}x^{n-1}h + {n \choose 2}x^{n-2}h^2 + \cdots + {n \choose {n-1}}x^{1}h^{n-1} + h^n}{h}\\ &= \lim_{h \to 0}\frac{h[{n \choose 1}x^{n-1} + {n \choose 2}x^{n-2}h + \cdots + {n \choose {n-1}}x^{1}h^{n-2} + h^{n-1}]}{h}\\ &= \lim_{h \to 0}\left[{n \choose 1}x^{n-1} + {n \choose 2}x^{n-2}h + \cdots + {n \choose {n-1}}x^{1}h^{n-2} + h^{n-1}\right] \end{flalign} \]

We know that \(n \choose 1\) is always \(n\), so we can simplify.

\[ \begin{flalign} \dv{}{x}x^n &= \lim_{h \to 0}\left[{n \choose 1}x^{n-1} + {n \choose 2}x^{n-2}h + \cdots + {n \choose {n-1}}x^{1}h^{n-2} + h^{n-1}\right]\\ &= \lim_{h \to 0}\left[nx^{n-1} + {n \choose 2}x^{n-2}h + \cdots + {n \choose {n-1}}x^{1}h^{n-2} + h^{n-1}\right] \end{flalign} \]

Now we can use direct substituion to find this limit.

\[ \begin{flalign} \dv{}{x}x^n &= \lim_{h \to 0}\left[nx^{n-1} + {n \choose 2}x^{n-2}h + \cdots + {n \choose {n-1}}x^{1}h^{n-2} + h^{n-1}\right]\\ &= nx^{n-1} + {n \choose 2}x^{n-2}(0) + \cdots + {n \choose {n-1}}x^{1}(0)^{n-2} + (0)^{n-1}\\ &= nx^{n-1} \end{flalign} \]

We know that if \(k\) is a positive integer and is less than \(n\), than \(n \choose k\) will also be a positive integer. In other words, the values of \(n \choose 2\), \(n \choose 3\), ... up to \(n \choose {n-1}\) will all be positive integers. This means that we can get rid of all the terms with a 0 in it since they will evaluate to 0 without any problems.

The power rule is cool, but it’s not the only rule we need if we want to differentiate polynomials like \(2x^2 + 7x + 4\). Here are some more basic derivative rules that will come in handy.

First of all, the derivative of any constant is 0. That’s because if you have a function that always gives a constant value, it isn’t changing at all, meaning that the rate of change, or derivative, has to always be 0.

This is the graph of \(f(x) = 2\). 2 is a constant, so the derivative of \(f(x)\) is 0. You can tell because the graph has a slope of 0 everywhere (it’s a horizontal line).

This next rule deals with constant multiples of functions. It says that if you multiply a function by a constant \(k\), then the derivative of that function is also multiplied by \(k\).

For example, the derivative of \(2x^2\) is 2 times the derivative of \(x^2\). This makes sense because if you imagine taking the graph of \(x^2\) and doubling every \(y\)-value to get the graph of \(2x^2\), you would expect the slope at every point to also double.

Multiplying each point’s \(y\)-value by 2 doubles the slope at each point.

The constant \(k\) can also be -1. This means that for example, the derivative of \(-x^3\) is -1 times the derivative of \(x^3\).

This also means that the derivative of \(kx\) where \(k\) is a constant is \(k \cdot \dv{}{x}(x)\), or simply \(k\). For example, the derivative of \(5x\) is 5, and the derivative of \(\pi x\) is \(\pi\). After all, the slope of a function \(f(x) = kx\) is always \(k\) no matter what.

Finally, we have the sum and difference rules, which tell us how to find derivatives of multiple functions or terms being added or subtracted together:

For example, the derivative of \(2x^2 + 3x\) is the derivative of \(2x^2\) plus the derivative of \(3x\). We can then use the power rule and the multiplication by a constant rule to find those derivatives.

We will continue our journey of finding the derivatives of common functions. Now we’re going to try finding the derivatives of two trigonometric functions, \(\sin(x)\) and \(\cos(x)\). We can’t use any of our previous rules here, so we’re going to have to start from scratch.

This is the graph of \(\sin(x)\), where \(x\) is in radians. Try to estimate the slope of the function at some points!

Let’s estimate the slope of this function at some points. At \(x = 0\), it seems that the slope of \(\sin(x)\) is about 1. When the sine function reaches its maximum at \(x = \frac{\pi}{2}\), the slope is 0. At \(x = \pi\), the slope is down to about -1. And at \(x = \frac{3\pi}{2}\), the slope goes back to 0 as the sine function reaches its minimum.

This is the graph of \(\sin(x)\) with some tangent lines showing the slope of the function at certain points.

This table shows the derivative of \(\sin(x)\) at some points.

Does this table seem familiar to you? Let’s graph the slopes we’ve found so far:

Now this shape should look familiar. What is the derivative of \(\sin(x)\)? Try figuring it out yourself!

The derivative of \(\sin(x)\) is \(\cos(x)\)! It’s very interesting how these two trig functions are actually closely related in this way. (This is actually one reason radians are so important in math - this is only true when \(x\) is in radians!)

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Now let’s try to find the derivative of \(\cos(x)\). Let’s draw some tangent lines again and make a table:

Let’s graph the slopes once again:

It’s your turn to find the derivative of \(\cos(x)\). What does this function look like to you?

The derivative of \(\cos(x)\) is \(-\sin(x)\). Notice the negative symbol in front of \(\sin(x)\): \(\sin(x)\) and \(\cos(x)\) are not derivatives of each other!

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Use this button to view a formal proof of these derivatives:

In a previous math class, you might have heard of the mysterious number \(e\), an irrational number with an infinitely long decimal expansion, \(2.718281828459045...\)

But what exactly is this number \(e\) and why is it so important? You may have seen it being used in the continuously compounded interest formula, \(A = Pe^{rt}\). But \(e\) is much more useful than just compound interest. In fact, so many things related to exponential functions relate to \(e\), specifically the function \(e^x\).

One way to calculate \(e\) is with a limit:

Here’s a slider to show that this limit does indeed approach \(e\):

This definition is interesting, but how could this number defined in such a strange way possibly be so useful? The answer lies in differential calculus!

This is the graph of \(e^x\). We’re going to do the same thing we did with \(\sin(x)\) and \(\cos(x)\) in order to find its derivative. Try to find the slope at a few points, like \(x = -1\), \(x = 0\), and \(x = 1\).

Let’s graph the slopes now...

Wait, doesn’t that look very familiar? Try to figure out what the derivative of \(e^x\) is!

The derivative of \(e^x\) is... well, itself! This is fascinating and is one of the most important facts in calculus. This is one of the reasons why the constant \(e\) is so important in mathematics! It is the only base that when raised to the \(x\)th power creates a function that is its own derivative. In other words, \(e^x\) is the only function of the form \(a^x\) that is its own derivative. For example, the derivative of \(2^x\) is not \(2^x\), because 2 is not equal to \(e\).

Exponential functions are important in mathematics, but the function \(e^x\) is so important that it is sometimes referred to as simply “the exponential function” (or the “natural exponential function” to distinguish it from other exponential functions).

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You may also remember the natural logarithm \(\ln(x)\) from a previous math class. It’s simply a logarithm with a base of \(e\) (i.e. \(\ln(x) = \log_e(x)\)). This is also really important in calculus, and we’ll see why.

Let’s try differentiating the inverse of \(e^x\): the natural logarithm \(\ln(x)\). Here’s what its graph looks like:

Let’s find the slopes of some tangent lines:

And let’s graph the slope.

The derivative of \(\ln(x)\) is \(\frac{1}{x}\). It is the only logarithmic function in the form \(\log_b(x)\) with such a simple derivative of \(\frac{1}{x}\). (\(\ln(x)\) can be rewritten as \(\log_e(x)\).) This derivative should make sense because the rate of change of \(\ln(x)\) decreases as \(x\) increases. In other words, as \(x\) gets bigger, \(\ln(x)\) increases more and more slowly, which explains the derivative \(\frac{1}{x}\).

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Use this button to view a formal proof of these derivatives:

These next two rules are used for differentiating a product or quotient of two functions. They are known as the product rule and quotient rule respectively. (Remember, \(\class{green}{f'(x)}\) means the derivative of \(\class{red}{f(x)}\) and \(\class{purple}{g'(x)}\) means the derivative of \(\class{blue}{g(x)}\).)

Here’s an example: find the derivative of \(h(x) = x^2\sin(x)\). First, realize that \(h(x)\) is actually just two functions being multiplied together, \(x^2\) and \(\sin(x)\). Let’s call these functions \(\class{red}{f(x) = x^2}\) and \(\class{blue}{g(x) = \sin(x)}\). Then, break \(h(x)\) down into these two functions: \(h(x) = \class{red}{f(x)}\class{blue}{g(x)}\). Finally, use the product rule:

The next section will give an example of the quotient rule being used to find the derivative of \(\tan(x)\)!

The quotient rule gives us the ability to find the derivatives of the other four trig functions: \(\tan(x)\), \(\cot(x)\), \(\csc(x)\), and \(\sec(x)\)! All four of these functions can be rewritten as quotients involving \(\sin(x)\) and \(\cos(x)\):

As a reminder, here is the quotient rule:

Let’s use it to find the derivative of \(\tan(x)\), which can be rewritten as \(\frac{\sin(x)}{\cos(x)}\):

The same procedure can be used to find the derivatives of \(\cot(x)\), \(\csc(x)\), and \(\sec(x)\). The results are summarized in the table below:

This isn’t a lesson on its own, but rather an interactive demo I’ve created to help you understand a concept better.

• A secant line is a line that crosses through two points on a function. It describes the function’s average rate of change between those two points.
• A tangent line is a line that just touches a function at a single point. It describes the function’s instantaneous rate of change at that point.
• A derivative describes how fast a function is changing at any given \(x\)-value. Specifically, it describes the function’s slope (AKA instantaneous rate of change) at any point. A function’s derivative at a point is the slope of the tangent line that touches that point.
  • Example: the function \(f(x) = x^2\) has a derivative of \(f'(x) = 2x\). This means that at any \(x\)-value, \(f(x)\) is changing at a rate of 2 times that \(x\)-value. For example, at \(x = 2\), \(f(x)\) has a slope (instantaneous rate of change) of \(2 \cdot 2 = 4\).
• The derivative of a function \(f(x)\) is denoted by \(f'(x)\) or \(\dv{}{x}f(x)\). The derivative of a variable \(y\) with respect to \(x\) is denoted by \(\dv{y}{x}\). These notations can be used with other variables, not just \(x\) or \(y\).
• The slope of a secant line that goes through two points can be calculated with this expression. \(x\) is the \(x\)-coordinate of the first point on the function, and \(h\) is the second point’s \(x\)-coordinate minus the first point’s \(x\)-coordinate.
• \[ \frac{f(x+h)-f(x)}{h} \]
• The slope of a secant line can also be calculuated with this expression, where \(c\) is the \(x\)-coordinate of the second point on the function.
• \[ \frac{f(x)-f(c)}{x-c} \]
• The derivative, or slope of a tangent line, can be calculated by either of these limits:
• \[ \lim_{x \to 0}{\frac{f(x+h)-f(x)}{h}} \] \[ \lim_{x \to c}{\frac{f(x)-f(c)}{x-c}} \]
• A function is differentiable at a point if it has a derivative (instantaneous rate of change) at that point.
  • If a function is differentiable at a point, then it is definitely continuous at that point, but if a function is continuous at a point, then it may or may not be differentiable.
  • If a function is discontinuous at a point, it is also not differentiable at that point.
• The power rule states that the derivative of a function \(f(x) = x^n\) (where \(n\) is any real number) is \(f'(x) = nx^{n-1}\).
• Other basic derivative rules: (\(k\) can be any constant)
• \[ \dv{}{x}(k) = 0 \] \[ \dv{}{x}[k \cdot f(x)] = k \cdot \dv{}{x}f(x) \] \[ \dv{}{x}[-f(x)] = -\dv{}{x}f(x) \] \[ \dv{}{x}[f(x) + g(x)] = \dv{}{x}f(x) + \dv{}{x}g(x) \] \[ \dv{}{x}[f(x) - g(x)] = \dv{}{x}f(x) - \dv{}{x}g(x) \]
• Common derivatives:
• \[ \dv{}{x}\sin(x) = \cos(x) \] \[ \dv{}{x}\cos(x) = -\sin(x) \] \[ \dv{}{x}e^x = e^x \] \[ \dv{}{x}\ln(x) = \frac{1}{x} \]
• Product and quotient rules:
• \[ \dv{}{x}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x) \] \[ \dv{}{x}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \]
• Derivatives of other trig functions:
• \[ \dv{}{x}\tan(x) = \frac{1}{\cos^2(x)} = \sec^2(x) \] \[ \dv{}{x}\cot(x) = -\frac{1}{\sin^2(x)} = -\csc^2(x) \] \[ \dv{}{x}\csc(x) = -\frac{\cos(x)}{\sin^2(x)} = -\cot(x)\csc(x) \] \[ \dv{}{x}\sec(x) = \frac{\sin(x)}{\cos^2(x)} = \tan(x)\sec(x) \]

Khan Academy Link: Differentiation: composite, implicit, and inverse functions

All topics covered in Khan Academy:
Green underlined topics are topics at least partially covered on my website and red topics are topics not yet covered on my website. Note that even green topics might not be covered in full detail on my page.

1. The chain rule: introduction
2. The chain rule: further practice
3. Implicit differentiation
4. Differentiating inverse functions
5. Differentiating inverse trigonometric functions
6. Selecting procedures for calculating derivatives: strategy
7. Selecting procedures for calculating derivatives: multiple rules
8. Calculating higher-order derivatives
9. Further practice connecting derivatives and limits

Now it’s time to learn some advanced techniques that can help you differentiate more complicated functions. Here’s a function we don’t know how to differentiate yet: \(\sin^2(x)\)

The trick to finding the derivative of this function is actually to break it down into two functions! We need to notice that \(\sin^2(x) = [\sin(x)]^2\) can be expressed as a composition of two other functions. (As a refresher, function composition is when you have two functions nested inside each other, like \(f(g(x))\).)

In this case, if we define \(\class{red}{f(x) = x^2}\) and \(\class{blue}{g(x) = \sin(x)}\), then \(\class{red}{f(}\class{blue}{g(x)}\class{red}{)} = \class{red}{[}\class{blue}{\sin(x)}\class{red}{]^2}\). Let’s call this function \(h(x)\), so \(h(x) = [\sin(x)]^2\).

Now we’re going to do something a bit strange. Instead of finding the derivative of \(h(x)\) with respect to \(x\), we’re going to find the derivative of \(h(x)\) with respect to \(\sin(x)\). In other words, we will find how much \(h(x)\) changes as \(\sin(x)\) changes by a tiny amount.

We know that the derivative of \(x^2\) with respect to \(x\) is \(2x\). But what is the derivative of \([\sin(x)]^2\) with respect to \(\sin(x)\)? To solve this, let’s define a variable \(y\) that is equal to \(\sin(x)\). Then this is the same as asking what the derivative of \(y^2\) is with respect to \(y\), which is simply \(2y\). Writing that in terms of \(x\), we find that the derivative of \([\sin(x)]^2\) with respect to \(\sin(x)\) is \(2 \sin(x)\).

(When I write something like \(\displaystyle\dv{[\sin^2(x)]}{[\sin(x)]}\), I mean the derivative of \(\sin^2(x)\) with respect to \(\sin(x)\).)

What does this even mean? This means that if \(\sin(x)\) changes by a tiny amount \(h\), \([\sin(x)]^2\) will change by about \(2\sin(x)\cdot h\).

If we change \(x\) by some tiny amount \(\Delta x\), \(\sin(x)\) will change by some amount (I’m calling that amount \(h\)). When that happens, \(\sin^2(x)\) will also change by some amount (in this case, approximately \(2\sin(x)h\)). The ratio in change in \(\sin^2(x)\) to change in \(\sin(x)\) is about \(\frac{2\sin(x)h}{h} = 2\sin(x)\). This is because the derivative of \(\sin^2(x)\) with respect to \(\sin(x)\) is \(2\sin(x)\).

We haven’t arrived at the derivative of \(h(x)\) yet though, because \(2 \sin(x)\) is the derivative of \(h(x)\) with respect to \(\sin(x)\), not \(x\). To arrive at our final derivative, we need to multiply \(2 \sin(x)\) by the derivative of \(\sin(x)\) with respect to \(x\), which is simply \(\cos(x)\). That means that the derivative of \(h(x) = \sin^2(x)\) with respect to \(x\) is \(h'(x) = 2\sin(x)\cos(x)\).

This is an application of the chain rule, which can be written like this in Leibniz’s notation:

What does this mean? It means that to find the derivative of \(y\) with respect to \(x\), you find the derivative of \(y\) with respect to some other expression \(u\), then multiply it by the derivative of that expression \(u\) with respect to \(x\).

\(x\), \(y\), and \(u\) can be any variables you want. In this case, \(u\) doesn’t even have to be a variable: it can be an expression like \(\sin(x)\)!

In simpler terms, \(\dv{y}{u}\) tells us how much \(y\) changes when we change \(u\) by a tiny amount. \(\dv{u}{x}\) tells us how much \(u\) changes when we change \(x\) by a tiny amount. The chain rule tells us that \(\dv{y}{x}\), the amount that \(y\) changes when \(x\) changes by a tiny amount, is the product of these two amounts (\(\dv{y}{u}\) and \(\dv{u}{x}\)).

In our previous example, when we found the derivative of \(\sin^2(x)\) with respect to \(x\), here’s what we did:

We first found the derivative of \(\class{red}{\sin^2(x)}\) with respect to \(\class{green}{\sin(x)}\) (which is \(2\sin(x)\)), then we multiplied that by the derivative of \(\class{green}{\sin(x)}\) with respect to \(\class{blue}{x}\) (which is \(\cos(x)\)). We multiplied those two derivatives together to find that the derivative of \(\class{red}{\sin^2(x)}\) with respect to \(\class{blue}{x}\) is \(2\sin(x)\cos(x)\).

Let’s rewrite the chain rule in Lagrange’s notation. First, we’ll start with the chain rule written in Leibniz’s notation, then replace \(y\) with \(f(g(x))\) and \(u\) with \(g(x)\). We’ll also let \(h(x) = f(g(x))\).

Now we can directly convert this into Lagrange’s notation to get:

(Remember, \(h(x) = f(g(x))\).)

This version of the chain rule is more straightforward to use. Simply decompose your function \(h(x)\) into \(f(g(x))\), find the derivatives of the two functions \(f'(x)\) and \(g'(x)\), then substitute into the chain rule formula \(f'(g(x)) \cdot g'(x)\).

In the next section, we’re going to use the chain rule to differentiate functions in the form \(a^x\), where \(a\) is any positive base! You can also use the button below to view a proof of the quotient rule using the chain rule.

Now that we’ve learned the chain rule, we’ve unlocked the ability to differentiate \(a^x\) for any positive base \(a\). For example, we can now differentiate expressions like \(2^x\), \(10^x\), and \(\pi^x\). Can you figure out how?

Remember, we already know how to differentiate \(e^x\). Can you think of a way to use the chain rule to differentiate \(a^x\) for other bases of \(a\)?

The trick is to rewrite \(a^x\) in the form of \(e^\text{something}\). How do we do that? Try it yourself: as an example, try to rewrite \(2^x\) as \(e\) raised to the power of something.

Using what we know, we can also differentiate \(\log_a(x)\), where \(a\) is a positive base not equal to 1. Examples are \(\log_2(x)\), \(\log_{10}(x)\), and \(\log_{1/2}(x)\). Can you figure out how?

As a reminder, here is when you should use each of the derivative rules.

• Power rule: monomials (e.g. \(x^4\), \(3x^5\)) and polynomials (e.g. \(x^3 + 3x^2\)). The exponents can also be negative (e.g. \(x^{-3}\), \(3x^{-4} - 6x^{-6}\)).
• Product rule: the product of two functions (when one function is being multiplied by another).
  • Examples: \(\sin(x)\cos(x)\), \(e^x \cdot 2x^4\)
• Quotient rule: the quotient of two functions (when one function is being divided by another).
  • Examples: \(\displaystyle\frac{e^x}{2x^2}\), \(\displaystyle\frac{2}{\sin(x)}\)
• Chain rule: composite functions (when one function is nested inside another; i.e. the output of one function is the input of another function).
  • Examples: \(\ln(\sin(x))\), \(e^{\cos(x)}\), \(\displaystyle\frac{1}{x + 4}\)
  • A common alternative notation for \(e^x\) is \(\exp(x)\). If it helps, you can write \(e^{\text{[something]}}\) as \(\exp(\text{[something]})\) to make it easier to spot composite functions. For example, \(e^{\cos(x)}\) can also be written as \(\exp(\cos(x))\), which looks more like a composite function.
  • Fractions like \(\frac{1}{x+4}\) are also considered composite functions. If \(f(x) = \frac{1}{x}\) and \(g(x) = x+4\), then \(f(g(x)) = \frac{1}{x+4}\).

Here are some strategies to make differentiating complicated functions easier:

• Expanding a product: Sometimes we can expand a product so that we can use the power rule instead of the product rule.
  • Example: \(f(x) = (2x + 1)(x + 2)\) can be rewritten as \(f(x) = 2x^2 + 5x + 2\), which can then be differentiated using the power rule to get \(f'(x) = 4x + 5\). (It’s perfectly acceptable to use the product rule here, but using the power rule takes less work sometimes.)
• Simplifying a quotient: The same simplification can be done with fractions.
  • Example: \(f(x) = \frac{2x^5 - 6x^4}{2x^2}\) can be simplified down to \(f(x) = x^3 - 3x^2\) so that you don’t have to use the quotient rule. Now you can simply use the power rule on \(f(x)\) to arrive at \(f'(x) = 3x^2 - 6x\).
• Factoring a quotient: Factoring can also help with simplifiying fractions.
  • Example: \(f(x) = \frac{x^2 + 3x + 2}{x + 2}\) can be simplified down to \(f(x) = \frac{(x+2)(x+1)}{x + 2} = x + 1\). The derivative of this function is \(f'(x) = 1\). (However, you do have to keep in mind that \(f(x)\) and \(f'(x)\) are both undefined at \(x = -2\), since we would be dividing by zero in the original function.)
• Simplifying fractions and radicals: For some functions involving fractions and radicals, you also don’t have to use the quotient rule.
  • Example: You can rewrite \(f(x) = \frac{\sqrt{x}}{x^3}\) as \(f(x) = \frac{x^{1/2}}{x^3} = x^{-5/2}\), then use the power rule to find the derivative: \(f'(x) = -\frac{5}{2}x^{-7/2}\).
• Rewriting quotients as products: if you dislike using or memorizing the quotient rule, you can always rewrite quotients as products.
  • Example: \(f(x) = \frac{2x}{\sin(x)}\) can be rewritten as \(f(x) = 2x \cdot \frac{1}{\sin(x)} = 2x[\sin(x)]^{-1}\), which can then be differentiated with the product and chain rules. (Note: \([\sin(x)]^{-1} = \frac{1}{\sin(x)}\) is not to be confused with the inverse sine function \(\sin^{-1}(x) = \arcsin(x)\)).

Sometimes we will have to use multiple rules to differentiate a function. An example is \(h(x) = \cos(2x\ln(x))\). This function is both a composite function (it has a natural log inside the cosine function) and involves a product of two functions (\(2x\ln(x)\)), so we need to use both the product and chain rule here.

We can define \(f(x) = \cos(x)\) and \(g(x) = 2x\ln(x)\) so that \(h(x) = f(g(x)) = \cos(2x\ln(x))\). Then we could use the chain rule:

But we still need to find \(g'(x)\). To do that, we use the product rule:

Now we can finally put everything together.

Sometimes we will have a composite function involving three functions. An example of this that we’re going to differentiate is \(h(x) = \frac{1}{\cos(x^2)}\). In cases like these, we need to use the chain rule twice.

For the first application of the chain rule, we can set \(f(x)\) to the outermost function and \(g(x)\) to whatever’s inside that outermost function. So \(f(x) = \frac{1}{x} = x^{-1}\) and \(g(x) = \cos(x^2)\). Just to check, \(h(x) = f(g(x)) = \frac{1}{\cos(x^2)}\), which matches our original definition of \(h(x)\).

To find \(g'(x)\), we use the chain rule once again. The chain rule tells us that \(g'(x) = -\sin(x^2)\cdot 2x\) \(= -2x \sin(x^2)\).

So far we’ve learned how to differentiate many different types of functions and we’ve learned many different strategies to differentiate complex functions. So let’s try differentiating an equation for a circle. It shouldn’t be that hard, right?

How can we make sense of the derivative of the equation \(x^2 + y^2 = 1\)?

Wait! That isn’t even a function! How can we take the derivative of something that isn’t a function?

Well, circles can still have tangent lines, right? So we can use differentiation to find the slope of the line tangent to any point on this circle.

Even though \(x^2+y^2=1\) isn’t a function, we can still use differentiation to find the slope of the tangent line at certain points.

One way to differentiate this is to rewrite \(x^2+y^2=1\) as \(y = [\text{something}]\), then split it into two functions:

\[ x^2 + y^2 = 1\] \[ y^2 = -x^2 + 1\] \[ y = \pm\sqrt{-x^2 + 1}\] \[ \text{Function 1: } y = \sqrt{-x^2 + 1} \] \[ \text{Function 2: } y = -\sqrt{-x^2 + 1} \]

Then we could differentiate each of these functions using the chain rule. But doing all of this is a lot of work. There’s an easier way to do this, and it’s called implicit differentiation.

Before we dive into it, let’s first discuss where the name “implicit differentiation” comes from. A function is explicit when it is in the form \(y = [\text{something involving }x]\), like \(y = x^2\) or \(y = 3\cos(x)\). An explicit function explicitly tells you how to solve for \(y\) when you’re given a value for \(x\). However, when a function or relation is implicit, it’s not in the form \(y = [\text{something involving }x]\). Examples of implicit relations are \(x + y = 3\), \(\sin^2(x) + \cos(y) = 1.2\), and \(x^2 + y^2 = 1\). So our circle equation is an implicit relation.

Implicit differentation is when we differentiate an equation involving an implicit relation. The main idea behind implicit differentiation is to differentiate both sides of the equation with respect to \(x\). To differentiate \(x^2 + y^2 = 1\), we would start by taking the derivative of both sides:

Here we have an interesting situation. How do we differentiate \(y^2\) with respect to \(x\)? The derivative of \(y^2\) with respect to \(y\) is \(2y\), but we’re trying to differentiate with respect to \(x\).

In this situation, we can use the chain rule! The derivative of \(y^2\) with respect to \(x\) is equal to the derivative of \(y^2\) with respect to \(y\) multiplied by the derivative of \(y\) with respect to \(x\).

The derivative of \(y^2\) with respect to \(y\) is \(2y\), and the derivative of \(y\) with respect to \(x\) is what we’re trying to solve for in the first place! So we can write \(\dv{}{x}(y^2)\) as \(2y \cdot \dv{y}{x}\).

We’ve arrived at our derivative, but what does this mean? This means that the slope of the line tangent to any point \((x, y)\) on our circle is \(-\frac{x}{y}\). For example, the slope of the line tangent to \((0.6, 0.8)\) is \(-\frac{0.6}{0.8} = -\frac{3}{4}\).

The slope of the line tangent to \(\class{red}{(0.6, 0.8)}\) is \(-\frac{3}{4}\).

One of the most powerful things we can do with implicit differentiation is to find the derivatives of inverse functions. Let’s use it to find the derivatives of inverse trig functions.

We’re going to start with inverse sine (also known as arcsine). The trick to differentiating this is to turn an explicit equation into an implicit one.

(Note: \(\arcsin(x)\) is an alternative notation for \(\sin^{-1}(x)\). The \(\text{arc-}\) prefix represents the inverse of any trig function: \(\arccos(x) = \cos^{-1}(x)\) and \(\arctan(x) = \tan^{-1}(x)\).)

Remember that by the definition of inverse functions, \(\sin(\arcsin(x)) = x\). (This is because arcsine takes in a ratio of side lengths of a triangle and outputs an angle, and the sine function turns that back into that same ratio of side lengths.)

Let’s see what happens when we start with the equation \(y = \arcsin(x)\).

Now we can proceed with implicit differentiation by taking the derivative of both sides.

(The second step is an application of the chain rule.)

Remember however that we are trying to find the derivative of \(\arcsin(x)\), which should also be in terms of \(x\). Therefore, we need to rewrite \(\frac{1}{\cos(y)}\) in terms of \(x\). We can do this using the Pythagorean trig identity \(\sin^2(y) + \cos^2(y) = 1\).

Because we started with \(y = \arcsin(x)\), we know that \(\sin(y) = x\), so we can replace \(\sin^2(y)\) with \(x^2\).

We have two possibilities for what \(\cos(y)\) can equal. Let’s consider both of them for now. We know that our derivative \(\dv{y}{x}\) is equal to \(\frac{1}{\cos(y)}\), so let’s replace \(\cos(y)\) with its equivalent in terms of \(x\):

Well, a derivative can’t be both \(\frac{1}{\sqrt{1 - x^2}}\) and \(-\frac{1}{\sqrt{1 - x^2}}\) at the same time, so let’s check to see which one is correct.

The graph of \(\arcsin(x)\) is always increasing within its domain, so its derivative has to always be positive.

Our derivative has to always be positive. By the definition of the square root symbol, \(\sqrt{1 - x^2}\) cannot be negative. This means that \(\frac{1}{\sqrt{1 - x^2}}\) has to be positive for all values of \(x\) in its domain and \(-\frac{1}{\sqrt{1 - x^2}}\) always has to be negative. Because we are looking for a positive derivative, \(\frac{1}{\sqrt{1 - x^2}}\) is the correct derivative.

We can use the same process to find the derivative of inverse cosine.

Inverse tangent is a little different. We start as usual:

The way we simplify this is by using the trig identity \(\tan^2(y) + 1 = \sec^2(y)\), derived by taking \(\sin^2(y) + \cos^2(y) = 1\) and dividing both sides by \(\cos^2(y)\).

Therefore:

Let’s find the derivative of \(\arccot(x)\) now. The proof for this one looks very similar to how we found the derivative of \(\arctan(x)\).

\[ y = \arccot(x) \] \[ \cot(y) = x \] \[ \dv{x}\cot(y) = 1 \] \[ \dv{[\cot(y)]}{y} \cdot \dv{y}{x} = 1 \] \[ -\csc^2(y) \cdot \dv{y}{x} = 1 \] \[ \dv{y}{x} = -\frac{1}{\csc^2(y)} \]

We can divide both sides of the trig identity \(\sin^2(y) + \cos^2(y) = 1\) by \(\sin^2(y)\) to find that \(\csc^2(y) = 1 + \cot^2(y)\).

\[ \dv{y}{x} = -\frac{1}{1+\cot^2(y)} \] \[ \dv{y}{x} = -\frac{1}{1+x^2} \quad (\text{because } \cot(y) = x) \] \[ \dv{x}\arccot(x) = -\frac{1}{1+x^2} \]

Now it’s time for the derivative of \(\arccsc(x)\).

\[ y = \arccsc(x) \] \[ \csc(y) = x \] \[ \dv{x}\csc(y) = 1 \] \[ \dv{[\csc(y)]}{y} \cdot \dv{y}{x} = 1 \] \[ -\cot(y)\csc(y) \cdot \dv{y}{x} = 1 \] \[ \dv{y}{x} = -\frac{1}{\cot(y)\csc(y)} \] \[ \dv{y}{x} = -\frac{1}{x\cot(y)} \quad (\text{because } \csc(y) = x)\]

We know that \(\csc^2(y) = 1 + \cot^2(y)\). Let’s solve for \(\cot(y)\) in terms of \(\csc(y)\) because we can replace \(\csc(y)\) with \(x\).

\[ \csc^2(y) = 1 + \cot^2(y) \] \[ \csc^2(y) - 1 = \cot^2(y) \] \[ \cot(y) = \pm\sqrt{\csc^2(y) - 1} \] \[ \cot(y) = \pm\sqrt{x^2 - 1} \quad (\text{because } \csc(y) = x) \]

Now that we have an expression for \(\cot(y)\) in terms of \(x\), we can write our derivative purely in terms of \(x\).

\[ \dv{y}{x} = -\frac{1}{x\cot(y)} \] \[ \dv{y}{x} = -\frac{1}{\pm x\sqrt{x^2 - 1}} \]

The function \(\arccsc(x)\) is decreasing across its entire domain, so the derivative of \(\arccsc(x)\) always has to be negative. This means that the denominator \(\pm x\sqrt{x^2 - 1}\) always has to be positive. To ensure that, we must take the absolute value of \(x\).

The graph of \(\arccsc(x)\) is always decreasing, so its derivative is always negative.

\[ \dv{y}{x} = -\frac{1}{|x|\sqrt{x^2 - 1}} \] \[ \dv{x}\arccsc(x) = -\frac{1}{|x|\sqrt{x^2 - 1}} \]

The process for finding the derivative of \(\arcsec(x)\) is very similar.

\[ y = \arcsec(x) \] \[ \sec(y) = x \] \[ \dv{x}\sec(y) = 1 \] \[ \dv{[\sec(y)]}{y} \cdot \dv{y}{x} = 1 \] \[ \tan(y)\sec(y) \cdot \dv{y}{x} = 1 \] \[ \dv{y}{x} = \frac{1}{\tan(y)\sec(y)} \] \[ \dv{y}{x} = \frac{1}{x\tan(y)} \quad (\text{because } \sec(y) = x)\]

Using the trig identity \(\sec^2(y) = 1 + \tan^2(y)\), we can solve for \(\tan(y)\) in terms of \(x\):

\[ \sec^2(y) = 1 + \tan^2(y) \] \[ \sec^2(y) - 1 = \tan^2(y) \] \[ \tan(y) = \pm\sqrt{\sec^2(y) - 1} \] \[ \tan(y) = \pm\sqrt{x^2 - 1} \quad (\text{because } \sec(y) = x) \]

Now we can rewrite \(\dv{y}{x}\) in terms of just \(x\).

\[ \dv{y}{x} = \frac{1}{x\tan(y)} \] \[ \dv{y}{x} = \frac{1}{\pm x\sqrt{x^2 - 1}} \]

Because \(\arcsec(x)\) is increasing across its entire domain, its derivative has to always be positive, so we have to take the absolute value of \(x\) to ensure the derivative is positive.

The graph of \(\arcsec(x)\) is always increasing, so its derivative is always positive.

\[ \dv{y}{x} = \frac{1}{|x|\sqrt{x^2 - 1}} \] \[ \dv{x}\arcsec(x) = \frac{1}{|x|\sqrt{x^2 - 1}} \]

Summary of additional inverse trig derivatives:

\[ \dv{x}\arccot(x) = -\frac{1}{1+x^2} \] \[ \dv{x}\arccsc(x) = -\frac{1}{|x|\sqrt{x^2-1}} \] \[ \dv{x}\arcsec(x) = \frac{1}{|x|\sqrt{x^2-1}} \]

---

Summary of this section:

The potential of implicit differentiation isn’t just limited to inverse trig functions: using it, we can find the derivative of the inverse of any function! Here’s how.

What it means for two functions \(f(x)\) and \(g(x)\) to be inverses is that \(f(g(x)) = g(f(x)) = x\). So let’s set \(g(f(x))\) equal to \(x\) and use implicit differentiation.

If \(f(x)\) and \(g(x)\) are inverses, then...

We’ve found a rule to differentiate a function if we know the derivative of its inverse! We can also use this formula to differentiate the inverse of a function we know.

Here is this formula using inverse function notation, where \(f^{-1}(x)\) is the inverse of \(f(x)\):

\[ \dv{x}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))} \]

Here’s an example: the inverse of \(e^x\) is \(\ln(x)\). Let’s say we forgot what the derivative of \(\ln(x)\) was. Luckily, we can still figure it out if we know the derivative of its inverse, \(e^x\)! Just use the formula with \(f(x) = \ln(x)\) and \(g(x) = e^x\):

Here’s a problem that our formula can help out with:

Problem: The function \(g(x)\) is the inverse of the function \(f(x) = x^3 + x\). The point \((2, 1)\) lies on the curve of \(g(x)\). What is the slope of the line tangent to \(g(x)\) at the point \((2, 1)\)?

We are trying to find the slope of the line tangent to \(g(x)\) at the point \((2, 1)\), which is \(g'(2)\). Your first instinct might be to find a formula for \(g(x)\), but we don’t actually need to do this. The power of our inverse derivative formula is that it allows us to come up with an expression for this derivative in terms of \(f'(x)\).

Because the point \((2, 1)\) is on the curve of \(g(x)\), we know that \(g(2) = 1\).

Now we can simply differentiate \(f(x)\), then evaluate the derivative at \(x = 1\) to get the value of \(g'(2)\)!

What happens if we find a function’s derivative, and then we take the derivative of that derivative? That gives us a second derivative.

For example, if we have the function \(f(x) = x^4\), we can differentiate it to get \(f'(x) = 4x^3\), then differentiate that again to get \(f''(x) = 12x^2\).

While a first derivative tells you how fast a function is changing, a second derivative tells you how fast that function’s derivative is changing.

The notation for a second derivative is \(f''(x)\) in Lagrange’s notation. But what about in Leibniz’s \(\dv{}{x}\) notation?

Well, a second derivative could be written as \(\dv{}{x}\left(\dv{}{x}f(x)\right)\), but that’s kind of a mess, so it’s shortened to \(\dv{^2}{x^2}f(x)\). The second derivative of \(y\) with respect to \(x\) is thus written as \(\dv{^2y}{x^2}\).

If we take the derivative of a second derivative, we get a third derivative, which describes how the second derivative changes. Taking the derivative of a third derivative gives us a fourth derivative, and so on.

Third derivatives and beyond are represented in Leibniz’s notation as \(\dv[3]{y}{x}\), \(\dv[4]{y}{x}\), and so on. In Lagrange’s notation, higher derivatives can be represented by \(f'''(x)\), \(f''''(x)\), and so on, but they are also sometimes written as \(f^{(3)}(x)\), \(f^{(4)}(x)\), etc. for better readability.

We can also find the second derivatives of implicit equations. We previously found that for the equation \(x^2 + y^2 = 1\), the derivative of \(y\) with respect to \(x\) was \(-\frac{x}{y}\).

Now we can take the derivative of both sides again to find the second derivative of \(y\) with respect to \(x\).

Here, we can use the quotient rule to find the derivative of the right side.

We know that \(\dv{y}{x} = -\frac{x}{y}\), so let’s substitute that in.

Here is a mysterious game about antimatter. I promise, it’s related to calculus and derivatives.

You have 10 antimatter.

You have 0 1st dimensions.

You have 0 2nd dimensions.

You have 0 3rd dimensions.

To start, try buying a 1st dimension for 10 antimatter. Who knows what will happen next?

This is a very primitive (and modified) version of the game Antimatter Dimensions by Hevipelle and others. If you enjoyed this demo, I recommend you check out the full game here!

Here’s a limit that seems hard to evaluate at first:

You could try algebraic manipulation, but getting an answer that way is really hard. Instead, you should try looking at it from a different perspective... (Hint: Why do you think this is in the “Advanced Differentiation” section and not the “Limits and Continuity” section? And what about the title of this section, “Derivatives in Disguise”?)

(Note: In the AP Calculus curriculum, L’Hôpital’s rule is taught in Unit 4 instead of Unit 3, but I’ve placed it here because it feels more logical.)

In the last section, we’ve seen one way we can use derivatives to evaluate limits. But that strategy only helps us in very specific cases when the limit just so happens to look like the limit definition of a derivative.

There is another rule involving derivatives that can help us more often, and that is L’Hôpital’s rule! It can be used to find tough limits where direct substitution doesn’t work.

For example, consider the limit \(\displaystyle\lim_{x \to 0}{\frac{\sin(x)}{x}}\) that we previously found using the squeeze theorem to be 1. Let’s see if L’Hôpital’s rule can be used to find this limit much more quickly.

L’Hôpital’s rule can be used whenever direct substitution gives a result of \(\frac{0}{0}\). In this case, substituting 0 for \(x\) gives \(\frac{\sin(0)}{0} = \frac{0}{0}\), so we can use the rule here.

Using L’Hôpital’s rule is simple: all we have to do is differentiate the numerator and denominator, then find the limit of that expression. In this case, differentiating both the numerator and denominator of \(\frac{\sin(x)}{x}\) gives \(\frac{\cos(x)}{1} = \cos(x)\).

Now, we just have to find the limit of \(\cos(x)\) as \(x\) approaches 0, which is simply \(\cos(0) = 1\). As we can see, using L’Hôpital’s rule gives us the same result as using the squeeze theorem!

(Note that we can’t use L’Hôpital’s rule to actually prove that \(\displaystyle\lim_{x \to 0}{\frac{\sin(x)}{x}} = 1\), since proving that the derivative of \(\sin(x)\) is \(\cos(x)\) requires you to figure out this limit first.)

More formally, L’Hôpital’s rule states that we have two differentiable functions \(f(x)\) and \(g(x)\), if \(\displaystyle\lim_{x \to c}f(x) = 0\) and \(\displaystyle\lim_{x \to c}g(x) = 0\), then \(\displaystyle\lim_{x \to c}\frac{f(x)}{g(x)} = \lim_{x \to c}\frac{f'(x)}{g'(x)}\) (if \(\displaystyle\lim_{x \to c}\frac{f'(x)}{g'(x)}\) exists).

Sometimes we will have to use the rule multiple times. Here’s an example:

Direct substitution gives us \(\frac{0}{0}\), so let’s use L’Hôpital’s rule to get:

Direct substitution still gives us \(\frac{0}{0}\), so we have to use the rule again, giving us:

Now we can use direct substitution.

It is possible to find this limit using trig identites and algebraic manipulation, but oftentimes L’Hôpital’s rule is a less messy alternative.

L’Hôpital’s rule can also be used for limits at infinity when using direct substitution gives an indeterminate form involving infinity, such as \(\frac{\infty}{\infty}\) or \(\frac{-\infty}{\infty}\). Here’s an example:

Both the numerator and denominator approach infinity as \(x\) approaches infinity, so we need to use L’Hôpital’s rule. Taking the derivative of both the numerator and denominator gives us:

Now we can evaluate the limit normally. As \(x\) approaches infinity, \(e^x\) approaches infinity, so \(\frac{1}{e^x}\) approaches 0, which is our limit.

L’Hôpital’s rule doesn’t just help us for limits that result in a \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form: it can also help us with other indeterminate forms.

Here are some other indeterminate forms you might encounter with limits:

• \(0 \cdot \infty\)
• \(\infty - \infty\)
• \(0^0\)
• \(1^\infty\)
• \(\infty^0\)

When you see a limit in one of these forms, you cannot immediately conclude that the limit equals a certain value! Here’s an example:

The inner expression \(1 + \frac{1}{x}\) approaches 1 as \(x\) approaches infinity, and the exponent \(x\) approaches infinity as \(x\) approaches infinity. Therefore, this is a limit in the form of \(1^\infty\). However, the limit is not equal to 1 - it is actually equal to \(e\) (by definition!)

Here’s another limit that is of the form \(1^\infty\):

We can find the value of this limit by using exponent properties:

As you can see, even though this is a \(1^\infty\) form, the limit equals \(e^2\). In conclusion, when we see a limit in this form, we cannot tell what it is going to equal without doing more work (and this applies for the other indeterminate forms).

Now let’s see how we can use L’Hôpital’s rule to help with some of these indeterminate forms.

Problem: Find the limit \(\displaystyle\lim_{x\to \infty}x^{1/x}\).

For this limit, the base approaches \(\infty\) as \(x\) approaches \(\infty\), and the exponent \(\frac{1}{x}\) approaches 0 as \(x\) approaches \(\infty\), so this is a limit of the indeterminate form \(\infty^0\). We can’t use L’Hôpital’s rule directly here, but there is a trick we can use to turn this into a form that allows us to use the rule. Let’s first call our limit \(L\):

Now we’re going to take the natural logarithm of both sides:

Now \(\ln(L)\) is in a form that we can use L’Hôpital’s rule on!

However, this is not our final answer. Remember that we took the natural logarithm of our original limit, so to find the original limit, we have to undo the logarithm by exponentiating this result:

In conclusion, \(\displaystyle\lim_{x\to \infty}x^{1/x} = 1\).

There is another way to differentiate complicated functions, and it’s called logarithmic differentiation. The concept is to take the natural logarithm of both sides, then use implicit differentiation.

Problem: Find the derivative of \(f(x) = \displaystyle\frac{x^5}{e^x\sin(x)}\).

We could use the product and quotient rules to differentiate this, but let’s instead try taking the logarithm of both sides first. This allows us to use logarithm properties to simplify the right side:

Now we can implicitly differentiate both sides without having to use the product or quotient rule! We do still have to use the chain rule to differentiate \(\ln(f(x))\).

We know from the original problem that \(f(x) = \frac{x^5}{e^x\sin(x)}\), so we can substitute that in.

Logarithmic differentiation can also help us differentiate functions where both the base and exponent have \(x\) in them.

Problem: Find the derivative of \(f(x) = x^x\).

One way to find this derivative is to rewrite \(x^x\) as \((e^{\ln(x)})^x = e^{x\ln(x)}\), then use the chain rule. Alternatively, we could also use logarithmic differentiation.

Here, we can use the product rule and the chain rule. Using the chain rule, the derivative of \(\ln(f(x))\) is \(\frac{f'(x)}{f(x)}\).

This isn’t a lesson on its own, but rather an interactive demo I’ve created to help you understand a concept better.

• The chain rule tells us how to differentiate a composite function in the form \(f(g(x))\).
• \[ \dv{x}f(g(x)) = f'(g(x)) \cdot g'(x) \] \[ \dv{y}{x} = \dv{y}{u} \cdot \dv{u}{x} \]
• Exponential and logarithmic derivatives:
• \[ \dv{x}a^x = a^x \ln(a) \] \[ \dv{x}\log_a(x) = \frac{1}{x\ln(a)} \]
• If we have the equation of a curve such as \(x^2 + y^2 = 1\), we can differentiate both sides of the equation using implicit differentation. This will require us to use the chain rule.
• Inverse trig derivatives:
• \[ \dv{x}\arcsin(x) = \frac{1}{\sqrt{1 - x^2}} \] \[ \dv{x}\arccos(x) = -\frac{1}{\sqrt{1 - x^2}} \] \[ \dv{x}\arctan(x) = \frac{1}{1 + x^2} \]
• Derivatives of inverse functions: if \(f(x)\) and \(g(x)\) are inverses, then:
• \[ f'(x) = \frac{1}{g'(f(x))} \]
• The derivative of a derivative is known as a second derivative, denoted by \(\dv[2]{x}f(x)\) or \(f''(x)\).
• L’Hôpital’s Rule: if using direct substitution on a limit yields \(\frac{0}{0}\) or both the numerator and denominator are infinite, you can take the derivative of both the numerator and denominator and the limit will stay the same.

Khan Academy Link: Contextual applications of differentiation

All topics covered in Khan Academy:
Green underlined topics are topics at least partially covered on my website and red topics are topics not yet covered on my website. Note that even green topics might not be covered in full detail on my page.

1. Interpreting the meaning of the derivative in context
2. Straight-line motion: connecting position, velocity, and acceleration
3. Rates of change in other applied contexts (non-motion problems)
4. Introduction to related rates
5. Solving related rates problems
6. Approximating values of a function using local linearity and linearization
7. Using L’Hôpital’s rule for finding limits of indeterminate forms*

We’ve been doing a lot of math related to derivatives, but it’s easy to get so focused on manipulating numbers and symbols that we forget what differentiation actually means. So let’s take a break from all of these differentiation techniques and stop to analyze the real-world meaning of derivatives.

I’ve said that derivatives describe how a function changes, but let’s go into some specific examples to make this concept clearer.

Here’s a button. Click it as fast as you can!

You have clicked this button 0 times.

You are currently clicking at a rate of 0 clicks per second.

You have been clicking this button for 0 seconds.

Your average clicking speed has been 0 clicks per second.

Knowing that derivatives represent how fast something is changing, in this situation, what do you think is the derivative of the number of clicks with respect to time? (Remember, “with respect to time” means “how does this variable change as time changes?”)

That would be your current clicking speed! Your clicking speed represents how fast the number of clicks is changing. The faster you click, the faster the number of clicks increases, the higher the derivative is. Because of this, we say that the current clicking speed is the derivative of the number of clicks with respect to time. In other words, as time changes, the number of clicks also changes, and the rate of change of the number of clicks as time changes is the clicking speed.

Don’t be fooled by the average clicking speed. Derivatives represent instantaneous rate of change: how fast something is changing in a single instant. Derivatives don’t describe average rates of change. In this case, the derivative of the number of clicks with respect to time represents how fast you’re clicking at any particular moment, not your average clicking speed over a span of time.

Notice that as you click the button, the current clicking speed changes a lot more than the average clicking speed. This is because unless you’re a robot, you won’t be clicking at exactly the same speed the entire time. Small variations in your click timings can have a large effect on your current clicking speed. This is because the current clicking speed describes instantaneous rate of change, which is more prone to change than the average rate of change, but also more accurately describes how fast you are clicking right now.

Notice that the unit for clicking speed is clicks per second. In general, the unit for a derivative is the unit for the dependent variable divided by the unit for the independent variable. (As a reminder, the independent variable is what you change, and the dependent variable is what changes as a result of the independent variable changing.) In this case, the independent variable is time (measured in seconds) and the dependent variable is the number of clicks, so the derivative is measured in clicks per second.

Let’s look at a different example. Let’s say that hypothetically, I spent some time building a website that explains calculus concepts to students and curious readers. Sounds familiar?

Let’s say that I graphed my progress over time and I found that the percentage of the website that was completed as time went on could be modeled by the equation \(p(t) = 0.5t^2 + 5t\) for \(0 \le t \le 10\). \(t\) is the number of weeks since I started and \(p(t)\) is the percentage of the website I was done with at that time, with 0 meaning no part of the website was done and 100 meaning the website was fully completed.

Problem: The function \(p(t) = 0.5t^2 + 5t\) models my website progress over time. What can this function tell us?

The percentage of the website done as time went on, modeled by the equation \(p(t) = 0.5t^2 + 5t\). (Note: This is fictitious and doesn’t accurately represent the development of the website you’re on right now.)

Being the curious person I am, I want to know how my development speed has changed over these 10 weeks. To do that, all I need to do is take the derivative of this function, and it will tell me my development speed at every point in time.

Using the power rule and sum rule, I quickly figure out that the derivative of this function is \(p'(t) = t + 5\).

This is the derivative of \(p(t)\), which describes how much the progress on my website changes over time. In other words, it describes how fast I work on my website as time passes.

From this graph of \(p'(t)\) alone, I can learn a lot. First of all, the line isn’t flat, meaning the speed that I made my website at was constantly changing. I notice that \(p'(t)\) is increasing, meaning that I’ve been working faster and faster as time passed. This makes sense because as I made more progress on my website, I became more and more familiar with website development, allowing me to work more efficiently.

The unit of \(p'(t)\) is percent per week, because the independent variable is time (measured in weeks) and the dependent variable is progress (measured in percent of the website completed). At \(t = 0\), \(p'(t)\) is 5, meaning that when I started working on the website, I was working at an instantaneous rate of 5% per week. At \(t = 10\), \(p'(t)\) is 15, meaning that by the time I finished my website 10 weeks after I started, I was completing 15% per week, 3 times faster than when I started!

I was only able to figure all of this out thanks to the power of differentiation. Differential calculus is useful because it can give you all sorts of insights into how things are changing, whether that is an object’s position, someone’s bank account, or in this case, the progress being made on something!

One more important application of derivatives is in economics, with the idea of marginal cost. Marginal cost is the amount of money it will cost a business to produce one more unit of a certain item.

If you have a function \(C(x)\) for the total cost to produce \(x\) units of an item, then taking the derivative of that function will tell you how fast that cost increases as the number of units increases: in other words, how much it costs to produce one more unit.

This information can help a business determine the optimal number of units to produce to maximize their profit. They might decide to stop producing at a certain point if they find that producing additional units would be too expensive (the derivative is too high).

It is summer and you are going on a road trip to your favorite vacation destination! As you sit in the car, the driver gets on the freeway, and you feel a burst of acceleration for a few seconds.

This is a classic example of a real-world situation that can be modeled with calculus! Notice that there are some related variables that are continuously changing in this story: the car’s position, the car’s speed, the car’s acceleration, etc. This calls for differential calculus!

Before we start our first problem, let’s get some definitions out of the way.

• Velocity describes the speed and direction of an object.
  • There are two types of velocity we can calculate:
    • Average velocity: the velocity of an object over a specified period of time, measured as an average. This can be calculated with the formula \(\frac{\Delta x}{\Delta t}\), where \(\Delta x\) is the change in position and \(\Delta t\) is the elapsed time.
    • Instantaneous velocity: the velocity of an object at a single point in time. This can be calculated with the formula \(\dv{x}{t}\), which is the derivative of position with respect to time.
  • In one-dimensional space, velocity can be positive or negative. Positive velocity typically means the object is moving to the right, and negative velocity means the object is moving to the left.
• Speed describes how fast an object is moving (ignoring direction).
  • Speed is the absolute value of an object’s velocity, and cannot be negative.
• Acceleration describes the rate of change of an object’s velocity.
  • Just like velocity, it is possible to calculate average and instantaneous acceleration. Instantaneous acceleration is the derivative of velocity with respect to time, or the second derivative of position with respect to time.
  • In one-dimensional space, an object is speeding up if its acceleration has the same sign as its velocity. An object is slowing down if its acceleration and velocity have different signs. For example:
    • An object with positive velocity and positive acceleration is speeding up.
    • An object with negative velocity and positive acceleration is slowing down. Be careful: positive acceleration does not necessarily mean an object is speeding up!

Problem: Let’s say the car is driving on a one-dimensional road. The car’s position can be modeled by \(f(t) = t^2 + 10t\) for \(0 \le t \le 10\), where \(f(t)\) represents position in meters and \(t\) represents time elapsed in seconds. What can we learn about the car’s movement?

This graph represents the car’s position over time.

Now let’s use calculus to break this down. If we take the derivative of position with respect to time, it tells us how fast our position is changing, which is our velocity! In one dimension, our velocity is our speed and direction described in one number. Positive velocity means we’re moving forwards and negative velocity means we’re moving backwards, and the absolute value of our velocity tells us our speed.

Using the power rule and sum rule, the derivative of \(f(t)\) is \(f'(t) = 2t + 10\). This means the velocity of our car at any given time is \(2t + 10\) meters per second, where \(t\) is the number of seconds that have elapsed.

This graph represents the car’s velocity over time.

By taking the derivative of our position, we can find that our car accelerated from 10 meters per second all the way up to 30 meters per second. We can also find our velocity at specific moments: at \(t = 3.5\) seconds, our velocity was \(f'(3.5) = 17\) meters per second.

What if we take it a step further and find the derivative of this derivative? That would describe how fast our velocity is changing, which is our acceleration. Assuming our velocity is positive, positive acceleration means we’re speeding up and negative acceleration means we’re slowing down.

Using the power rule again, the derivative of \(f'(t)\) is \(f''(t) = 2\). This is the second derivative of position with respect to time. What this means is that our acceleration (change in velocity) throughout the 10 seconds was constant at 2 meters per second per second, or 2 meters per second squared. Notice how the unit has changed once again because we’re describing how fast our velocity is changing, not how fast our position is changing. (Meters per second squared, or \(\text{m/s}^2\), is the unit of acceleration. You may have seen this unit previously in a physics class.)

This graph represents the car’s acceleration over time.

Problem: Here’s a more complicated example. The function \(f(t) = t^4 - 3t^3 + 3t\) represents the position of a particle over time relative to its starting position. What can differential calculus tell us about this particle’s motion?

This graph represents the particle’s position over time.

Taking the derivative of this gives us the particle’s velocity over time.

This graph represents the particle’s velocity over time. \(f'(t) = 4t^3 - 9t^2 + 3\)

What does this graph tell us? It tells us that at first, the particle was moving forwards (as indicated by its positive velocity), then it started moving backwards for a short time (when its velocity was negative), then it started moving forwards again. Notice that when its velocity is positive, its position is increasing, and when its velocity is negative, its position is decreasing.

In general, a negative derivative means that some variable is decreasing and a positive derivative means that some variable is increasing.

This graph represents the particle’s acceleration over time. \(f''(t) = 12t^2 - 18t\)

At first, the particle’s acceleration was negative, which means that it was slowing down (since its velocity was positive). But then the velocity reached 0 and started going negative, meaning the particle actually started speeding up at that point (negative velocity and negative acceleration means its speed is actually increasing). When its acceleration became positive, the particle started slowing down until its velocity became positive, after which it started speeding up.

Remember, acceleration only means something is speeding up if it has the same sign as the velocity (both acceleration and velocity are positive or both acceleration and velocity are negative).

In conclusion, the first derivative of an object’s position with respect to time is its velocity, and the second derivative of position with respect to time is acceleration.

Often times, two variables are dependent on a third variable that changes. There are many real-world problems that involve variables like these, and they are known as related rates problems. One example is a growing circle: both its area and its radius are dependent on time (how long has elapsed since the circle started growing).

Problem: Let’s say we have a circle with a radius of 2 meters, and its radius is currently increasing by 1 meter per second. How fast is its area growing right now? In other words, what is the derivative of the circle’s area with respect to time?

How fast is the circle’s area growing?

Both the area and radius of the circle can be represented as functions of time, as they both depend on how much time has elapsed. We’ll represent the area of the circle with the function \(A(t)\) and the radius with the function \(r(t)\). To answer our question, we need to find the value of \(A'(t)\) at this particular instant.

The area directly depends on the radius, so let’s relate the area and the radius of our circle by taking the equation \(A = \pi r^2\) and substituting our functions in.

We are trying to find the rate of change of the area, or \(A'(t)\), so we’re going to take the derivative of both sides next. Note that because we’re differentiating with respect to time (as \(A(t)\) and \(r(t)\) are functions of time), you’ll see \(\dv{}{t}\) instead of \(\dv{}{x}\).

The third step is an application of the chain rule. The derivative of \([r(t)]^2\) with respect to \(t\) is the derivative of \([r(t)]^2\) with respect to \(r(t)\) multiplied by the derivative of \(r(t)\) with respect to \(t\).

Now we have an equation for the area’s growth rate at any point in time \(t\) if we know \(r(t)\) and \(r'(t)\).

To solve our problem, we’ll call the current time \(t_0\) (when the circle’s radius is 2 meters and it’s growing by 1 meter per second). We know that the radius of the circle is currently 2 meters, so \(r(t_0) = 2\). The radius is growing by 1 meter per second, so \(r'(t_0) = 1\). Now all we have to do is substitute \(r(t_0)\) and \(r'(t_0)\) into our equation to find \(A'(t_0)\).

What this means is that our circle’s area is currently growing at a speed of \(4\pi\) square meters per second! Note that this speed is itself increasing as the circle gets bigger, since the radius \(r(t)\) is increasing at a rate of 1 meter per second. But at this instant, the instantaneous rate of change of the circle’s area is \(4\pi\) square meters per second.

Here is a simulation of the circle. Pay attention to what happens at the time \(t_0\)!

Here’s a more complicated related rates problem.

Problem: Imagine we have two moving cars, Car A and Car B, and we want to know how fast the distance between them is shrinking. Car A and Car B are heading towards the same perpendicular intersection. Car A, which is currently 200 meters from the intersection, is moving 15 meters per second south. Car B, which is currently 300 meters from the intersection, is moving 10 meters per second west. How fast is the distance between the cars changing at this instant?

For related rates problems, it helps to assign variables to all of the values in the situation. Let’s make a list of what we know:

• The intersection is perpendicular
• Car A is 200m from the intersection
• Car B is 300m from the intersection
• Car A is moving at 15 m/s towards the intersection
• Car B is moving at 10 m/s towards the intersection

We’ll say that \(A(t)\) is Car A’s distance from the intersection, \(B(t)\) is Car B’s distance from the intersection, and \(D(t)\) is the distance between the two cars at any time \(t\). Once again, we’ll call the current time \(t_0\), meaning that \(A(t_0) = 200\) and \(B(t_0) = 300\). \(D(t_0)\) will tell us how far the cars are apart right now (at time \(t_0\)).

It also helps to make a diagram of the situation as it will make it much easier to figure out what to do next. Here’s a diagram of this situation:

Recalling the meaning of a derivative, \(A'(t_0)\) represents how fast Car A’s distance from the intersection is changing now. Because Car A is moving towards the intersection, this distance is decreasing, so \(A'(t_0)\) is negative. The magnitude of \(A'(t)\) tells you the car’s speed. In this case, \(A'(t_0) = -15\) because the car is moving at 15 m/s towards the intersection. Likewise, \(B'(t_0) = -10\).

Now let’s set up an equation. Looking at the diagram, we could use the Pythagorean Theorem here (since the intersection is perpendicular and thus forms a right angle). At any point in time, \([A(t)]^2 + [B(t)]^2 = [D(t)]^2\).

We’re ready to start solving for our answer. First, we need to know what variable we’re even solving for. The answer asks for how fast the distance \(D(t)\) between the cars is decreasing at this moment \(t_0\), so that would be \(D'(t_0)\). It’s important to always keep in mind which value the problem asks you to solve for! Then, use the equations you have to find an expression for this value.

In this case, we can differentiate our equation \([A(t)]^2 + [B(t)]^2 = [D(t)]^2\) to get an equation for \(D'(t)\). Once again, we have to use the chain rule.

We’re talking about the current time \(t_0\), so we can replace all instances of \(t\) with \(t_0\), then substitute the values we know.

We can find \(D(t_0)\) with our original Pythagorean equation.

We can ignore the negative solution for \(D(t_0)\) since distance is always positive. Now we can put everything together.

The distance between the cars is currently decreasing at a rate of about 16.641 meters per second.

If a function is differentiable, it possesses a very interesting property: if you zoom in on it enough, it will approach a straight line! This property is known as local linearity.

This function \(f(x) = x^2\) looks like a curve, but if you zoom in enough...

It looks like a line. This property actually comes in very useful, because it means that if we have a differentiable function, we can approximate a small region of it as a linear function.

The green line closely approximates the red function in this region.

How do we find this line though? It’s simple: it’s simply the line tangent to a point on the function! In this example, the green line is the tangent line to the function \(f(x) = x^2\) at \(x = 2\).

Using this tangent line, we can approximate values of \(f(x)\) for values of \(x\) near 2. For example, if we wanted to approximate \(f(1.9)\), we could find the \(y\)-value of the point on the tangent line with an \(x\)-value of 1.9.

Problem: How do we approximate \(f(1.9)\) using local linearity?

The derivative of \(f(x) = x^2\) is \(f'(x) = 2x\), so the slope of the tangent line at \(x = 2\) is \(2 \cdot 2 = 4\). Using point-slope form, the equation of the tangent line is \(y - 4 = 4(x - 2)\). Plugging in \(x = 1.9\) into this equation and solving for \(y\) gives us our approximation for \(f(1.9)\).

The true value of \(f(1.9)\) is \(1.9^2 = 3.61\), which is very close to our approximation of 3.6.

This technique is especially useful if our function is more complicated and its value at a certain point is hard to calculate. In fact, computers often use this property of local linearity to quickly calculate values of certain functions such as square roots. (You can learn more about this in the next section about Newton’s method!)

Here’s an example problem involving square roots:

Problem: Approximate the value of \(\sqrt{26}\) using a linear approximation.

Let’s say that you didn’t have a calculator and wanted to calculate the square root of 26. How would you do that?

We know that the square root of 25 is 5, so therefore the square root of 26 will be close to 5. But how can we get a more precise answer?

We can use the local linear approximation of \(f(x) = \sqrt{x}\) at \(x = 25\), since \(f(25)\) and \(f'(25)\) are easy to calculate. In addition, 25 is relatively close to 26, so our linear approximation will give us an accurate value for \(f(26)\).

Therefore, the tangent line to \(f(x) = \sqrt{x}\) at \(x = 25\) is:

What’s nice about this tangent line is that we can easily evaluate what \(y\) is at \(x = 26\). This is the value of \(y\) at \(x = 26\):

Therefore, \(\sqrt{26} \approx 5.1\), which is really close to the true value of about 5.09902.

Did you know that calculus was used by clever programmers to make the first-person shooter game Quake III Arena run faster? It’s true, and the secret involves a calculus technique known as Newton’s method!

Before I explain what Newton’s method is, let’s first talk about the problem that led Newton’s method to be implemented in this game in the first place.

Because Quake III Arena was a 3D game, it had to do many graphics calculations per second. The problem is that the game was released in 1999 when computers were much slower than they are today. The calculations were often too much for computers in the 1990s to handle, so it was important that each calculation could be done as fast as possible.

One calculation that was done frequently was calculating the reciprocal of the square root of a number. Algebraically, given a number \(x\), the number \(\frac{1}{\sqrt{x}}\) needed to be calculated. (If you’re interested in the details, this calculation was used to convert vectors of any magnitude to unit vectors, which are vectors of magnitude 1.)

The reciprocal square root \(\frac{1}{\sqrt{x}}\) was often used in 3D rendering, such as the lighting calculations required to render this cube.

Numbers in computers are stored in binary, a number system that uses only the digits 0 and 1. Programmers in the 1990s came up with a very clever way to approximate \(\frac{1}{\sqrt{x}}\) by manipulating the binary representation of numbers in computers. This trick worked very well and computers could perform the trick extremely quickly, but there was a problem.

The approximation that this algorithm produced was not exact: sometimes it could be off by more than 3% of the true value of \(\frac{1}{\sqrt{x}}\). The programmers wanted a way to get a more accurate result while still keeping the algorithm fast. Their problem was essentially:

Problem: Given an approximation of \(\frac{1}{\sqrt{x}}\) for a certain value of \(x\), how can we make our approximation better without doing many calculations (i.e. without directly calculating \(\frac{1}{\sqrt{x}}\))?

Here’s an example: let’s say we’re trying to calculate \(\frac{1}{\sqrt{2}}\) as a decimal, and our best guess for its value is 1. How could we get a closer estimate?

One way we can do this is by using Newton’s method, named after Isaac Newton. To use it, we need to set up a function \(f(x)\) that has a zero at \(\frac{1}{\sqrt{2}}\) (i.e. \(f(\frac{1}{\sqrt{2}}) = 0\)). One function that meets this property is \(f(x) = \frac{1}{x^2} - 2\).

The function \(f(x) = \frac{1}{x^2} - 2\) has a zero at \(x = \frac{1}{\sqrt{2}}\).

Our goal is to find the approximate \(x\)-coordinate of this zero (since we want to find an approximation for \(\frac{1}{\sqrt{2}}\)). Let’s start off by plotting the point on the function corresponding to our initial guess \(x = 1\):

The point \((1, -1)\) corresponds to our initial guess \(x = 1\).

Using this point \((1, -1)\), how can we come up with a better guess of one of the zeros of this function? In other words, what calculations can we do to get closer to \(x = \frac{1}{\sqrt{2}}\)? (Hint: we can use the idea of local linearity!)

The trick is to draw the line tangent to the function at \(x = 1\), which will give us a linear approximation of the curve \(f(x)\). Here’s what that looks like:

The tangent line to \(f(x)\) can give us a better approximation!

If we look at the zero (\(x\)-intercept) of the tangent line, its \(x\)-coordinate is slightly closer to \(x = \frac{1}{\sqrt{2}}\) than our starting guess \(x = 1\). This means that the tangent line’s zero is a better approximation of the function’s zero \(x = \frac{1}{\sqrt{2}}\) than our original guess of \(x = 1\).

We can solve for the zero of the tangent line. Using point-slope form, the equation of a line that passes through \((1, -1)\) is \(y + 1 = m(x - 1)\). To find the slope of the tangent line, we need to differentiate \(f(x)\) first.

The slope of the tangent line is \(f'(1)\), which is equal to \(-\frac{2}{1^3} = -2\). This means the equation of the tangent line is \(y + 1 = -2(x - 1)\). To find the zero of this tangent line, we want to find where \(y = 0\), so we will substitute \(y = 0\) and solve for \(x\).

This means the zero of the tangent line (which is our improved estimate for \(\frac{1}{\sqrt{2}}\)) is \(x = \frac{1}{2}\). This is slightly closer to \(x = \frac{1}{\sqrt{2}}\) than our initial guess \(x = 1\)!

The beauty of Newton’s method is that we can start this process all over again to get a more accurate result. If we draw the line tangent to \(f(x)\) at \(x = \frac{1}{2}\), we can get an even better approximation of \(\frac{1}{\sqrt{2}}\).

Let’s find a general equation for the line tangent to the function \(f(x) = \frac{1}{x^2} - 2\) at any point \(x = a\). We will start off with point-slope form:

The tangent line passes through the point \((a, f(a))\), so let’s substitute that in for \((x_1, y_1)\).

The slope of the tangent line is \(f'(a)\), so we will substitute that for \(m\).

We know that \(f(x) = \frac{1}{x^2} - 2\) and \(f'(x) = -\frac{2}{x^3}\).

Now we have an equation for the line tangent to \(f(x)\) at \(x = a\). If we plug in \(x = \frac{1}{2}\) and solve for the zero of that line (by setting \(y = 0\)), we can get a better approximation of \(\frac{1}{\sqrt{2}}\).

If we’re going to use Newton’s method multiple times, it’s useful to write an explicit equation for the next more accurate \(x\)-value in terms of the current \(x\)-value (i.e. a function that takes in an approximation and gives a more accurate approximation). To do that, we can just solve for \(x\) in our general tangent line equation.

We want to set \(y = 0\) to find the zero of the tangent line:

Given an initial approximation \(a\), this formula will output a better approximation \(x\) of the value of \(\frac{1}{\sqrt{2}}\). For example, plugging in \(a = \frac{1}{2}\) will give us \(x = \frac{5}{8}\). If we then plug in \(a = \frac{5}{8}\), we get an even better approximation \(x = \frac{355}{512} \approx 0.693359\).

What’s neat about this equation is that it only has multiplication and subtraction, meaning that it’s very easy for calculators and computers to use it. (\(\frac{3}{2}\) can be stored as the constant 1.5 and \(a^2\) is just \(a\) times \(a\).)

Each time we use this formula to get a better approximation, we call it an iteration. The more iterations we do, the closer we get to \(\frac{1}{\sqrt{2}}\), as shown by this table:

The true value of \(\frac{1}{\sqrt{2}}\) is about 0.707107.

Let’s say that instead of approximating \(\frac{1}{\sqrt{2}}\), we want to approximate \(\frac{1}{\sqrt{b}}\) for any value of \(b\). What formula could we use to do that?

The function \(f(x) = \frac{1}{a^2} - b\) has a root at \(\frac{1}{\sqrt{b}}\), so if we can find the zeros of the line tangent to that curve, we can find a formula that will allow us to calculate the reciprocal square root of any number \(b\).

We can figure out the tangent line to this curve just like we did before:

By setting \(y = 0\), we can find the zero of this tangent line:

This was the formula that Quake III Arena used to calculate the reciprocal square root quickly! With one iteration of Newton’s method, it was able to get the maximum error of its algorithm down from over 3% to under 0.2%.

The full algorithm involving both the binary manipulation and Newton’s method is known as Fast Inverse Square Root, and it remains a famous example of a creative algorithm in computer science.

Want to learn more about Fast Inverse Square Root? Luckily, I’ve written an entire page about it!

Now let’s try to generalize Newton’s method for any function \(f(x)\). The slope of the tangent line to the point \((a, f(a))\) for any function \(f(x)\) is \(f'(a)\).

By substituting this point and the slope of the tangent line into the point-slope formula, we can get a general expression for the zero of the tangent line.

Sometimes you will see Newton’s method written like this:

In this equation, \(x_{n+1}\) is the next approximation of one of the zeros of \(f(x)\), which is usually a better approximation than \(x_n\).

The idea behind Newton’s method is to approximate a function \(\class{blue}{f(x)}\) with a tangent line, then find the \(x\)-intercept of that tangent line to approximate one of the zeros of the function. The zero of the tangent line is \(x = a - \frac{f(a)}{f'(a)}\).

In the previous section, we used a local linear approximation to approximate \(\sqrt{26}\). Now let’s take it a step further and use Newton’s method to get an even better approximation.

Problem: Approximate \(\sqrt{26}\) using Newton’s method.

The first step to using Newton’s method is to find a function \(f(x)\) that has a zero at \(\sqrt{26}\). One function that works is \(f(x) = x^2 - 26\).

Second, we need an initial guess for \(\sqrt{26}\). We know that \(\sqrt{26}\) is close to 5, so 5 works as a good initial guess here.

Now let’s use Newton’s method. The formula is:

For the function \(f(x) = \sqrt{x}\), this becomes:

We will set \(x_0\) as our initial guess, in this case 5.

This value of \(x_1\) is closer to \(\sqrt{26}\) than our original guess. We can get an even better approximation by repeating the process with \(x_1\) instead of \(x_0\).

Let’s do this one more to get an even better approximation \(x_3\).

Notice how the two approximations \(x_2\) and \(x_3\) are nearly identical. This means that we can be confident that the square root of 26 is close to \(x_3\), or about 5.099020.

Not only does Newton’s method allow us to approximate numbers, it also lets us find approximate solutions to some equations that were previously unsolvable!

Problem: Find an approximate solution to the equation \(x = \cos(x)\).

To use Newton’s method, we need an equation in the form \(f(x) = 0\). By subtracting \(x\) from both sides, we can get our equation into this form.

Now we can use the Newton’s method formula to find where the function \(\cos(x) - x\) equals zero!

First, we need an initial guess for the solution of \(x = \cos(x)\). If we do a quick sketch, we can find that the lines \(y = x\) and \(y = \cos(x)\) intersect somewhere between \(x = 0\) and \(x = \frac{\pi}{2}\). So let’s use the midpoint \(x_0 = \frac{\pi}{4}\) as our initial guess.

The solution to \(x = \cos(x)\) is somewhere in between \(x = 0\) and \(x = \frac{\pi}{2}\).

Here is the Newton’s method work:

Our first approximation \(x_1\) is about 0.739536. If we run through this same process to get \(x_2\), we find that \(x_2\) is about 0.739085. Doing one more iteration to get \(x_3\) once again gives us a value of about 0.739085.

Because we got two approximations in a row \(x_2\) and \(x_3\) that agree to 6 decimal places, we can be confident that the solution to \(x = \cos(x)\) is very close to 0.739085.

Sometimes, Newton’s method will fail to work. Its failure might be caused by certain functions (for example, it can’t be used to find the zero of \(f(x) = \sqrt[3]{x}\)), or it might fail if the initial approximation is too far off. In addition, sometimes the sequence of approximations never converges to a value or you will run into a division by zero (if \(f'(x_n) = 0\)).

• The derivative of a function represents a function’s instantaneous rate of change, or how fast and in what direction it is changing at any \(x\)-value.
• The derivative of an object’s position is its velocity (speed and direction), and the derivative of an object’s velocity is its acceleration.
• If you zoom into the graph of a differentiable function, it will eventually look similar to a straight line. This is known as local linearity.
  • You can use the line tangent to a differentiable function at a point \(x = a\) to approximate the value of that function for values of \(x\) near \(a\).

Khan Academy Link: Applying derivatives to analyze functions

All topics covered in Khan Academy:
Green underlined topics are topics at least partially covered on my website and red topics are topics not yet covered on my website. Note that even green topics might not be covered in full detail on my page.

1. Using the mean value theorem
2. Extreme value theorem, global versus local extrema, and critical points
3. Determining intervals on which a function is increasing or decreasing
4. Using the first derivative test to find relative (local) extrema
5. Using the candidates test to find absolute (global) extrema
6. Determining concavity of intervals and finding points of inflection: graphical
7. Determining concavity of intervals and finding points of inflection: algebraic
8. Using the second derivative test to find extrema
9. Sketching curves of functions and their derivatives
10. Connecting a function, its first derivative, and its second derivative
11. Solving optimization problems
12. Exploring behaviors of implicit relations

It’s road trip time again! (You can probably tell that I like road trip examples!)

This time, I’ve driven all the way from San Francisco to Los Angeles, a 380-mile drive. I completed my trip in 6 hours and 20 minutes, putting my average speed at 60 miles per hour.

Then one of my math friends, who is also studying calculus, tells me that since my average speed throughout the trip was 60 miles per hour, at some point during the drive, I must have been driving at exactly 60 miles per hour.

It turns out my friend actually used a calculus theorem, the mean value theorem, to make that conclusion!

The mean value theorem states that if a function is continuous in the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then at some point on that function between \(x = a\) and \(x = b\), the slope of that function is equal to the average rate of change between \(x = a\) and \(x = b\).

In my road trip example, let’s say \(f(t)\) represents the distance I’ve traveled so far as a function of time. Because my average speed was 60 miles per hour, the average rate of change of this function across the entire trip is 60 miles per hour.

What the mean value theorem tells me is that at some point on the trip (at some time between \(t = \text{0 minutes}\) and \(t = \text{6 hours 20 minutes}\)), my instantaneous speed was exactly 60 miles per hour.

More formally, the mean value theorem states that if \(f(x)\) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there is a value \(c\) in \((a, b)\) such that:

This is the average rate of change of \(f(x)\) over the interval \([a, b]\).

Here’s a more theoretical example. Problem: Consider the function \(f(x) = x^3 + 3\). What does the mean value theorem tell us about this function between \(x = 0\) and \(x = 2\)?

In order to use the mean value theorem in this case, we need to make sure that \(f(x)\) is continuous over \([0, 2]\) and differentiable over \((0, 2)\). \(f(x)\) is continuous and differentiable everywhere, so we can use the mean value theorem here.

The average rate of change of \(f(x)\) between \(x = 0\) and \(x = 2\) is the slope of the secant line between those two points.

The secant line tells us the average rate of change between \(x = 0\) and \(x = 2\).

We can calculate the slope of the secant line with the rise over run formula, which will tell us the function’s average rate of change over the interval \([0, 2]\).

The mean value theorem tells us that at some point in this interval \((0, 2)\), this function’s instantaneous rate of change will be equal to the average rate of change, which is 4 in this case. In other words, at some value of \(x\) between 0 and 2, the function’s derivative is equal to 4. Let’s see if we can find this point.

The derivative of \(f(x) = x^3 + 3\) is \(f'(x) = 3x^2\). Let’s set that derivative equal to 4 and solve for \(x\).

One of the \(x\)-values, \(\frac{2}{\sqrt{3}}\), is in the interval \((0, 2)\), and that is the point we are looking for! At \(x = \frac{2}{\sqrt{3}}\), the slope of the tangent line is exactly 4. The mean value theorem guaranteed that we would find at least one of these points within the interval.

At \(x = \frac{2}{\sqrt{3}}\), the slope of the tangent line is exactly the same as the slope of the secant line. Because of this, the tangent line is parallel to the secant line. If the continuity and differentiability requirements for the mean value theorem are met, we will always be able to find a tangent line within an interval that is parallel to the secant line through the interval’s endpoints.

The special case of the mean value theorem where \(f(a) = f(b)\) is known as Rolle’s theorem. Rolle’s theorem states that if \(f(x)\) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there is a point \(c\) in \((a, b)\) where \(f'(c) = 0\).

After my road trip, my calculus-obsessed friend tells me that sometime during the trip, my car reached a maximum speed (i.e. the car was moving faster at that point in time than it was during the rest of the trip). Once again, there is a calculus theorem for this, and it’s known as the extreme value theorem. Before we get into the details, let’s talk about the extrema of functions.

One of the most important uses of derivatives is using them to identify the extrema of functions (their maxima and minima). Let’s review what those are first.

(Extrema, maxima, and minima are the plurals of extremum, maximum, and minimum respectively. The term “extrema” refers to both maxima and minima.)

A global (or absolute) maximum is simply the highest \(y\)-value a function reaches across its entire domain. A global minimum is the lowest \(y\)-value a function reaches. A local (or relative) maximum is a point on the function that is higher than the points nearby, and a local minimum is a point that is lower than all nearby points.

The red point is both a local and global minimum. This function has no global maximum because it approaches positive infinity at both ends.

Time for another value theorem! The extreme value theorem states that if a function is continuous over a closed and bounded interval (an interval that includes its endpoints and whose endpoints are finite), it has a maximum and minimum within that interval. This might sound obvious, but it’s an important theorem nevertheless.

Why does the function need to be continuous for the extreme value theorem to apply? Well, consider this discontinuous function.

Problem: What is the function’s maximum value in the interval \([-2, 2]\)?

You might say it’s 4, but there is no \(x\)-value for which \(f(x) = 4\) because of the hole at \((0, 4)\). You also might say a value very close to 4, but that also wouldn’t work. For example, if I said that the maximum value was 3.9, that would be incorrect because the function reaches values higher than 3.9. The same thing goes for 3.99, 3.999, etc. If I name any number less than 4, I can find another number greater than it that the function reaches.

What this means is that this function has no maximum in the interval \([-2, 2]\) because there is no reasonable value that could be the maximum (it can’t be 4, it can’t be less than 4, and it definitely can’t be greater than 4).

Another example that shows why the extreme value theorem only works on continuous functions is the function \(f(x) = \frac{1}{x}\). Think about it: what is its minimum and maximum value in the interval \([-1, 1]\)?

Discontinuous functions might not have extrema within a closed interval. But if a function is continuous, it will definitely have a maximum and minimum within any closed interval (that’s what the extreme value theorem says).

Local extrema have a very special property: If a function is differentiable at a local minimum or maximum, its tangent line will be horizontal at that extremum. That means that the derivative at an extremum will be 0 (if it is defined at all).

The function has a derivative of 0 at both its local minimum and local maximum. This is shown by the horizontal tangent lines.

Note that just because a function has a local extremum doesn’t guarantee that its derivative at that point is 0. Sometimes the derivative is undefined, like in this example:

This function is not differentiable at its maximum because it has a sharp turn.

And sometimes that extremum is just at the edge of a function’s domain:

This function has a minimum, but it is at the endpoint of its domain, so its derivative isn’t necessarily 0.

In addition, just because a function’s derivative is 0 at some point doesn’t necessarily mean that the function has an extremum at that point. Sometimes, the function can flatten out like this:

This function has a horizontal tangent line at a point that isn’t a maximum or minimum.

So let’s say that we want to use a function’s derivative to find its local maxima and minima. How would we do that?

The first thing to notice is that if a function has an extremum somewhere, its derivative is either 0 or undefined (if the extremum isn’t at the edge of a function’s domain; we’re not going to worry about that case for now).

So clearly there’s something special about these points where the derivative is 0 or undefined. We call these points critical points: points in the domain of a function \(f(x)\) where \(f'(x)\) is either 0 or undefined. These are points that could potentially be local maxima or minima.

Importantly, critical points must be in the original function’s domain. Even if a function’s derivative is undefined at a point, it’s not a critical point unless the original function is defined at that point.

If a function has a local extremum that is not on the edge of its domain, that point will be a critical point, but not all critical points are extrema.

Finding critical points is simple: just take the derivative of a function and find where its derivative is undefined or equal to 0.

Example problem: find the critical points of \(f(x) = \sqrt[3]{x} - x\).

First we find the derivative of this function. \(f(x)\) can be rewritten as \(f(x) = x^{1/3} - x\), allowing us to use the power rule next.

Now we find the \(x\)-values for which \(f'(x)\) is 0 or undefined. First, let’s find the undefined points first. \(f'(x)\) is undefined when the denominator in the fraction, \(3 \cdot \sqrt[3]{x^2}\), is zero:

That gives us our first critical point. Now let’s find where the derivative is equal to zero:

Now we have to confirm that all three of our potential critical points are in the original function’s domain. \(f(x) = \sqrt[3]{x} - x\) is defined for all real numbers, so all three of our points, \(x = 0\), \(x = \frac{1}{\sqrt{27}} \approx 0.192\), and \(x = -\frac{1}{\sqrt{27}} \approx -0.192\), are critical points.

Now we know that the function’s local extrema can only happen on these three points. However, we would still need to do some additional checking to verify that these points are actually local extrema and whether they are minima or maxima.

The derivative tells us the rate of change of a function. But importantly, it also tells us what direction a function is changing in. If a function’s derivative is positive, that function is increasing, and if the derivative is negative, the function is decreasing.

The blue function is the derivative of the red function. I’ve shaded the areas where the red function is increasing and decreasing. Notice how the function’s derivative is positive when the function is increasing and negative when the function is decreasing.

Differential calculus allows us to find these increasing and decreasing intervals easily just by taking the derivative of a function and finding where it is positive and negative.

Problem: Let’s say we want to find where the function \(f(x) = -x^3 + x\) is increasing and decreasing.

First, we take the derivative of this function to get \(f'(x) = -3x^2 + 1\).

Now we need to find where this derivative \(f'(x)\) is positive and negative. To do this, we can utilize the function’s critical points, points where the derivative is 0 or undefined. These are places where the derivative can change sign, from positive to negative or negative to positive.

The critical points for this function are \(\frac{1}{\sqrt{3}}\) and \(-\frac{1}{\sqrt{3}}\). That means that we can split the function into three regions:

• \(x < -\frac{1}{\sqrt{3}}\)
• \(-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}\)
• \(x > \frac{1}{\sqrt{3}}\)

The sign of the derivative will stay the same within each of these regions. Now we just need to figure out what the sign of each region is. We can do that by plugging in one \(x\)-value from each region (shown in the second column of the table below) into the derivative and seeing whether it’s positive or negative.

What this means is that the derivative is negative when \(x < -\frac{1}{\sqrt{3}}\) or \(x > \frac{1}{\sqrt{3}}\), and positive when \(-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}\).

The function is increasing when its derivative is positive, so the function’s increasing interval is \(-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}\). Its decreasing intervals are the places where the derivative is negative: \(x < -\frac{1}{\sqrt{3}}\) and \(x > \frac{1}{\sqrt{3}}\).

We’ve seen that critical points are points that could potentially be local extrema. But how do we tell if a critical point is a maximum, minimum, or neither? Let’s go back to this diagram from the last section:

The blue function is the derivative of the red function.

Look at the local maximum of the red function. The function is increasing before it reaches the maximum and decreasing after. Its derivative is positive before it reaches the maximum and negative after.

Now look at the function’s local minimum. We see the exact opposite: the function is decreasing before it reaches the minimum (its derivative is negative), then the function is increasing after (its derivative is positive).

Whenever a function hits a critical point, one of three things can happen:

• Its derivative goes from positive to negative. This means that the function hit a local maximum.
• Its derivative goes from negative to positive. This means that the function hit a local minimum.
• The sign of the derivative doesn’t change. This means that the function did not reach a local extremum.

Let’s go back to the example in the previous section with the function \(f(x) = -x^3 + x\). Here’s the table from the previous section showing the sign of the derivative in each interval:

We can actually use this table to find the local extrema of this function! At \(x = -\frac{1}{\sqrt{3}}\), the derivative changes from negative to positive, meaning that the function reaches a local minimum. At \(x = \frac{1}{\sqrt{3}}\), the derivative goes from positive to negative, meaning that the function reaches a local maximum.

By plugging in \(x = -\frac{1}{\sqrt{3}}\) into \(f(x) = -x^3 + x\), we can find that \(f(x)\) has a local minimum value of \(-\frac{2\sqrt{3}}{9} \approx -0.385\). Similarly, we can find that the local maximum value is \(\frac{2\sqrt{3}}{9} \approx 0.385\).

The function’s local minimum occurs at \((-0.577, -0.385)\) and its local maximum occurs at \((0.577, 0.385)\).

Note that even if a function’s derivative swaps signs, it’s only a local extremum if the function is defined at that point. Be sure to verify that the function is defined at a point before concluding that it is a local extremum.

We can now find the local minima and maxima of a function, which tells us where a function is greater than or less than all other points around it. But it would be even more useful if we could find global extrema: the lowest or highest point a function reaches within an entire interval.

Problem: Let’s say we want to find the global extrema of the function used in our previous examples, \(f(x) = -x^3 + x\), within the interval \([-1, 2]\).

There are two possibilities for where each global extremum could be:

1. On one of the endpoints of the interval.
2. In between the endpoints of the interval. In this case, the global extremum also has to be a local extremum, meaning it has to occur on a critical point.

Examples of extrema on the endpoints of an interval.

An example of an extremum within the endpoints of an interval. Notice how the extremum lands on a critical point of the function, since the derivative is zero at this point.

This means that to find the global extrema of a function within an interval, we need to check the function at the endpoints of the interval and at the function’s critical points.

We’ve already figured out the function’s critical points: \(x = -\frac{1}{\sqrt{3}}\) and \(x = \frac{1}{\sqrt{3}}\). So we just have to check the value of the function at these two points along with the endpoints of the interval, \(x = -1\) and \(x = 2\). These are our candidates for global extrema.

The largest \(f(x)\) value in this table is 0.385, which corresponds to an \(x\)-value of \(\frac{1}{\sqrt{3}}\). Therefore, 0.385 is the global maximum of \(f(x)\) within the interval \([-1, 2]\), which occurs at \(x = \frac{1}{\sqrt{3}} \approx 0.577\). The smallest \(f(x)\) value is -6, which corresponds to an \(x\)-value of 2. Therefore, the global minimum within this interval is -6, which occurs at \(x = 2\).

The global maximum of \(f(x)\) within the interval \([-1, 2]\) occurs around \((0.577, 0.385)\) while the global minimum within this interval occurs at \((2, -6)\).

This tells us how to find global extrema over an interval, but how do we find global extrema across a function’s entire domain? That will tell us how low or how high a function’s value ever gets.

The first thing we have to do is to consider the end behavior of a function: what happens to \(f(x)\) as \(x\) approaches positive infinity and negative infinity.

What is the global extrema of the function \(f(x) = -x^3 + x\)? Well, \(f(x)\) approaches \(-\infty\) as \(x\) approaches \(\infty\) and \(f(x)\) approaches \(\infty\) as \(x\) approaches \(-\infty\). What this means is that \(f(x)\) has no global minimum or maximum because the function can take on an arbitrarily high or low value.

Let’s try a different function instead. Problem: What are the global extrema of \(f(x) = x \ln(x)\) across its entire domain?

To find the global extrema of this function across its entire domain \(x > 0\), we once again need to find its critical points first.

Differentiating the function with the product rule gives us \(f'(x) = \ln(x) + 1\). Let’s find the critical points where \(f'(x) = 0\).

Now let’s find the function’s increasing and decreasing intervals using the critical points.

Let’s imagine starting at the left endpoint of the domain of \(f(x)\), \(x = 0\). Our function will keep decreasing until it hits a local minimum at \(x = \frac{1}{e}\), then it will keep increasing forever. This means that \(x = \frac{1}{e}\) must be where \(f(x) = x \ln(x)\) hits a global minimum, since its value never goes lower than that. The function’s value at its global minimum is \(\frac{1}{e}\ln(\frac{1}{e}) = \frac{1}{e} \cdot -1 = -\frac{1}{e}\). Thus, the global minimum of \(f(x)\) occurs at \((\frac{1}{e}, -\frac{1}{e}) \approx (0.368, -0.368)\).

Does \(f(x)\) have a global maximum? Because \(f(x)\) approaches infinity as \(x\) approaches infinity, \(f(x)\) does not have a global maximum as it can reach any arbitrarily high value.

The global minimum of \(f(x) = x \ln(x)\) occurs around \((0.368, -0.368)\). There is no global maximum because \(f(x)\) grows without bound as \(x\) approaches infinity.

When a differentiable function has a local minimum, its curve usually looks something like this: (By “differentiable”, I mean differentiable at the local minimum.)

And when a differentiable function has a local maximum, it typically looks like this:

How would you describe these shapes? The local minimum looks like a U shape and the local maximum looks like an upside-down U (\(\cap\)) shape. What is a more mathematical way of describing these shapes?

Let’s look at the first shape, the local minimum shape. Starting from the left side, at first, the function is decreasing very quickly, but then it decreases less quickly until eventually it hits the local minimum and starts increasing faster and faster. This means that its derivative is constantly increasing.

The blue function is the derivative of the red function. Notice how the derivative is constantly increasing.

The opposite is true when we have a local maximum. The function increases, but it increases slower and slower until it eventually starts decreasing faster and faster. The derivative of the function is constantly decreasing in this case.

The blue function is the derivative of the red function. Notice how the derivative is constantly decreasing.

Mathematically, we call the U shape concave up and the \(\cap\) shape concave down. In other words, a function is concave up when its derivative is increasing and a function is concave down when its derivative is decreasing. Whether a function is concave up or concave down at a certain point is called its concavity.

We can also express concavity in terms of a function’s second derivative. A function is concave up when its second derivative is positive and concave down when its second derivative is negative.

What this means is that if we want to find where a function is concave up and down, we just need to analyze the sign of its second derivative. Let’s analyze the concavity of the example function we’ve used in the last few lessons, \(f(x) = -x^3 + x\).

Problem: When is the function \(f(x) = -x^3 + x\) concave up and concave down?

First, we have to find the second derivative of this function.

Now we need to find out when this second derivative is negative. The process we’re going to use is very similar to finding the increasing and decreasing intervals of a function. After all, we’re trying to find where our first derivative is increasing and decreasing!

To do this, we first need to find the places where the second derivative is 0 or undefined. These are points where the concavity (or sign of the second derivative) could potentially switch. A function’s derivative could potentially switch from increasing to decreasing or vice versa at these points.

I’m going to call these points “candidate inflection points”. Finding these is just like finding critical points, except we’re dealing with the second derivative instead of the first derivative.

In this case, \(f''(x) = 0\) when \(x = 0\), and there are no points where \(f''(x)\) is undefined. Now let’s find the sign of the second derivative before and after this candidate inflection point of \(x = 0\). We’ll do that by plugging in an \(x\)-value less than 0 and another \(x\)-value greater than 0 into \(f''(x) = -6x\).

Our function is concave down where \(f''(x)\) is negative and concave up where \(f''(x)\) is positive. This means that \(f(x)\) is concave up when \(x < 0\) and concave down when \(x > 0\).

The point at which a function switches concavity is known as an inflection point. In this case, \(x = 0\) is the only inflection point of \(f(x)\).

This graph shows when the function \(f(x) = -x^3 + x\) is concave up and concave down. The switch in concavity occurs at the inflection point \((0, 0)\).

This slider shows what happens as \(x\) changes. Notice what happens to \(f'(x)\) and \(f''(x)\) when \(x = 0\).

I mentioned at the beginning of the last section that when a differentiable function has a local minimum, it is concave up, and when a function has a local maximum, it is concave down. Relating concavity to local extrema like this means that we can use a function’s concavity to identify whether a function’s local extremum is a minimum or maximum!

As a reminder, a differentiable function is concave up around a local minimum and concave down around a local maximum.

If we know a function has a local extremum somewhere and it is concave up around that point, then the extremum must be a local minimum. If the function is concave down around the point, it must be a local maximum.

In fact, just knowing that a function has a critical point somewhere and that the function is concave up around that point is enough to ensure that the critical point is a local minimum. (The same is true with concave down and local maxima.) If we could describe this observation mathematically, we could create a fast way of identifying when a critical point is a local minimum or maximum.

Mathematically, “concave up” means that a function’s second derivative is positive and “concave down” means that the second derivative is negative. Because of this, we’re going to create a test involving the second derivative, and we’re going to call it the second derivative test. This test will help us determine if a local extremum is a minimum or maximum.

A critical point occurs when a function’s first derivative is 0 or undefined. If the first derivative is undefined, then the second derivative will also be undefined, so we’re not going to worry about critical points where the first derivative is undefined because the second derivative won’t be able to help us in that case.

This means our test will only be useful when the first derivative of a function is 0. In addition, since this test is based on the second derivative, the second derivative must exist at this critical point where the first derivative is 0.

If we use our test at a point \(x = c\), the two conditions to use our test are:

• \(f'(c) = 0\)
• \(f''(c)\) exists

If both conditions are met, all we have to do to use the second derivative test is to check whether \(f''(c)\) is positive or negative. There are three possibilities:

• If \(f''(c)\) is positive (i.e. \(f(x)\) is concave up at \(x = c\)), then the point \((c, f(c))\) is a local minimum.
• If \(f''(c)\) is negative (i.e. \(f(x)\) is concave down at \(x = c\)), then the point \((c, f(c))\) is a local maximum.
• If \(f''(c) = 0\), then the test is inconclusive - it doesn’t tell us anything.

Here’s an example: consider the function \(f(x) = x^4 + x^3\). Problem: How can we use the second derivative test to find the local extrema of this function?

With some work, we can find that the function’s first derivative \(f'(x) = 4x^3 + 3x^2\) is equal to 0 at \(x = 0\) and \(x = -\frac{3}{4}\). The second derivative \(f''(x) = 12x^2 + 6x\) is defined for all \(x\)-values, so we can use the second derivative test for both of these points. We just need to plug in \(x = 0\) and \(x = -\frac{3}{4}\) into \(f''(x)\).

We’ve found that a local minimum happens at \(x = -\frac{3}{4}\), since the second derivative is positive (the function is concave up). The function’s value at this point is \(-\frac{27}{256} \approx -0.105\).

Note that we can’t determine for sure using the second derivative test what happens at \(x = 0\) because it is inconclusive. There could be a local minimum, maximum, or neither. We also can’t guarantee that there isn’t a local extremum at \(x = 0\).

For the function \(f(x) = x^4 + x^3\), a local minimum occurs at about \(\class{red}{(-0.75, -0.105)}\). This makes sense since the function is concave up at \(x = -0.75\). The second derivative test is inconclusive for the point \(\class{blue}{(0, 0)}\), and it turns out there isn’t a local extremum there (although that isn’t guaranteed whenever the second derivative test is inconclusive).

Let’s summarize everything we’ve learned about the connection between functions and their derivatives using graphs.

If we have a graph of a function’s first derivative, \(f'(x)\), what does it tell us about \(f(x)\)?

What does the graph of \(f'(x)\) tell us about \(f(x)\)?

We’ll start with the basics. By the meaning of a derivative, when \(f'(x)\) is positive, \(f(x)\) is increasing. When \(f'(x)\) is negative, \(f(x)\) is decreasing. When \(f'(x)\) crosses the \(x\)-axis from negative to positive, that’s a local minimum for \(f(x)\), and when \(f'(x)\) crosses from positive to negative, that’s a local maximum. If all you know is that \(f'(x) = 0\) at an extremum point, then you can use the second derivative test to identify if it is a local minimum or maximum.

When \(f'(x)\) is increasing, that means that \(f(x)\) is concave up, and when \(f'(x)\) is decreasing, \(f(x)\) is concave down. In addition, when \(f'(x)\) has a local extremum, \(f(x)\) has an inflection point. This is because an inflection point occurs when the derivative switches from decreasing to increasing or vice versa, which happens when the derivative has a local extremum.

The intervals where \(f(x)\) is increasing are highlighted in green, and the intervals where \(f(x)\) is decreasing are highlighted in red. Notice what happens to \(f(x)\) when \(f'(x)\) crosses the \(x\)-axis and when \(f'(x)\) reaches its minimum.

What about the second derivative of \(f(x)\)? The second derivative, \(f''(x)\), tells us about the concavity of \(f(x)\). When \(f''(x)\) is positive, \(f(x)\) is concave up, and when \(f''(x)\) is negative, \(f(x)\) is concave down. When \(f''(x)\) crosses the \(x\)-axis, that’s an inflection point for \(f(x)\).

This graph is shaded according to the sign of \(f''(x)\). Pay attention to the behavior of \(f(x)\) when \(f''(x)\) is positive, when \(f''(x)\) is negative, and when \(f''(x)\) crosses the \(x\)-axis.

Here’s a table that summarizes all this information:

Calculus gives us the power to graph complicated functions by hand using what we know about derivatives! And no, we’re not going to just plug in a bunch of \(x\)-values and plot them. We’re going to use differential calculus to make our lives easier!

We’ll be exploring how we can use calculus to graph a polynomial function. To do this, we’re going to use derivatives to analyze some of the properties of this function.

Problem: How can we graph the function \(f(x) = \frac{1}{3}x^3 + x^2 - 3x\) without a graphing calculator?

You will need to click the “Previous Step” and “Next Step” buttons below to move through the steps of graphing this function. As you do this, more details will appear on the graph below.

← Previous Step

→ Next Step

We start with a blank graph. The first thing we need to do to analyze this function is find the function’s critical points: points that could potentially be a local minimum or maximum.

← Previous Step

→ Next Step

The function’s critical points are found by setting \(f'(x)\) to 0 (\(f'(x)\) will never be undefined since it’s a polynomial).

\[ f(x) = \frac{1}{3}x^3 + x^2 - 3x \] \[ f'(x) = x^2 + 2x - 3 \] \[ x^2 + 2x - 3 = 0 \] \[ (x+3)(x-1) = 0 \] \[ x = 1, \, x = -3 \]

Let’s plot those critical points in our graph. We find the \(y\)-values of them simply by plugging them into our function \(f(x)\). In this case, the points are \((-3, 9)\) and \((1, -\frac{5}{3})\). Click “Next Step” to view the critical points on the graph.

← Previous Step

→ Next Step

Now we’re also going to find candidate inflection points, as those points are also interesting. To do that, we find the points where the second derivative \(f''(x)\) is equal to 0 (it will never be undefined in this case).

\[ f'(x) = x^2 + 2x - 3 \] \[ f''(x) = 2x + 2 \] \[ 2x + 2 = 0 \] \[ x = -1 \]

Here, the candidate inflection point is \((-1, \frac{11}{3})\). We’re going to graph that point!

← Previous Step

→ Next Step

In order to know what our function does around our critical points, we are going to analyze the function’s concavity at these points. To do that, we find what the second derivative is at those points.

\[ f''(x) = 2x + 2 \] \[ f''(-3) = -4 \text{ (negative)} \] \[ f''(1) = 4 \text{ (positive)} \]

This shows us that \(f(x)\) is concave up around \(x = 1\) and concave down around \(x = -3\). Using the second derivative test, this means that \(x = 1\) is a local minimum and \(x = -3\) is a local maximum. Let’s draw that on our graph.

← Previous Step

→ Next Step

Now we’re going to focus on the candidate inflection point at \(x = -1\). To check if it actually is an inflection point, we need to check the concavity on either side of it. Luckily, we already know that \(f(x)\) is concave down at the local maximum \(x = -3\) and concave up at the local minimum \(x = 1\), so we know that the function does indeed change concavity at \(x = -1\).

Let’s find the slope of \(f(x)\) at this inflection point so we can accurately draw the curve around it.

\[ f'(x) = x^2 + 2x - 3 \] \[ f'(-1) = -4 \]

\(f'(-1)\) is -4, so the slope of \(f(x)\) at the inflection point \(x = -1\) is -4.

← Previous Step

→ Next Step

Now all that’s left is to complete the curve as best as we can using what we know! The curve should get steeper and steeper as \(x\) increases to \(\infty\), because the graph is concave up for \(x \gt -1\). Likewise, the curve should get steeper in the negative direction as \(x\) decreases to \(-\infty\), because the graph is concave down for \(x \lt -1\). Click “Next Step” to view the completed graph.

← Previous Step

→ Next Step

This is a graph of \(f(x) = \frac{1}{3}x^3 + x^2 - 3x\). Using differential calculus the way we just did, we can draw a decently accurate version of this graph by hand!

Let’s say you run a business, and naturally you want to maximize the profit that you’re making. Believe it or not, calculus is a good tool for this!

Profit is simply revenue minus costs, so your profit can be modeled by the function \(P(x) = R(x) - C(x)\).

To maximize our profits, we just need to find where this function is the highest! In math speak, this means finding the global maximum of \(P(x)\) within its domain \(x \ge 0\).

To do that, we find the critical points of \(P(x)\). We take the derivative \(P'(x)\) and set it equal to 0:

There are no points where \(P'(x)\) is undefined. We can ignore the negative value of \(x\) since the domain of \(P(x)\) is \(x \ge 0\) (you can’t sell negative books, after all). So our only relevant critical point occurs at \(x \approx 4.859\).

Let’s make sure that this point is a global maximum first. We’re going to use the second derivative test to do this.

The second derivative is negative at the critical point \(x \approx 4.859\), so it’s a local maximum! There are no other critical points, which means that this point must also be a global maximum (as the function doesn’t go to infinity as \(x\) goes to infinity). This means that our maximum profit occurs at \(x \approx 4.859\), so we can maximize our profit by producing and selling 4,859 books. Doing this generates a profit of \(P(4.859) \approx 25.035\) thousand dollars, or $25,035.

The maximum profit occurs at the point \(\class{green}{(4.859, 25.035)}\), representing 4,859 books sold for a $25,035 profit.

Let’s move on to a more complicated example.

Problem: I want to create a flyer for this calculus website, and I need to have 72 square inches of text on a piece of paper with 1-inch margins. To save money and resources (because I want to be environmentally friendly), I want to find the smallest piece of paper that meets these requirements (the piece of paper with the minimum area).

What is the minimum area of this piece of paper? What are the dimensions of the piece of paper with the minimum area?

Let’s define \(w_p\) as the width of the paper and \(w_t\) as the width of the text on the paper. We’ll also define \(h_p\) as the height of the paper and \(h_t\) as the height of the text.

Because of the two 1-inch margins on the left and right, the width of the text is 2 inches less than the width of the paper. The same can be said about the height of the text and paper. So \(w_t = w_p - 2\) and \(h_t = h_p - 2\). We know that the area of the text is 72 square inches, so \(w_t \cdot h_t = 72\).

We want to minimize the area of the paper, which is \(w_p \cdot h_p\). To optimize this function, we need to write it in terms of one variable. Let’s choose \(w_p\) as that one variable, meaning that we need to rewrite \(h_p\) in terms of \(w_p\).

Now we find the critical points of \(A(w_p)\).

\(A'(w_p)\) is undefined at \(w_p = 2\), but a paper with a width of 2 inches would have a text width of 0 inches, so this can’t be our answer. We can ignore the critical point \(w_p = -\sqrt{72} + 2\) since it’s negative.

That leaves us with one critical point, \(w_p = \sqrt{72} + 2\). Let’s use the second derivative test to confirm this is a local minimum:

If we plug any value of \(w_p\) greater than 2 into \(A''(w_p)\), then both the numerator and denominator will be positive, meaning that \(A''(w_p)\) is positive for all values of \(w_p > 2\). This means that our critical point \(w_p = \sqrt{72} + 2\) is a local minimum. By checking the end behavior of \(A(w_p)\), we can confirm that it is also a global minimum across the domain \(w_p > 2\).

This means that our minimum paper area occurs with a paper width of \(\sqrt{72} + 2\) inches. Let’s find the paper height now using our expression for \(h_p\) in terms of \(w_p\):

It turns out the optimal paper height is also \(\sqrt{72} + 2\) inches. This means the paper with the minimum area that meets our requirements is a square with dimensions \(\sqrt{72} + 2\) inches by \(\sqrt{72} + 2\) inches (about 10.485 by 10.485 inches).

We’re going to take a deeper dive into implicit differentiation and see how we can analyze the properties of implicit curves.

Problem: Consider the relation \(x^2 + xy + y^2 = 1\). Let’s say we want to know all the spots where the tangent line to this curve is horizontal. How could we do that?

Well, a horizontal tangent line just means that the derivative \(\dv{y}{x}\) is zero. So we need to differentiate this equation, then find the locations where \(\dv{y}{x} = 0\). (If you need to review implicit differentiation, you can do so here.)

This equation for \(\dv{y}{x}\) now tells us the slope of the tangent line at any point \((x, y)\) on the curve. Now we just need to find when this slope is equal to 0.

\(\dv{y}{x}\) will be equal to 0 when the numerator \(2x + y\) is equal to 0 but not the denominator \(x + 2y\) (since \(\frac{0}{0}\) is undefined). This means that to find when the derivative is 0, we set the numerator equal to 0.

We will first write \(y\) in terms of \(x\) so that we can rewrite our implicit equation fully in terms of \(x\).

Now we can solve for \(x\) normally.

We have two possible \(x\)-values where the tangent line could be horizontal. Let’s focus on the first one, \(x = \frac{1}{\sqrt{3}}\). We can use the equation \(y = -2x\) we figured out earlier to find the \(y\)-value for this point.

This means that the tangent line to the point \((\frac{1}{\sqrt{3}}, -\frac{2}{\sqrt{3}})\) could be horizontal. If we plug in the \(x\) and \(y\)-values back into the expression for \(\dv{y}{x}\), we can confirm that this does not cause a division by zero, showing that the tangent line is indeed horizontal. Doing the same for the other possible \(x\)-value, we find that the tangent line is also horizontal at \((-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}})\).

The tangent lines are horizontal at the points \((\frac{1}{\sqrt{3}}, -\frac{2}{\sqrt{3}}) \approx (0.577, -1.155)\) and \((-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}) \approx (-0.577, 1.155)\).

Problem: Now let’s say we want to know when the tangent line to the curve is vertical.

A vertical tangent line occurs when the denominator of our expression for \(\dv{y}{x}\) is zero and the numerator is not zero. In other words, if plugging in \(x\) and \(y\) into our equation \(\dv{y}{x} = -\frac{2x + y}{x + 2y}\) gives a result \(\frac{a}{0}\) where \(a\) is nonzero, then the tangent line to \((x, y)\) is vertical.

The first step is to find values of \(x\) and \(y\) where the denominator \(x + 2y = 0\). This is a very similar process to what we did when finding horizontal tangent lines.

Now that we have our possible \(x\)-values, we can find the corresponding \(y\)-values using our equation \(y = -\frac{x}{2}\). For the first \(x\)-value \(x = \frac{2}{\sqrt{3}}\):

This gives us our first point \((\frac{2}{\sqrt{3}}, -\frac{1}{\sqrt{3}})\). Doing the same for our other \(x\)-value, we get the point \((-\frac{2}{\sqrt{3}}, \frac{1}{\sqrt{3}})\).

We should plug these points back into the equation for \(\dv{y}{x}\) to make sure that the numerator is not also zero. Doing that, we can confirm that the tangent line is indeed vertical at these two points.

The tangent lines are vertical at the points \((\frac{2}{\sqrt{3}}, -\frac{1}{\sqrt{3}}) \approx (1.155, -0.577)\) and \((-\frac{2}{\sqrt{3}}, \frac{1}{\sqrt{3}}) \approx (-1.155, 0.577)\).

• The mean value theorem states that if a function \(f(x)\) is continuous over an interval \([a, b]\) and differentiable over \((a, b)\), there is a point \(c\) in \((a, b)\) where \(f'(c)\) equals the average rate of change of \(f(x)\) from \(x = a\) to \(x = b\).
• \[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
• A local minimum (or maximum) is when a point on a function is lower (or higher) than all points nearby. A global minimum (or maximum) is a function’s lowest (or highest) point in its entire domain. An extremum (plural: extrema) is a minimum or maximum.
• The extreme value theorem states that if a function \(f(x)\) is continuous over an interval \([a, b]\), \(f(x)\) has a maximum and minimum within that interval.
• Critical points are points where a function’s derivative is 0 or undefined and the original function is defined.
• A function is increasing when its derivative is positive and decreasing when its derivative is negative. You can find a function’s increasing and decreasing intervals by splitting it into regions using its critical points and finding the sign of the derivative at one point in each region.
• A function hits a local minimum when its derivative goes from negative to positive at a critical point and hits a local maximum when its derivative goes from positive to negative at a critical point.
• Concavity is whether a function is concave up or concave down. A function is concave up if its derivative is increasing (its second derivative is positive) and concave down if its derivative is decreasing (its second derivative is negative).
• An inflection point is where a function switches concavity. Inflection points happen when a function’s first derivative hits a local extremum or the second derivative crosses the \(x\)-axis.
• The second derivative test allows us to find local minima and maxima. For a point \(x = c\) on a function \(f(x)\), if \(f'(c) = 0\) and \(f''(c)\) exists, we can use the second derivative test. The second derivative test says that:
  • If \(f''(c)\) is positive, \(x = c\) is a local minimum.
  • If \(f''(c)\) is negative, \(x = c\) is a local maximum.
  • If \(f''(c)\) is zero, the test is inconclusive.

Khan Academy Link (Calc AB): Integration and accumulation of change

Khan Academy Link (Calc BC): Integration and accumulation of change

All topics covered in Khan Academy:
Green underlined topics are topics at least partially covered on my website and red topics are topics not yet covered on my website. Note that even green topics might not be covered in full detail on my page.

1. Exploring accumulations of change
2. Approximating areas with Riemann sums
3. Riemann sums, summation notation, and definite integral notation
4. The fundamental theorem of calculus and accumulation functions
5. Interpreting the behavior of accumulation functions involving area
6. Applying properties of definite integrals
7. The fundamental theorem of calculus and definite integrals
8. Finding antiderivatives and indefinite integrals: basic rules and notation: reverse power rule
9. Finding antiderivatives and indefinite integrals: basic rules and notation: common indefinite integrals
10. Finding antiderivatives and indefinite integrals: basic rules and notation: definite integrals
11. Integrating using substitution
12. Integrating functions using long division and completing the square
13. (Calc BC only) Using integration by parts
14. (Calc BC only) Integrating using linear partial fractions
15. (Calc BC only) Evaluating improper integrals

Let’s say that I go on a long walk for 3 hours. I walk at 2 miles per hour (mph) for the first hour, 4 mph for the second hour, and 3 mph for the third hour. How many miles have I walked in total?

We know that \(\text{speed} = \text{distance}/\text{time}\), so we can rearrange that equation to get \(\text{distance} = \text{speed} \cdot \text{time}\). For example, if we walk at 2 miles per hour for 1 hour, the distance we’ve traveled is \(\text{2 miles per hour} \cdot \text{1 hour} = \text{2 miles}\).

So all we have to do to solve this problem is to multiply the speeds at which we’ve walked at by their respective durations.

Now we just sum up the distances to get the total distance of 9 miles.

Let’s use graphs to represent this situation. Here is a graph of our speed over time:

In the first hour, I traveled at 2 mph, and my distance over that hour was \(\text{2 mph} \cdot \text{1 hour} = \text{2 miles}\). We can represent that distance as the area under the graph from \(x = 0\) to \(x = 1\), or more precisely, the area between the graph and the \(x\)-axis from \(x = 0\) to \(x = 1\). (From now on, whenever I say “area under the graph” or “area under the function”, I really mean the area between the graph and the \(x\)-axis.)

The area under the graph from \(x = 0\) to \(x = 1\) represents how far I’ve walked in the first hour.

We can do the same for the other 2 hours:

The total distance I’ve traveled is the sum of all of these areas. What this means is that the distance I’ve walked over these 3 hours can be represented as the area under the function that represents my walking speed from \(x = 0\) to \(x = 3\).

What we essentially did is split the area under the function into 3 rectangles, find the area of each rectangle using the formula \(\text{area} = \text{length} \cdot \text{width}\), then sum the areas of all the rectangles.

The distance I’ve walked, 9 miles, is known as an accumulation of change. This is because the distance I’ve traveled really just represents how much my position has changed over time. Over 3 hours, I’ve accumulated a total of 9 miles of position change (i.e. distance).

In fact, for any time \(t\) between 0 and 3 hours, the total distance I’ve walked from time 0 to \(t\) is equal to the area under the graph from \(x = 0\) to \(x = t\).

Here’s a harder problem where my speed is constantly changing.

Problem: My walking speed \(t\) hours after I start is modeled by the function \(f(t) = 0.5t + 2\). How many miles did I walk in the first 3 hours?

You might think that we could just find the area under the function again. But first, we need to check if this idea of finding the area actually gives us the right answer in situations where the speed is continuously changing.

Since we don’t know for sure how to calculate the distance traveled when our speed is constantly changing, we’re going to approximate the situation so that we can convert it to something more similar to our first distance problem (which we know how to solve).

Our speed in the first hour varied between 2 and 2.5 mph, but let’s assume that we walked at a constant speed of 2 mph the entire first hour, the lower bound of our speed during this interval. During the second hour, our speed was between 2.5 and 3 mph, so let’s assume that we walked at 2.5 mph for the second hour. Finally, we’re going to use a speed of 3 mph for the third hour.

This simplified model of the situation gives us a graph that looks like this:

Our simplified model is represented by the blue function.

We can then find the distance we traveled by finding the area under the blue function as we did before. We can do this by splitting the area under the blue function into 3 rectangles (1 rectangle for each hour), then sum their areas.

This approximation gives us an area of 7.5, meaning that according to this approximation, we’ve walked 7.5 miles in 3 hours.

However, this model is not perfect. In the first hour, our speed started at 2 mph and increased all the way up to 2.5 mph, while our simplified model assumes a speed of 2 mph the entire time, which is lower than the true speed by as much as 0.5 mph. This means that our model underestimates the true distance we traveled.

What happens if we use smaller intervals for our approximation? For example, instead of assuming a constant speed for every hour, we could assume a constant speed for every half hour. So we assume a speed of 2 mph for the first 30 minutes, then a speed of 2.25 mph for the next 30 minutes, and so on.

Our new model approximates the actual speed much better than the old model. If we find the area under the function, we get a value of 7.875 miles, which is closer to the true distance than our old approximation of 7.5 miles.

Now our approximation becomes better since we’re not underestimating our speed by as much. At any point in time, our approximation is at most 0.25 mph below the true speed (instead of 0.5 mph with our old approximation). If we find the area under the blue function now, we will get a distance even closer to the actual distance traveled.

As we use smaller and smaller intervals, the error of our approximation will approach zero, getting closer and closer to the true distance we traveled. If we take the limit of our calculated distance as the size of these intervals approaches zero, we will get the true distance!

As our approximation gets better and better, eventually our graph will look something like this:

As our speed approximation gets better and better, it will eventually become indistinguishable from the actual speed.

And when we find the area under the approximated function (which gets closer and closer to the area under the actual function as our approximation gets better), it looks like this:

What this means is that the distance traveled really is equal to the area under the function, even if the function is constantly changing! So the question of “how many miles have I walked in 3 hours?” is the same as “what is the area under the function between \(t = 0\) and \(t = 3\)?”

In this case, we can find the area under the function by splitting it up into two shapes we know how to calculate the area of: a triangle and a rectangle.

The area under the function is equal to the area of the triangle plus the area under the rectangle.

The area of the triangle is \(\frac{1}{2}bh = \frac{1}{2} \cdot 3 \cdot 1.5 = 2.25\) and the area of the rectangle is \(3 \cdot 2 = 6\), so the total area under the function is \(2.25 + 6 = 8.25\). This means we walked a total of 8.25 miles in 3 hours.

This process of finding the area under a curve is known as integration. (Note: Functions are often called “curves” even if they aren’t actually curved.) Specifically, when we find the area under a function over a specific interval, it’s called a definite integral.

It’s important to realize that integration just tells you how much something has changed (the net change) and not the final conditions. For example, let’s say that the function \(f(t) = 2t\) shows you the rate in which water is being poured into a bucket.

The area under the curve between \(t = 0\) and \(t = 3\) is 9 liters, so you might jump to the conclusion that there are 9 liters of water in the bucket after 3 minutes. However, this is not necessarily true! What if there was already water in the bucket when we started adding water at \(t = 0\)? All we can conclude is that 9 liters of water were added to the bucket in 3 minutes. Always make sure to take these initial conditions into account when applying integration to real life.

Sometimes we can find a definite integral by splitting up the area under a curve into multiple shapes that we know how to find the areas of (e.g. a rectangle and a triangle). That’s what we did in the previous example. However, most of the time, it’s not that simple. Take this problem for example:

Problem: What is the area under the curve \(f(x) = x^2\) from \(x = 0\) to \(x = 2\)?

What is the area of the shaded region under \(f(x) = x^2\)?

In this case, we don’t have a simple formula to find the area of this shape. Instead, we need to go back to using rectangles to approximate the area. Let’s try splitting up this region into 4 rectangles that approximate the area under the function. This is known as a Riemann sum.

The 4 rectangles approximate the area under the red function.

Notice how we determine the height of each rectangle. We’re using the value of the function at the right boundary of each rectangle to determine its height. This is known as a right Riemann sum. But this isn’t the only way to make an approximation like this! We could have also used the left boundary of each rectangle or even the midpoint of each rectangle to determine its height.

To find the area of these rectangles, we find the width and height of each of them, then sum them up. We are splitting up the region between \(x = 0\) and \(x = 2\) into 4 rectangles, so the width of each rectangle is \(\frac{2}{4} = 0.5\).

The total area of the rectangles comes out to be 3.75. However, as you can see in the diagram, our rectangles overestimate the true area by a decent amount, so the true area is less than 3.75.

To make our approximation better, let’s use more rectangles. We’re going to use 8 rectangles instead of 4 this time.

Using 8 rectangles instead of 4 approximates the area under the function better.

Let’s find the area of the rectangles:

(Note: The areas are rounded to 4 decimal places.)

The total area of the rectangles is 3.1875. This value is much more accurate to the true value than our previous 4-rectangle approximation, as we aren’t overestimating by as much now.

Use this slider to explore what happens to the area as we use more and more rectangles:

As we add more rectangles, the area approaches the true area under the function. In this case, the area approaches \(\frac{8}{3} \approx 2.667\). We can make our area approximation arbitrarily close to \(\frac{8}{3}\) by adding more rectangles, so we define \(\frac{8}{3}\) as the area under \(f(x) = x^2\) from \(x = 0\) to \(x = 2\) (the definite integral of \(f(x) = x^2\) from \(x = 0\) to \(x = 2\)).

Let’s try a different type of Riemann sum, where the rectangles’ heights are determined by the value of the function at the left boundary of each rectangle. This is known as a left Riemann sum. Let’s try it with 4 rectangles first.

Note: The leftmost rectangle has a height of 0, so it’s invisible.

The total area in this case is 1.75. This approximation underestimates the true area under the function.

To make our approximation better, we will once again double the number of rectangles from 4 to 8.

The total area comes out to be 2.1875. This still underestimates the true area, but not as much as the previous 4-rectangle approximation.

Explore what happens as the number of rectangles increases:

The area still approaches \(\frac{8}{3}\). It doesn’t matter if we use a left Riemann sum or a right Riemann sum, as the number of rectangles approaches infinity, the area approaches the same value!

Now left and right Riemann sums and aren’t that accurate, especially if we’re using a smaller number of rectangles. A more accurate type of Riemann sum is a midpoint Riemann sum. Instead of using the function’s value at the left or right boundary of each rectangle to determine its height, we use the function’s value at the midpoint of each rectangle. Here’s what it looks like with 4 rectangles:

The midpoint Riemann sum more accurately approximates the area than the left or right Riemann sums in this case.

The total sum of the rectangles’ areas is 2.625, which is much closer to the actual area (\(\frac{8}{3}\)) than the left or right Riemann sum with 4 rectangles. Let’s bump it up to 8 rectangles now:

The total area is 2.65625, even closer to the true area of \(\frac{8}{3}\).

Notice how it takes much fewer rectangles to get an accurate answer with a midpoint Riemann sum than with a left or right Riemann sum.

One last thing we can do to get a good approximation is to use trapezoids instead of rectangles. This is known as a trapezoidal sum or trapezoidal rule. Let’s see it in action with 4 trapezoids:

To find the area of each trapezoid, we use the formula \(A = \frac{a + b}{2} \cdot h\), where \(a\) and \(b\) are the side lengths of the trapezoid’s bases and \(h\) is the trapezoid’s height. However, in a trapezoidal sum, the trapezoids are flipped on their sides, so \(h\) is actually the width of the trapezoid and \(a\) and \(b\) are the heights of each side.

The total area is 2.75. This approximation is more accurate than the left and right Riemann sums, but it’s not as accurate as the midpoint approximation with 4 rectangles. This is because in this case, the trapezoids consistently overestimate the area.

This is what the trapezoidal approximation looks like with 8 trapezoids:

The total area is 2.6875. Again, not as accurate as the midpoint sum, but more accurate than the left and right Riemann sums. Just like with rectangular sums, trapezoidal sums get more accurate the more trapezoids you have.

Finally, here’s a slider that shows you how all four types of Riemann sums behave as the number of rectangles/trapezoids increases.

In this case, the left Riemann sum underestimates the actual area and the right Riemann sum overestimates it. However, this is only true because \(f(x) = x^2\) is always increasing from \(x = 0\) to \(x = 2\). If the function was always decreasing over the interval we were looking at, the opposite would be true: the left Riemann sum would overestimate the area and the right Riemann sum would be an underestimation.

In each of our examples, all of the rectangles/trapezoids have the same width, but that isn’t necessary in a Riemann sum; it’s just the simplest and most common way to set up Riemann sums.

Often in math, the solution to a problem requires us to sum a lot of numbers. For example, what is the sum of the integers from 1 to 100? We’re not going to focus on the answer, but rather how mathematicians would write this problem down.

Your first thought might to just write something like \(1 + 2 + 3 + \text{...} + 100\), and that would work, but there are a few problems with that.

Not only is that notation clunky and long, but it’s also imprecise - it’s fair to assume that the “...” covers all integers from 4 to 99, but it’s not guaranteed unless you clarify it somewhere else.

What would be more precise is if we said “the sum of all integers from 1 to 100, inclusive”, but that’s a lot of words, and it would be nice if we could describe this with only symbols.

The notation that mathematicians have decided on is known as sigma notation (or summation notation). In this notation, to describe the sum of all integers from 1 to 100, we would write this:

The sigma symbol \(\sum\) indicates that we are doing a summation. The \(\class{red}{n}\) under the sigma symbol is just a variable that we will use within the sigma notation expression. This variable is known as an index, and it can be any letter we want. The \(\class{blue}{1}\) is what value the variable \(\class{red}{n}\) starts at. We will increase \(\class{red}{n}\) by 1 each time, and the \(\class{green}{100}\) on top means that we stop after \(\class{red}{n}\) reaches 100 (we still include \(n = 100\)).

This means that we will cycle through all values of \(\class{red}{n}\) from \(\class{blue}{1}\) to \(\class{green}{100}\), including 1 and 100. The expression after the sigma symbol, in this case \(\class{purple}{n}\), tells us what we are summing up. In this case, we are cycling through \(n = 1\) to \(100\), and we are adding \(\class{purple}{n}\) each time.

The end result is that we add up all integer values of \(n\) from 1 to 100, or \(1 + 2 + 3 + \text{...} + 100\).

This is a simple example, but sigma notation is much more useful when we do more complicated things. For example, we can use sigma notation to describe the sum of the first 100 perfect squares:

In this case, we are still going from \(n = 1\) to \(100\), but we are squaring \(n\) each time and adding that to the running total.

We can also have other variables in a sigma notation expression - in that case, the sum is in terms of those other variables.

            
total = 0
for n in range(1, 100 + 1):
  total += n ** 2
print(total)
            
          

            
var total = 0;
for (var n = 1; n <= 100; n++) {
  total += n ** 2;
}
console.log(total);
            
          

            
total = 0
for n in range(a, b + 1):
  total += f(n)
print(total)
            
          

            
var total = 0;
for (var n = a; n <= b; n++) {
  total += f(n);
}
console.log(total);
            
          

How does any of this relate to integration? Well, the Riemann sum involves taking the sum of a bunch of areas, so this is the perfect opportunity to use sigma notation!

Let’s go back to our previous example of finding the area under \(f(x) = x^2\) from \(x = 0\) to \(x = 2\). We’ll use a right Riemann sum with 4 rectangles.

How can we use sigma notation to express the sum of the rectangles’ areas?

Because there are 4 rectangles, our Riemann sum will index from 1 to 4. We’ll let \(j\) be our index and \(A(j)\) denote the area of the \(j\)th rectangle. (In other calculus texts, you’ll typically see \(i\) used as the index of a Riemann sum, but I’m using \(j\) to avoid confusion with imaginary numbers.) This gives us the following sigma expression:

Now we need to find an expression for \(A(j)\), the area of the \(j\)th rectangle. Because we’re splitting up 2 units of the \(x\)-axis into 4 equal parts, the width of each rectangle is \(\frac{2}{4} = \frac{1}{2}\).

What about the height of each rectangle? Because this is a right Riemann sum, the height of each rectangle is the function evaulated at the right side of each rectangle. We’ll call the \(x\)-value at the right endpoint of the \(j\)th rectangle \(x_j\). So the function evaluated at the right side of the \(j\)th rectangle is \(f(x_j)\). This is the height of the \(j\)th rectangle.

This means that for the \(j\)th rectangle, the area is \(\text{width} \cdot \text{height} = \frac{1}{2}f(x_j)\). So we can rewrite our sigma expression like this:

Now all that’s left is to find an expression for \(x_j\) and \(f(x_j)\). The right boundary of the first rectangle is at \(x = \frac{1}{2}\). The right boundary of the second rectangle is at \(x = 1\), the third is at \(x = \frac{3}{2}\), and so on. \(x_j\) increases by \(\frac{1}{2}\) each rectangle, so the right edge of the \(j\)th rectangle is at an \(x\)-value of \(x_j = \frac{1}{2}j\).

Let’s plug this expression for \(x_j\) into our sigma expression:

Evaluating this sum, we get an answer of 3.75, which is the combined area of our 4 rectangles.

Now let’s generalize this process so that we can use it to find the area under any function over any interval using a right Riemann sum with any number of rectangles. First, we’re going to define some variables.

We’ll let \(f(x)\) be our function, \([a, b]\) be the interval we’re finding the area under, and \(n\) be the number of rectangles of equal width. We’ll call the width of each rectangle \(\Delta x\) and the right endpoint of the \(j\)th rectangle \(x_j\).

The height of the \(j\)th rectangle is \(f(x_j)\), and the width of each rectangle is \(\Delta x\). This means that the area of the \(j\)th rectangle is \(f(x_j)\cdot\Delta x\). The sum of the areas of all the rectangles is:

To find \(\Delta x\), we take the width of the interval we are finding the area under and divide it by the number of rectangles. The width of the interval is \(b - a\), so the width of each rectangle is \(\Delta x = \frac{b-a}{n}\).

For example, if we are finding the area under a function from \(x = \class{red}{0}\) to \(x = \class{blue}{2}\) with 4 rectangles, \(\Delta x\) will equal \(\frac{\class{blue}{2} - \class{red}{0}}{\class{green}{4}} = \frac{1}{2}\).

Now we need to find a formula for \(x_j\). Let’s consider the right endpoints of the first few rectangles. The 1st rectangle has a right endpoint of \(x_1 = a + \Delta x\), the 2nd rectangle’s right endpoint is \(x_2 = a + 2\cdot\Delta x\), and so on. This means the \(j\)th rectangle has a right edge at \(x_j = a + j\cdot\Delta x = a + j \cdot \frac{b-a}{n} \).

Here are some example values for \(x_j\) if we’re finding the area under a curve from \(x = 0\) to \(x = 2\) with 4 rectangles:

Substituting our new formulas for \(\Delta x\) and \(x_j\), we get:

This is what a right Riemann sum looks like as a sigma expression. But what about a left Riemann sum?

The width of each rectangle, \(\Delta x\), stays the same, but the height of each rectangle is determined by the function evaluated at its left boundary instead of its right boundary. How does that change the sum?

Well, we defined \(x_j\) as the \(j\)th rectangle’s right boundary. The important thing to notice is that the left boundary of the 2nd rectangle is the right boundary of the 1st rectangle, the left edge of the 3rd rectangle is the right edge of the 2nd, and so on.

This means that the height of the 2nd rectangle is \(x_1\), the height of the 3rd is \(x_2\), and so on. In general, the height of the \(j\)th rectangle is \(x_{j-1}\) in a left Riemann sum. (This works for the 1st rectangle if you imagine a “zeroth rectangle” to the left of it! The right edge of the “zeroth rectangle” is the left edge of the 1st rectangle.)

Writing this in sigma notation gives us these expressions:

Now let’s move on to the trapezoidal rule. How could we write a trapezoidal sum in sigma notation?

How could we concisely represent the sum of the areas of the trapezoids?

As a reminder, the area of a trapezoid is \(\frac{A+B}{2}\cdot h\), where \(A\) and \(B\) are the lengths of the bases of the trapezoid and \(h\) is the “height” of the trapezoid (which in a trapezoidal sum is actually the width of each trapezoid).

When we find the area under a curve from \(x = a\) to \(x = b\), we are dividing \(b-a\) units of the \(x\)-axis into \(n\) trapezoids. This means the width of each trapezoid \(h\) is \(\frac{b-a}{n}\). We’ll call the right endpoint of the \(j\)th trapezoid \(x_j\) and the left endpoint of the 1st trapezoid \(x_0\).

How would we write the area of the first trapezoid? The left side of that trapezoid lies on \(x = x_0\), so the height of the left side is \(f(x_0)\). The right side of the trapezoid lies on \(x = x_1\), so its height is \(f(x_1)\).

This means that the area of the first trapezoid is:

The area of the second trapezoid is very similar. The left edge of the trapezoid lies on \(x = x_1\) and the right edge has an \(x\)-coordinate of \(x = x_2\), so the heights of the trapezoid’s edges are \(f(x_1)\) and \(f(x_2)\) respectively. The area of the 2nd trapezoid is:

We can continue the pattern. The third trapezoid will have area:

This means that if we are finding the combined area of \(n\) trapezoids in a trapezoidal sum, the sum will look like this:

We can actually simplify this a little further to make it easier for us to calculate this by hand.

We know that \(h = \frac{b-a}{n}\), so let’s make that substitution:

In summary, a right Riemann sum in summation notation looks like this:

A left Riemann sum looks like this:

And a trapezoidal sum looks like this:

If \([a, b]\) is the interval we are finding the area under, and \(n\) is the number of rectangles we are using, then for Riemann sums and the trapezoidal rule:

When solving for definite integrals (the area under a function over a specific interval), it’s important to remember that a single Riemann sum is an approximation and doesn’t give us the actual area that we are trying to find. However, we can use Riemann sums in a different way to get the exact area.

Remember that Riemann sums get more accurate the more rectangles you use. In other words, as the number of rectangles gets larger and larger, the error of our area approximation approaches 0.

So if we want that error to be exactly 0, we have to use infinitely many rectangles. But how do we do that? We can’t just plug in \(n = \infty\) into our Riemann sum formula. Instead, we have to use a limit!

Let’s go back to our example of finding the area under \(f(x) = x^2\) from \(x = 0\) to \(x = 2\). We will use a right Riemann approximation with \(n\) rectangles, and we will use a limit to describe what happens as \(n\) approaches infinity.

The right Riemann sigma expression looks like this:

The interval that we are finding the area under is \([0, 2]\), so \(a = 0\) and \(b = 2\). We’ll keep \(n\) as a variable for now.

Now we want to take the limit as \(n\) approaches \(\infty\) in order to get the exact area under the curve (the exact definite integral)!

It doesn’t matter if we use a right or a left Riemann sum, because both of them approach the true area as \(n\) approaches \(\infty\).

Let’s see if we can generalize this process. We’ll go back to the general form of a right Riemann sum:

We want to take the limit as \(n\) approaches infinity:

Now this gives us a good definition of the definite integral. However, it’s still long and has a lot of symbols, so mathematicians have created an alternate shorter notation for this. It looks like this:

The sigma symbol \(\sum\) becomes an integral symbol \(\int\) to signify that we’re taking a definite integral (the sum of infinitely many rectangles). In addition, because \(\Delta x\) becomes infinitely small (\(\Delta x\) approaches 0), it is replaced with the symbol \(\dd{x}\) instead. Since the integral sign \(\int\) already implies we are taking the limit as the number of rectangles approaches infinity, we can get rid of the \(\displaystyle\lim_{n \to \infty}\).

The \(\dd{x}\) in the definite integral expression clarifies what variable we are integrating over (in this case \(x\)). If we had a function \(f(t)\), its definite integral would be \(\int_a^b f(t)\dd{t}\) instead.

The variables \(a\) and \(b\) on the top and bottom of the integral sign tell us the bounds of integration: the interval of the function we are finding the area under. For example, if we were finding the area under \(f(x) = x^2\) over the interval \([\class{red}{0}, \class{blue}{2}]\), we would write the integral like this:

We can convert from this integral notation into sigma notation and vice versa. Here’s an example: the area under \(f(x) = \sin(x)\) from \(x = \class{red}{1}\) and \(x = \class{blue}{3}\).

Now we have to find \(\Delta x\) and \(x_j\).

Now we plug these values into the sigma expression for a definite integral:

Now that we know that a definite integral gives us the area under a function over a specific interval, let’s explore the properties of these integrals so that we can solve them more easily.

First, we’ll start simple. What is the area under any function \(f(x)\) from \(x = a\) to \(x = a\)? The lower and upper bounds are the same.

What is the area of the red region with a width of 0?

This is essentially asking what the area of a rectangle with a width of zero is, which is obviously 0. Therefore, no matter what \(f(x)\) is or what the value of \(a\) is, the area under \(f(x)\) from \(x = a\) to \(x = a\) is always 0.

What happens to the area under a function if we multiply that function by a constant? For example, let’s define \(g(x) = 2f(x)\). How does the area under \(g(x)\) compare to the area under \(f(x)\) over an interval \([a, b]\)?

The area under \(g(x) = 2f(x)\) is twice the area under \(f(x)\).

This isn’t only true when we’re multiplying by 2. If we multiply a function \(f(x)\) by any constant \(k\), the definite integral over any interval will also be multiplied by \(k\).

What if instead of multiplying \(f(x)\) by a constant, we added another function instead? In other words, if we have two functions \(f(x)\) and \(g(x)\), what is the definite integral of \(f(x) + g(x)\)? (When I say something like “the definite integral of \(f(x)\)”, I really mean the definite integral of \(f(x)\) over any interval.)

The area under \(\class{green}{f(x) + g(x)}\) is equal to the area under \(\class{red}{f(x)}\) plus the area under \(\class{blue}{g(x)}\).

The same thing is true with subtraction: the definite integral of \(f(x) - g(x)\) is equal to the definite integral of \(f(x)\) minus the definite integral of \(g(x)\).

To get some intuition for this next property, imagine your favorite shape. If we split that shape into two parts, that doesn’t change the total area of the shape - the total area of the shape is just the sum of the two parts.

We can split definite integrals similarly. If we have a definite integral from \(x = a\) to \(x = b\), and we split it at \(x = c\), then the total area from \(x = a\) to \(x = b\) is the sum of the two parts. Those two parts are the definite integral from \(x=a\) to \(x=c\) and the definite integral from \(x=c\) to \(x=a\).

The definite integral from \(x = a\) to \(x = c\) plus the definite integral from \(x = c\) to \(x = b\) is equal to the definite integral from \(x = a\) to \(x = b\).

It turns out this property works for any values of \(a\), \(b\), and \(c\), even if \(c\) isn’t in between \(a\) and \(b\)!

So far, when we’ve talked about definite integrals, we’ve used examples where the upper bound \(b\) is greater than or equal to the lower bound \(a\). But believe it or not, that doesn’t have to be the case! Let’s see what happens with this integral:

In this case, \(a = 2\) and \(b = 0\). This doesn’t seem to make any sense! However, let’s look closer at the definition of a definite integral.

Remember, \(\Delta x = \frac{b-a}{n} \). If \(b\) is less than \(a\), all that means is that \(\Delta x\) becomes negative! What this means is that when we calculate the area of each rectangle \((f(x_j) \cdot \Delta x)\) in our summation, we get the negative of the actual area each time. This means that swapping the bounds \(a\) and \(b\) of a definite integral swaps the signs of the integral (i.e. multiplies it by -1).

Finally, what if we’re finding the definite integral of a function, but the function goes below the \(x\)-axis? It doesn’t make sense to talk about the “area under a curve” when that curve is below the \(x\)-axis.

Well, imagine I have a function that describes my velocity in miles per hour, \(f(t) = -3\). Let’s say that positive velocity means I’m walking to the right and negative velocity means I’m going to the left. What this means is that because my velocity is always -3, I’m walking to the left at 3 miles per hour.

If we find the definite integral of this function over say, \(t = 0\) to \(t = 2\), that will give us the total change in my distance over the first 2 hours. Because I’m walking to the left at 3 miles per hour, after 2 hours, I will have moved 6 miles to the left. Because negative values mean left, the definite integral should be -6.

In this case, this change in position can still be modeled by an area on the graph; we just need to be careful. Instead of looking at the area under the function, we are looking at the area above the function (between the function and the \(x\)-axis). In addition, because our function is below the \(x\)-axis, we need to count it as negative area.

The area between \(\class{red}{f(t)}\) and the \(x\)-axis over \(t = 0\) to \(t = 2\) is 6. However, because the function is below the \(x\)-axis, when we take the definite integral, we must consider this area to be negative. So \(\int_0^2 f(t)\dd{t} = -6\). This makes sense because since our velocity is negative, our change in position that is represented by the definite integral also has to be negative.

Let’s use some of these properties to find the value of a definite integral!

Problem: What is the value of the definite integral \(\displaystyle\int_{3}^{-2} 22(2x + 3)\dd{x}\)?

First, because the upper bound is less than the lower bound, let’s swap the bounds of the integral, remembering to add a negative sign before the integral:

Here, we could distribute the 22 inside the integral, but we could also just take the constant 22 out of the integral because of the multiplication by constant property.

We can then split the integral into two using the addition property. (This isn’t required to evaluate the integral, but I want to show that the property works.)

Now let’s evaluate \(\int_{-2}^3 2x\dd{x}\).

To find this integral, we can split it up into two integrals: one from -2 to 0 and another from 0 to 3.

The integral from -2 to 0 is the area of a triangle with a base of 2 and a height of 4. The area of that triangle is \(\frac{1}{2}bh = \frac{1}{2} \cdot 2 \cdot 4 = 4\). However, because this triangle is below the \(x\)-axis, we need to make this area negative, so the area is -4.

The integral from 0 to 3 is the area of a triangle with a base of 3 and height of 6. This area is \(\frac{1}{2} \cdot 3 \cdot 6 = 9\).

Now we need to find \(\int_{-2}^3 3\dd{x}\). This area is just a rectangle with width \(3 - (-2) = 5\) and height 3, so the area is 15.

The width of this rectangle is \(3 - (-2) = 5\) and the height is 3. Therefore, \(\int_{-2}^3 3\dd{x} = 15\).

Now we can finally put everything together to solve our integral.

We managed to show that \(\int_{3}^{-2} 22(2x + 3)\dd{x} = -440\) just using integral properties and a bit of geometry!

We’ve learned one way to find definite integrals: to find the limit of a Riemann sum as the number of rectangles approaches infinity. There are two ways to evaluate this: you can do a lot of confusing algebraic manipulation with summations, or you can estimate the limit with a computer. The first method is outside the scope of this website, so let’s try the second method here.

We’re going to find the definite integral of \(\ln(x)\) from \(x = 1\) to \(x = 100\) using a right Riemann sum. Explore what happens as the number of rectangles in the Riemann sum increases:

Now this isn’t the best way to calculate a definite integral. First of all, it takes a lot of calculations and processing power to get an accurate result. It might have taken a few seconds for your device to calculate the Riemann sum with 100 million rectangles. We’re lucky that computers are this powerful nowadays - this would have been impossible even a century ago!

We could have gotten better results with a midpoint or trapezoidal sum, but even then, it would be a pain to do this by hand before computers existed.

The other problem is that this method doesn’t give us an exact answer. In this case, the Riemann sum with a billion rectangles gives us an area of about 361.5170188. For reference, the actual area is about 361.5170186.

Luckily, there is a much, much better way to calculate definite integrals, and you might be surprised that it is actually related to differentiation. Let’s explore the connection between integrals and derivatives!

For the umpteenth time on this website, we’re going to use an example where we’ve traveled from one place to another. Let’s say that our distance at time \(t\) is modeled by the function \(d(t)\) and our speed at time \(t\) is \(v(t)\).

If we only had our distance function \(d(t)\), we could get our speed function \(v(t)\) by differentiating \(d(t)\).

However, let’s say we only had our speed function \(v(t)\). How do we find the total distance we’ve traveled at any given time?

If we want to know how far we’ve traveled by the end of the 2nd hour, we can find the definite integral of \(v(t)\dd{t}\) from \(t = 0\) to \(2\). In general, we can find the distance at any time \(x\) by finding the integral of \(v(t)\dd{t}\) from \(t = 0\) to \(x\). So we can recreate our distance function using just \(v(t)\):

\(d(x)\) gives us the distance we’ve traveled by time \(x\). \(v(t)\) gives us our speed at time \(t\).

What does all of this mean? Well, the derivative of our distance function gives us our speed function, and in a sense, the integral of our speed function gives us our distance function. This must mean that differentiation and integration are related in some way - they are essentially inverses in this case!

This is what the fundamental theorem of calculus tells us. It’s also known as the first fundamental theorem of calculus (because there’s a second part to it that I’ll cover in a future lesson). I’ll describe what the theorem states in more detail now.

To use the fundamental theorem of calculus, we need two things: a function \(f\) and an interval \([a, b]\) that the function is continuous over. We’ll only focus on this interval of the function for now, and not the entire function. As an example, we’ll use \(f(x) = 3x^2\) as our function and \([0, 1]\) as our interval.

This is the function \(f(x) = 3x^2\). We’re going to focus on the interval \([0, 1]\), but we could choose any interval that \(f(x)\) is continuous over.

Then we’re going to define a function \(F(x)\) that tells us the area under the curve from \(a\) to \(x\). The value of \(x\) must be in the interval \([a, b]\). Symbolically, that looks like this:

(We’re using \(t\) as the variable in \(f(t)\dd{t}\) because we’re already using \(x\) as the upper bound of the integral.)

In our example, \(F(x)\) would be defined as this:

For example, \(F(0.5)\) would give the area under the curve from \(x = 0\) to \(x = 0.5\).

Now, what is the derivative of this area function \(F(x)\)? Remember, the derivative tells you the rate of change of \(F(x)\) as you change \(x\) by an infinitesimal amount.

Let’s imagine increasing \(x\) by a tiny amount \(\Delta x\). How will that affect the value of \(F(x)\)?

Increasing \(x\) by \(\Delta x\) increases the area \(F(x)\) by the area of the green shaded region. That region has an area of approximately \(f(x) \cdot \Delta x\).

It can be proved that as \(\Delta x\) gets smaller and smaller, the area added must approach \(f(x) \cdot \Delta x\). What does this tell us?

We are trying to find the derivative of the area \(F(x)\) with respect to \(x\). Normally, a derivative tells you the slope of a function at any point: the ratio of the change in \(y\) to the change in \(x\) as the change in \(x\) approaches 0. However, in this case, because \(F(x)\) is measuring area, the derivative of \(F(x)\) is the ratio of the change in area to the change in \(x\) (\(\Delta x\)) as \(\Delta x\) approaches 0. This means that as \(\Delta x\) approaches zero, the rate of change (derivative) of \(F(x)\) must be \(\frac{f(x) \cdot \Delta x}{\Delta x} = f(x)\)!

In other words, the instantaneous rate of change of the area \(F(x)\) as you change \(x\) is equal to the function \(f\) evaluated at \(x\).

In conclusion, the derivative of \(F(x) = \int_0^x 3t^2\dd{t}\) is simply \(F'(x) = f(x) = 3x^2\). All we have to do is take the expression inside the integral and replace the variable \(t\) with \(x\). The derivative and integral essentially cancel out each other!

Notice how the value of \(F'(x)\) is always the same as the value of \(f(x)\)!

More generally, the fundamental theorem of calculus states that if we have an interval \([a, b]\) that \(f\) is continuous over, then the derivative of \(F(x) = \int_a^x f(t)\dd{t}\) for all \(x\) in the interval \([a, b]\) is simply \(F'(x) = f(x)\).

Let’s try another example. Problem: What is the derivative \(F'(x)\) of the function \(\displaystyle F(x) = \int_3^x \sqrt[3]{e^t + t}\dd{t}\)?

Well, \(f(t) = \sqrt[3]{e^t + t}\) is continuous for all values of \(t\), so we can use the fundamental theorem of calculus here. To get the derivative, all we have to do is replace all instances of \(t\) with \(x\) and remove the integral, so our answer is \(F'(x) = \sqrt[3]{e^x + x}\).

Here’s a harder function to differentiate:

Problem: Find the derivative \(F'(x)\) of the function \(\displaystyle F(x) = \int_0^{\sin(x)} t^2\dd{t}\).

Now we have \(\sin(x)\) instead of \(x\) as the upper bound of the definite integral. How do we solve this? We can use the chain rule here!

Let’s define a new area function with an upper bound of \(x\), \(g(x) = \int_0^x t^2\dd{t}\). This means that \(g'(x) = x^2\) by the fundamental theorem of calculus. We’ll also define \(h(x) = \sin(x)\).

The key realization to make here is that \(g(h(x)) = g(\sin(x)) = F(x)\). So differentiating \(F(x)\) is the same as differentiating \(g(h(x))\). We can do that with the chain rule.

So now we know the deep connection between derivatives and integrals. In the next few sections, we’ll discover how we can use this connection to calculate the value of definite integrals faster than ever!

Why is the fundamental theorem of calculus true? It’s one of the most important theorems you learn in a calculus class, so it’s important to understand why it’s true.

Let’s start with any continuous function \(f(x)\) and its corresponding area function \(F(x) = \int_0^x f(t) \dd{t}\). We want to show that the derivative of this area function \(F'(x)\) is equal to the original function \(f(x)\).

To do this, we need to remember what the derivative actually means. The derivative tells us “as we change the input of a function by a tiny amount \(h\), how much does the output change?” So in this case, we’re asking, as we change \(x\) by a tiny amount, how much does the area function \(F(x)\) change? In other words, how much does the area under \(f(x)\) from 0 to \(x\) change when we change \(x\) by a tiny amount?

More formally, we want to find this expression:

The value of \(F(x+h)\) can be written as the integral \(\int_0^{x+h}f(t) \dd{t}\), and \(F(x)\) can be written as \(\int_0^x f(t)\dd{t}\). Therefore, the limit becomes:

We can use integral properties to rewrite this difference of integrals as simply \(\int_x^{x+h} f(t)\dd{t}\). This integral represents the area added when we increase \(x\) by a tiny amount \(h\).

Between \(t = x\) and \(t = x+h\), the function \(f(t)\) must obtain a maximum and minimum within that interval according to the extreme value theorem (remember, \(f(t)\) is continuous). We’ll call this minimum \(m\) and the maximum \(M\).

Therefore, the integral \(\int_x^{x+h} f(t)\dd{t}\) must be in between \(hm\) and \(hM\).

This diagram shows that \(hm \le \int_x^{x+h} f(t)\dd{t} \le hM\).

We can divide this inequality by \(h\) to get that \(\frac{1}{h}\int_x^{x+h} f(t)\dd{t}\) must be in between \(m\) and \(M\).

Intuitively, as \(h\) approaches 0, the values of \(m\) and \(M\) must both approach \(f(x)\), so using the squeeze theorem, we can find that \(\frac{1}{h}\int_x^{x+h} f(t)\dd{t}\) must also approach \(f(x)\). This means that the derivative of our area function, \(F'(x)\), must equal \(f(x)\).

How can we make this argument more rigorous? We can use the mean value theorem for integrals.

If a function \(f(x)\) is continuous over the interval \([a, b]\), then by the extreme value theorem, it obtains a minimum value \(m\) and maximum value \(M\) over this interval.

Therefore, for any continuous function \(f(x)\), this inequality holds:

\(m(b-a)\) is the area of a rectangle with width \(b-a\) and height \(m\). Because \(m\) is the minimum value of \(f(x)\) on \([a, b]\), the integral \(\int_a^b f(x) \dd{x}\) must be greater than this. Similarly, the integral must be less than \(M(b-a)\), since \(M\) is the maximum value of \(f(x)\) on \([a, b]\).

Dividing by \(b-a\), we get:

Therefore, because \(f(x)\) is continuous over \([a, b]\), by the intermediate value theorem, there must be a value \(c\) in \([a, b]\) such that \(f(c) = \frac{1}{b-a}\int_a^b f(x) \dd{x}\).

Let’s go back to our proof of the fundamental theorem of calculus. If we replace \(a\) with \(x\) and \(b\) with \(x+h\), we get that by the mean value theorem, there must be a value \(c\) in the interval \([x, x+h]\) such that \(f(c) = \frac{1}{h}\int_x^{x+h} f(t)\dd{t}\). Therefore, \(F'(x)\) can be written as follows:

Note that \(c\) must be in the interval \([x, x+h]\). As \(h\) approaches 0, this interval \([x, x+h]\) becomes smaller and smaller, and so since \(c\) is in this interval, \(c\) must approach \(x\). This also means that \(f(c)\) must approach \(f(x)\). Therefore, \(F'(x) = f(x)\), which is the fundamental theorem of calculus!

We’ve seen that we can analyze a function using its derivative. Since integrals are the inverse of derivatives, we can also analyze a function using its integral! We just have to remember that it works in reverse.

Let’s say we have a continuous function \(f(x)\) and we define a function \(F(x) = \int_a^x f(t) \dd{t}\). Because of the fundamental theorem of calculus, we know that \(F'(x) = f(x)\). Knowing this, we can analyze \(F(x)\) using \(f(x)\) and vice versa.

In this graph, \(\class{red}{F(x)} = \int_0^x \class{blue}{f(t)}\dd{t}\). In words, \(F(x)\) gives us the area under \(f(x)\) from 0 to \(x\). The background is colored according to the sign of \(f(x)\).

If \(f(x)\) is positive, then \(F(x)\) is increasing, because if a function’s derivative is positive, that means the original function is increasing. Another way to think of it is that if \(f(x)\) is positive, the area under \(f(x)\) (which is given by \(F(x)\)) will increase as \(x\) increases. Likewise, if \(f(x)\) is negative, then \(F(x)\) is decreasing.

Whether \(f(x)\) is increasing or decreasing determines the concavity of \(F(x)\). If \(f(x)\) is increasing, \(F(x)\) is concave up, and if \(f(x)\) is decreasing, \(F(x)\) is concave down.

If \(f(x)\) crosses the \(x\)-axis from negative to positive, \(F(x)\) reaches a local minimum, and if \(f(x)\) crosses from positive to negative, \(F(x)\) reaches a local maximum. A local extremum in \(f(x)\) signifies an inflection point in \(F(x)\).

The key is to remember that \(f(x)\) is the derivative of \(F(x)\) and to use what you know about derivatives to make conclusions about \(f(x)\) and \(F(x)\).

A few lessons ago, I said that the fundamental theorem of calculus would lead to a faster way of finding definite integrals. We’re going to get to that now! In this lesson, we will explore how we can use the fundamental theorem of calculus to find the definite integral \(\int_{0.2}^{0.8} 3x^2\dd{x}\) without having to deal with Riemann sums.

Problem: How can we evaluate the definite integral \(\displaystyle\int_{0.2}^{0.8} 3x^2\dd{x}\) without using the limit of a Riemann sum?

Imagine we want to find the definite integral from \(a\) to \(b\) of a continuous function \(f\), or \(\int_a^b f(t)\dd{t}\). How can we do this?

First, let’s define an interval \([c, d]\) that the interval \([a, b]\) is contained within. In order for us to proceed, \(f(x)\) must be continuous over this interval \([c, d]\).

We are trying to find the area under the curve from \(a\) to \(b\). In this case, the function is \(f(x) = 3x^2\) and the interval \([a, b]\) is \([0.2, 0.8]\). The interval \([c, d]\), in this case \([0, 1]\), must be continuous and must contain the interval \([a, b]\).

Let’s focus on the area under the curve between \(c\) and \(b\). Due to the properties of definite integrals, we know this is equal to the area between \(c\) and \(a\) plus the area between \(a\) and \(b\).

We can rearrange this to solve for the area under the curve between \(a\) and \(b\), which is what we want to figure out.

In words, this means that the definite integral of \(f(t)\) from \(a\) to \(b\) is equal to the definite integral from \(c\) to \(b\) minus the definite integral from \(c\) to \(a\).

Now let’s define a function \(F(x)\) that gives the area under the curve from \(c\) to \(x\). This means that \(F(x) = \int_c^x f(t)\dd{t}\). We can rewrite the equation using this function:

This equation is known as the second fundamental theorem of calculus! But before we go into more detail about it, let’s talk about antiderivatives.

Because of the first fundamental theorem of calculus, we know that \(f(x)\) is the derivative of \(F(x)\). This means that \(F(x)\) is an antiderivative of \(f(x)\). An antiderivative of a function \(f(x)\) is a function that has a derivative of \(f(x)\). For example, \(x^2\) is an antiderivative of \(2x\) because the derivative of \(x^2\) is \(2x\). Finding antiderivatives is like finding derivatives but in reverse.

It turns out that in the equation for the second fundamental theorem of calculus, \(F(x)\) can be any antiderivative of \(f(x)\). In other words, \(F(x)\) can be any function that has a derivative of \(f(x)\).

For example, let’s find the definite integral of \(3x^2\) from 0.2 to 0.8, or \(\int_{0.2}^{0.8} 3x^2\dd{x}\). We can use the second fundamental theorem of calculus to come up with this expression:

All we have to do now is find what \(F(x)\) is. I said that \(F(x)\) could be any antiderivative of \(f(x)\). In this case, \(f(x) = 3x^2\). What is a function \(F(x)\) whose derivative is \(3x^2\)?

Your first thought might be that \(F(x) = x^3\), since it has a derivative of \(3x^2\). However, can you come up with another function with a derivative of \(3x^2\)?

It turns out there are infinitely many functions with a derivative of \(3x^2\)! For example, \(x^3 + 1\) has a derivative of \(3x^2\), and so does \(x^3 - 10\), \(x^3 + \pi\), and \(x^3 - 20e\). In fact, any function that is defined as \(x^3 + \text{[a constant]}\) is an antiderivative of \(3x^2\). This is because the derivative of a constant is zero.

Instead of writing \(\text{[a constant]}\), we just write \(C\) for short. So all antiderivatives of \(3x^2\) can be summarized with the expression \(x^3 + C\), where \(C\) is any constant. This \(C\) is known as the constant of integration.

This expression, \(x^3 + C\), is known as the indefinite integral of \(3x^2\). (Note: I will sometimes use the terms antiderivative and indefinite integral interchangeably.)

The notation for an indefinite integral is similar to the definite integral, except without any bounds:

Now we’re finally ready to solve the integral we’ve been looking at, \(\int_{0.2}^{0.8} 3x^2\dd{x}\). We can set \(F(x)\) to be any antiderivative of \(3x^2\), so let’s set \(F(x) = x^3\). Now we can plug into the equation we found previously with the second fundamental theorem of calculus:

That was a lot easier than using a Riemann sum! Sometimes you’ll see one of these notations as shorthand for \(F(0.8) - F(0.2)\):

I’ll be using the first notation from now on. Here’s one more example of this notation:

Notice how the \(C\) gets canceled out: this means that we can ignore \(C\) when calculating definite integrals with the second fundamental theorem of calculus.

In the last lesson, I gave an example of an indefinite integral: the indefinite integral of \(3x^2\) is \(x^3 + C\). But we only figured this out because we knew that the derivative of \(x^3\) is \(3x^2\). Is there a more algorithmic way to find this indefinite integral that we could also use to find, say, the indefinite integrals of \(x^{10}\) or \(x^{1000}\)?

Well, we know that the derivative of \(x^3\) is \(3x^2\) because of the power rule, and we also know that indefinite integrals are antiderivatives: finding an indefinite integral is like finding a derivative but in reverse. So what if we could just apply the power rule in reverse?

As a refresher, the power rule requires you to multiply by the exponent and then decrease the exponent by 1. Here’s an example on how to find the derivative of \(x^5\):

We know that the derivative of \(x^5\) is \(5x^4\), so the indefinite integral of \(5x^4\) is \(x^5 + C\). Let’s see if we can find this indefinite integral of \(5x^4\) by applying the power rule in reverse. To do that, we need to undo each step and apply the rules in reverse order. This means increasing the exponent by 1, then dividing by the new exponent. Here’s what that looks like:

As you can see, doing this actually reverses the differentiation process and gives us the correct indefinite integral! The antiderivative of \(5x^4\) is \(x^5 + C\), and we can find it by applying the power rule in reverse.

Now we just need a way to generalize this process. Luckily, since the steps are so simple, there is a formula for it:

This is known as the reverse power rule. The reverse power rule works for all real number exponents except for -1. Here are some examples:

You can verify these antiderivatives by differentiating them: you will get the original function back if you differentiate an antiderivative. For example, the derivative of \(-\frac{1}{2x^2} + C\) is indeed \(\frac{1}{x^3}\).

Note: the reverse power rule cannot be used to integrate \(x^{-1} = \frac{1}{x}\). (Try it yourself: what happens?)

Now we know how to find the indefinite integrals of simple expressions like \(x^2\) and \(x^{-5}\), but to find the antiderivatives of more complicated expressions such as \(3x^2\) or \(4x^3 + x\), we need to learn about some indefinite integral properties.

The first rule is that if you have two functions added or subtracted together within an indefinite integral, you can split it up into two indefinite integrals being added or subtracted.

In addition, if you have a constant within an indefinite integral, you can take it out of the integral:

Notice how these properties are very similar to the properties of definite integrals and derivatives. Now that we know these properties, we can integrate more functions! Here is an example:

Remember that \(C\) doesn’t represent a specific number and is just shorthand for all possible constants. Because of this, we can combine the two \(C\)’s into one since two arbitrary constants added together is just another arbitrary constant. In addition, \(C\) multiplied or added by any constant remains \(C\).

Here is one more interesting integral:

Problem: \(\displaystyle\int \frac{x^2(3x)}{x^5}\dd{x}\)

How do we solve this? We don’t know of a product or quotient rule for integrals. Well, all we have to do is simplify the expression!

Note that there isn’t a straightforward product rule or quotient rule for integrals. In other words, there is no easy formula for these integrals:

To integrate functions in this form, we will need to use some more advanced techniques that I will go over later in this unit.

Now let’s explore the indefinite integrals of other common functions, specifically \(\frac{1}{x}\), \(e^x\), \(\cos(x)\), and \(\sin(x)\).

First, let’s tackle \(\frac{1}{x}\). Remember that an indefinite integral represents the antiderivatives of a function: functions that when differentiated give back the original function. So finding the indefinite integral of \(\frac{1}{x}\) is just asking, “what functions have a derivative of \(\frac{1}{x}\)?”

You might think that the indefinite integral of \(\frac{1}{x}\) is \(\ln(x) + C\), but this isn’t a perfect answer. It’s true that the derivative of \(\ln(x) + C\) is \(\frac{1}{x}\). However, the original function \(\frac{1}{x}\) is defined for all \(x\)-values except for 0, but \(\ln(x)\) is only defined for positive \(x\)-values. This means that the antiderivative \(\ln(x) + C\) only covers positive values of \(x\). Can you think of another antiderivative that covers the entire domain of \(\frac{1}{x}\)?

The function \(\ln(x)\) is undefined for negative values of \(x\), even though \(\frac{1}{x}\) is defined for those values. Can you think of a better indefinite integral of \(\frac{1}{x}\) that covers both positive and negative values of \(x\)?

Can you find the formula for the antiderivative in this graph?

The antiderivative is \(\ln(|x|) + C\). I’ll explain why we use an absolute value expression here.

Absolute value functions can be rewritten as piecewise functions. In this case, \(\ln(|x|) + C\) can be rewritten as:

\[ \ln(|x|) + C = \begin{cases} \ln(x) + C & x \gt 0\\ \ln(-x) + C & x < 0 \end{cases} \]

(This function is undefined for \(x = 0\).)

If we take the derivative of both cases, we get:

\[ \begin{flalign} \dv{}{x}[\ln(|x|) + C] &= \begin{cases} \frac{1}{x} & x \gt 0\\ -1 \cdot \frac{1}{-x} & x < 0 \end{cases}\\ &=\begin{cases} \frac{1}{x} & x \gt 0\\ \frac{1}{x} & x < 0 \end{cases}\\ &= \frac{1}{x}, \, x \ne 0 \end{flalign} \]

As you can see, the derivative of \(\ln(|x|) + C\) is always \(\frac{1}{x}\), no matter if \(x\) is negative or positive. So \(\ln(|x|) + C\) is the best expression to represent the indefinite integral of \(\frac{1}{x}\).

(Note: The indefinite integral of \(\frac{1}{x}\) can either be written as \(\int \frac{1}{x} \dd{x}\) or \(\int \frac{\dd{x}}{x}\).)

Now I’m going to work out the indefinite integrals of \(e^x\), \(\cos(x)\), and \(\sin(x)\). Try to find them yourself first!

This isn’t a lesson on its own, but rather an interactive demo I’ve created to help you understand a concept better.

I’ve already given an example of how to solve for definite integrals using the second fundamental theorem of calculus. As a refresher, the second fundamental theorem of calculus says that the definite integral from \(a\) to \(b\) of a function \(f(x)\) is \(F(b) - F(a)\), where \(F\) is the indefinite integral of \(f\). In this lesson, I’m going to be showcasing more examples of using indefinite integrals to find definite integrals. Here’s the first example:

Problem: \(\displaystyle \int_3^{12} \frac{5x^2 + 6x^3}{x^4} \dd{x}\)

To solve this, we first need to find the indefinite integral of the function \(f(x) = \frac{5x^2 + 6x^3}{x^4}\), or \(F(x)\):

Then, we find \(F(b) - F(a)\):

Here’s another example of a definite integral, this time involving a piecewise function.

Problem: \(f(x)\) is defined as a piecewise function: \(f(x) = \begin{cases} x^2 & x \le 0\\ \sin(x) & x \gt 0 \end{cases} \)   . What is \(\displaystyle\int_{-1}^\pi f(x)\dd{x}\)?

In order to find this definite integral, we need to split it up into two parts: one for each part of the piecewise function. We need to evaluate the definite integral of \(f(x)\) for \(x \le 0\) separately from the definite integral for \(x \gt 0\).

Finally, here’s an example of a definite integral of an absolute value function.

Problem: \(\displaystyle\int_{-1}^1 |2x-1|\dd{x} \)

To solve this, we need to rewrite the absolute value function \(f(x) = |2x-1|\) as a piecewise function. \(2x-1\) is non-negative when \(x \ge \frac{1}{2}\), so \(|2x-1|\) simply equals \(2x-1\) when \(x \ge \frac{1}{2}\). However, when \(2x-1\) is negative, the absolute value will turn it positive, so \(|2x-1| = -(2x-1)\) when \(x \lt \frac{1}{2}\).

Now we can solve the definite integral the same way we would with a piecewise function: by splitting it into two parts.

We now know how to find the indefinite integrals of basic functions, but we still need to learn a few more integration techniques in order to integrate more complicated functions. I’m going to give you a few functions, and my challenge for you is to figure out their antiderivatives!

We can also use \(u\)-substitution to solve definite integrals, but we have to be a bit more careful. There are two ways to solve a definite integral with \(u\)-substitution. Let’s say we have this definite integral:

Problem: \(\displaystyle \int_1^2 2(2x+1)^4 \dd{x} \)

The first strategy we can use is to find the indefinite integral first, then use that to evaluate the definite integral.

Now that we have the indefinite integral, we can use it to find the definite integral.

The second way to solve for this definite integral doesn’t require us to find the indefinite integral in terms of \(x\), but it does require us to be more careful with the bounds of integration. First, we perform the \(u\)-substitution directly on the definite integral:

Notice how even though our integral is now in terms of \(u\), the bounds of integration are still in terms of \(x\). That’s because we can’t just directly switch from the bounds being in terms of \(x\) to being in terms of \(u\); the variable that the bounds are associated with needs to remain the same if we’re keeping their numerical values the same.

To correctly switch the bounds to being in terms of \(u\), we need to find what values of \(u\) correspond to the bounds \(\class{green}{x = 1}\) and \(\class{purple}{x = 2}\). We set \(u = 2x+1\), so the lower bound becomes \(2(\class{green}{1}) + 1 = 3\) and the upper bound becomes \(2(\class{purple}{2}) + 1 = 5\).

Now we can find the indefinite integral in terms of \(u\) and then use it to evaluate the definite integral.

Before we integrate more functions, there’s something very important we need to keep in mind with indefinite integrals. To demonstrate, here’s a fairly simple integral:

Problem: Evaluate \(\displaystyle\int\frac{1}{2x}\dd{x}\).

There are two ways to solve this integral using the techniques we know. First, we could simply factor out \(\frac{1}{2}\) from the integral:

Alternatively, we could use \(u\)-substitution with \(u = 2x\):

We got a different answer this time! Does this mean that one of these answers is wrong?

Remember that we can check if an indefinite integral is correct by differentiating it to see if we get back the original function. So let’s do that for both of our answers:

The derivative of both of these expressions are the same! So what’s going on?

To find out what’s going on, let’s use logarithm properties to simplify the second expression:

Notice that \(\frac{1}{2}\ln(2)\) is just a constant, so we can say that \(\frac{1}{2}\ln(2) + C \) is an arbitrary constant. Therefore, we can simplify \(\frac{1}{2}\ln|x| + \frac{1}{2}\ln(2) + C\) to \(\frac{1}{2}\ln|x| + C\), which is the first expression!

What happened here is that the expressions \(\frac{1}{2}\ln|x|\) and \(\frac{1}{2}\ln|2x|\) differ by just a constant (in this case \(\frac{1}{2}\ln(2)\)), so both \(\frac{1}{2}\ln|x| + C\) and \(\frac{1}{2}\ln|2x| + C\) are valid ways to express the antiderivative of \(\frac{1}{2x}\).

The two expressions \(\class{red}{\frac{1}{2}\ln|x|}\) and \(\class{blue}{\frac{1}{2}\ln|2x|}\) differ by a constant, so they are both valid ways to express \(\int\frac{1}{2x}\dd{x}\).

Remember that the constant of integration \(C\) expresses all possible constants in one symbol, so it is possible to come up with multiple ways to express the same antiderivative that differ by a constant!

From now on, I’m going to show you how to integrate lots of different rational functions. But first, let’s get the basics down with a few examples using \(u\)-substitution that aren’t too difficult.

Problem: Evaluate \(\displaystyle\int\frac{1}{4x+5}\dd{x}\).

We want to get this integral to the form \(\int\frac{1}{u}\dd{u}\), since we know how to integrate that. We can do that by setting \(u\) to the denominator.

We can generalize this process to any integral of the form \(\int\frac{B}{ax+b}\dd{x}\), where \(B\), \(a\), and \(b\) are constants.

Problem: Evaluate \(\displaystyle\int\frac{B}{ax+b}\dd{x}\).

Here’s a problem involving a linear function divided by a quadratic function.

Problem: Evaluate \(\displaystyle\int\frac{x}{x^2+10}\dd{x}\).

Notice that the derivative of the denominator is \(2x\), which is twice the numerator. We can easily manipulate the numerator to equal \(2x\) so we can perform the substitution \(u = x^2 + 10\).

We can get rid of the absolute value bars in our answer because \(x^2+10\) is always positive.

This strategy will work whenever we have an integral of the form \(\int\frac{Ax}{ax^2+c}\dd{x}\). But what if we just have a constant in the numerator instead?

Problem: Evaluate \(\displaystyle\int\frac{1}{x^2+10}\dd{x}\).

This time the numerator isn’t a constant multiple of the derivative of the denominator, so we can’t get this integral into the form \(\int\frac{1}{u}\dd{u}\). Instead, we need to use the fact that \(\int\frac{1}{u^2+1}\dd{u} = \arctan(u) + C\), since the derivative of \(\arctan(u)\) is \(\frac{1}{u^2+1}\).

However, we have \(x^2 + 10\) in the denominator, not \(x^2 + 1\). We can fix this by dividing both the numerator and denominator by 10.

Remember that we want to get our integral into the form \(\int\frac{1}{u^2+1}\dd{u}\). To do that, let’s rewrite \(\frac{x^2}{10}\) as \((\frac{x}{\sqrt{10}})^2\), because then we can use the substitution \(u = \frac{x}{\sqrt{10}}\).

Here, we want to use the substitution \(u = \frac{x}{\sqrt{10}}\), but that requires us to have \(\frac{1}{\sqrt{10}}\) in the numerator (since \(\dd{u} = \frac{1}{\sqrt{10}}\dd{x}\)).

We can finally perform the substitution now!

This strategy will work for integrals of the form \(\int\frac{B}{ax^2 + c}\dd{x}\), where \(a\) and \(c\) are positive.

Finally, one last problem:

Problem: Evaluate \(\displaystyle\int\frac{x+1}{x^2+10}\dd{x}\).

For this integral, we can split it up into two:

We already did these two integrals previously, so let’s substitute the answers we found:

Now we know how to integrate functions of the form \(\int\frac{Ax+B}{ax^2 + c}\dd{x}\), where \(a\) and \(c\) are positive! In these cases, we split the integral up into two integrals, then solve them individually.

There are some more strategies we can use to find indefinite integrals of rational functions when the other methods don’t work. The first strategy is to use polynomial long division to simplify an expression. This will work if the degree of the numerator is greater than or equal to the degree of the denominator.

Problem: Evaluate \( \displaystyle\int \frac{4x^2 + 4x - 1}{2x - 1}\dd{x} \).

We need to simplify \(\frac{4x^2 + 4x - 1}{2x - 1}\) down to something we can integrate more easily. However, \(4x^2 + 4x - 1\) is not factorable, so we need to use polynomial long division:

We find that \(\frac{4x^2 + 4x - 1}{2x - 1} = \class{red}{2x + 3} + \frac{\class{blue}{2}}{\class{green}{2x - 1}}\). Now we can integrate that more directly:

To integrate \(\frac{2}{2x-1}\) in the last step, I used \(u\)-substitution:

(Alternatively, you can use the formula in the last section: \(\int\frac{B}{ax+b}\dd{x} = \frac{B}{a}\ln|ax+b| + C\).)

The other strategy I’ll be going over is to use the derivatives of \(\arcsin(x) = \sin^{-1}(x)\) and \(\arctan(x) = \tan^{-1}(x)\) along with the “completing the square” technique from algebra. As a reminder, here are their derivatives:

Here’s our first indefinite integral that can be solved this way:

Problem: Evaluate \( \displaystyle\int \frac{1}{x^2 + 4x + 20}\dd{x}\).

(Note: Sometimes you will see integrals like this where the numerator is 1 written as \(\int \frac{\dd{x}}{x^2 + 4x + 20}\) instead.)

We can get this expression to look like the derivative of \(\arctan(x)\). We first need to complete the square in the denominator to turn it into the form \((x + a)^2 + b^2\). Here’s how:

We want the denominator to be in the form \(1 + [\text{something}]^2\) (so our integral looks like the \(\arctan\) derivative), so we multiply both the numerator and denominator by \(\frac{1}{4^2}\) to get it into our desired form.

We want to use \(u\)-substitution here with the goal of turning the integral expression into \(\int \frac{1}{1+u^2}\dd{u}\). The best value to use for \(u\) here is \(\frac{x+2}{4}\), meaning that \(\dd{u} = \frac{1}{4}\dd{x}\). So we need to get a \(\frac{1}{4}\) in the numerator in order to use \(u\)-substitution. We can do that by taking out a factor of \(\frac{1}{4}\) from the integral.

We know the derivative of \(\arctan(u) = \frac{1}{1 + u^2}\), so the antiderivative of \(\frac{1}{1 + u^2}\) is \(\arctan(u) + C\).

If you don’t want to do all of this work, you can first complete the square, then use this formula:

Here’s how we would solve our integral with this formula:

Another very similar integral problem is this:

Problem: \( \displaystyle\frac{1}{\sqrt{-x^2 - 6x - 5}} \dd{x}\)

This is very similar to our last problem except we have a square root in the denominator. This means we’re going to need to transform this into something similar to the derivative of \(\arcsin(x)\), which is \(\frac{1}{\sqrt{1-x^2}}\). We can do this by completing the square again.

Now we have \(\sqrt{4 -[\text{something}]^2}\) in the denominator, but we want the constant before \([\text{something}]^2\) to be 1 since the derivative of \(\arcsin(x)\) has \(\sqrt{1 - x^2}\) in the denominator. We can fix this by dividing both the numerator and denominator by \(\sqrt{4}\) (which in this case is just 2).

Now we can use \(u\)-substitution to get the integral expression into the form \(\int\frac{1}{1-\sqrt{x}}\dd{x}\).

Since we know that the derivative of \(\arcsin(u)\) is \(\frac{1}{\sqrt{1 - u^2}}\), we can integrate \(\int \frac{1}{\sqrt{1 - u^2}}\dd{u}\) to get \(\arcsin(u) + C\).

Once again, there is a formula for integrals like these. It looks like this:

Here’s how we would use this formula for this integral:

At this point, you’ve probably realized that integrating most functions is not as straightforward as differentiation. For example, there is no straightforward formula for integrating a product or quotient of functions; instead you need to use techniques like \(u\)-substitution or integration by parts (which I will go over in the next section).

However, these integration techniques require specific circumstances to be used. For example, \(u\)-substitution can typically only be used when you can make the derivative of an inner function appear within the integral.

For example, the integral \(\int xe^{-x^2} \dd{x}\) can be performed using the substitution \(u = -x^2\). But what about the integral \(\int e^{-x^2} \dd{x}\)? Without the extra \(x\) in the front, we can’t perform a \(u\)-substitution any more.

In fact, there is no way to integrate \(e^{-x^2}\) and get a result in terms of elementary functions, which are familiar functions like \(\sin(x)\), \(\cos(x)\), \(\ln(x)\), polynomials, etc. (Note: This doesn’t mean that \(e^{-x^2}\) doesn’t have an antiderivative. It’s just that the antiderivative can’t be calculated exactly using only elementary functions.)

Even in these cases, we can still use Riemann sums to approximate the definite integrals of these types of functions. In addition, sometimes we define special functions that equal certain definite integrals of these functions.

For example, one of these special functions is known as the error function (denoted by \(\erf(x)\)), and it’s defined as this integral:

Using this definition of the error function, we can express the integral \(\int e^{-x^2} \dd{x}\) as \(\frac{\sqrt{\pi}}{2}\erf(x) + C\).

Other examples of functions that have nonelementary integrals are:

• \(\frac{\sin(x)}{x}\)
• \(\frac{\cos(x)}{x}\)
• \(\sin(x^2)\)
• \(\cos(x^2)\)
• \(\frac{1}{\ln(x)}\)
• \(\frac{e^x}{x}\)

Just like with the error function, many of these nonelementary integrals have their own functions. For example, the sine integral function \(\operatorname{Si}(x)\) is defined as:

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We’ve seen that \(u\)-substitution is how we reverse the chain rule, but is there a way to reverse the product rule for derivatives? Let’s experiment with this to see if we can discover a new way to find indefinite integrals.

Let’s start with the product rule formula:

Then we can integrate both sides and use the fundamental theorem of calculus so that the derivative on the left side cancels out.

Let’s try solving for one of the integrals on the right side.

Now this formula is very interesting because we essentially have a formula to integrate a product of a function and the derivative of another function. Using this formula is known as integration by parts. This formula might not seem that useful at first because there’s still an integral on the right side, but let’s experiment and see what we can do with it.

Problem: Evaluate \(\displaystyle\int xe^x \dd{x}\).

Here we have two functions multiplied by each other, so we can use integration by parts. Remember that in the integration by parts formula we start off with a function (\(f(x)\)) multiplied by the derivative of another function (\(g'(x)\)). Here, we can set \(f(x) = x\) and \(g'(x) = e^x\) so that \(f(x)g'(x) = xe^x\). This is our starting point for integration by parts.

Here we need to find the expressions for \(f'(x)\) and \(g(x)\). \(f'(x)\) is simple: it’s just the derivative of \(f(x) = x\), which is \(f'(x) = 1\). \(g(x)\) is a little more complicated: we know that \(g'(x)\), the derivative of \(g(x)\), is \(e^x\). So to find \(g(x)\), we need to find the antiderivative of \(g'(x) = e^x\), which is \(e^x + C\). For integration by parts, we can use any antiderivative of \(g'(x)\), so we can set \(C = 0\), essentially getting rid of it.

Don’t forget to add \(+C\) at the end of the final integral!

As you can see, integration by parts did work in this case! The trick is that the integral on the right side, \(\int \class{green}{1}\class{purple}{e^x} \dd{x}\), is something we can easily solve. If used properly, integration by parts can take a hard integral and express it in terms of an easier integral.

When using integration by parts, you want to set \(f(x)\) and \(g'(x)\) very carefully. The key is to set \(f(x)\) to a function that becomes simpler when differentiated, so that the integral on the right side, \(\int f'(x)g(x) \dd{x}\), is easier to evaluate. In the previous example, if we set \(f(x) = e^x\) and \(g'(x) = x\), the integral on the right side would actually become harder to evaluate!

Sometimes we can use integration by parts in unexpected places.

Problem: Evaluate \(\displaystyle\int \ln(x) \dd{x}\).

Here, we don’t have two functions multiplied together, but what we can do here is set \(g'(x) = 1\), since any function is itself multiplied by 1.

And sometimes we even have to use integration by parts multiple times!

Problem: Evaluate \(\displaystyle\int x^2\sin(x) \dd{x}\).

We can’t solve the integral \(\int \class{green}{2x}[\class{purple}{-\cos(x)}] \dd{x}\) any other way, so we have to use integration by parts again.

Now we can substitute this result into our expression for \(\int x^2\sin(x) \dd{x}\).

Remember that subtracting \(C\) is the same as adding \(C\) since \(C\) is just any arbitrary constant, which can be positive or negative.

Finally, here’s one last example where we need to do something slightly different.

Problem: Evaluate \(\displaystyle\int e^x\cos(x) \dd{x}\).

Notice that neither \(e^x\) nor \(\cos(x)\) gets simpler when you differentiate it. In this case, we can choose either of them to be \(f(x)\) and the other to be \(g'(x)\). Once again, for this integral, we need to use integration by parts twice.

Now we have a very interesting situation: we have \(\int e^x\cos(x) \dd{x}\) on both sides of the equation. In this case, we can actually just add that integral to both sides and solve for it.

There is a shorter form of the integration by parts formula, and it looks like this:

Here, \(u\) replaces \(f(x)\), \(v\) replaces \(g(x)\), \(\dd{u}\) replaces \(f'(x) \dd{x}\), and \(\dd{v}\) replaces \(g'(x)\dd{x}\).

Here is an example using this notation:

Problem: Evaluate \(\displaystyle\int xe^{-x} \dd{x}\).

This content is covered in AP^® Calculus BC but not in AP^® Calculus AB.

From algebra, you might remember having to add two fractions with different denominators. In these cases, you have to rewrite both fractions with a common denominator, then add the numerators. Here’s an example:

But how would we do this in reverse? In other words, if we had a fraction like \(\frac{3x + 7}{x^2 + 5x + 6}\), how would we split it up into two fractions (like \(\frac{1}{x+2}\) and \(\frac{2}{x+3}\))?

We’ll be exploring this in our next integral example.

Problem: \(\displaystyle\int \frac{24x + 3}{8x^2 + 2x - 3} \dd{x}\)

Because the denominator here has a degree that is 1 higher than the numerator, we will need to decompose this fraction into two. The first step to doing that is to factor the denominator.

Now, we can express this fraction as the sum of two: one with a denominator of \(4x+3\) and another with a denominator of \(2x-1\). We don’t know what their numerators are yet, so we’ll call them \(A\) and \(B\) for now.

We can express the sum on the right as the sum of two fractions with a common denominator.

Now we can multiply both sides by the common denominator \((4x + 3)(2x - 1)\) to get rid of the denominators in the equation.

From here, there are two strategies we can use to find \(A\) and \(B\).

Strategy 1: Solving a system of equations

We can isolate the \(x\) terms on the right side and the constants.

Here, we can see that \(2A + 4B\) must equal 24 and \(-A + 3B\) must equal 3. We can use this knowledge to set up and solve a system of equations to solve for \(A\) and \(B\).

Therefore, \(A = 6\) and \(B = 3\).

Strategy 2: Using a shortcut

Let’s look at the equation we had before:

This equation is true for any value of \(x\) we plug in. Notice that if we set \(x = \frac{1}{2}\), then the \(2x-1\) term will equal zero. This means that the variable \(A\) will disappear from the equation, allowing us to easily solve for \(B\).

We can use the same strategy to solve for \(A\), but this time we want the \(4x+3\) term to be zero. To do that, we can use the value \(x = -\frac{3}{4}\).

We get the same values for \(A\) and \(B\) with this strategy: \(A = 6\) and \(B = 3\).

Now that we know what \(A\) and \(B\) are, we can complete the partial fraction decomposition and solve the integral as usual!

This content is covered in AP^® Calculus BC but not in AP^® Calculus AB.

Normally when we find definite integrals, the bounds are finite numbers. For example, this is what the integral of \(e^{-x}\) from \(x = 0 \) to \(x = 2\) looks like:

This is a visual representation of \(\int_0^2 e^{-x}\dd{x}\).

But what if we wanted to find all of the area to the right of \(x = 0\)? In other words, what if we wanted to find the area of this function from \(x = 0\) to infinity?

What if we wanted to find the entire area right of \(x = 0\)? (The shaded region has an infinite width; it keeps going on forever to the right of \(x = 4\).)

How is that even possible? Shouldn’t a definite integral that goes on forever always have an infinite area?

Well, not always. Believe it or not, it’s actually possible for an area like this to be finite. (Want to learn more about how adding up infinitely many things can result in a finite sum? Check out my section on the sums of infinite series.)

How will we calculate this integral? We can see what happens as we change the upper bound, \(b\), to larger and larger numbers.

As we increase the upper bound, the integral area approaches 1 (and never exceeds it). This means that if we were to extend the upper bound to infinity, it would make sense to say, in a sense, that the area under \(x = 0\) to \(x = \infty\) is 1.

We can mathematically prove that the area approaches 1 as \(b\) approaches infinity. What we are trying to find is the limit of the integral as \(b\) approaches infinity, which is written as:

Using the fundamental theorem of calculus, we can rewrite the definite integral. (The indefinite integral of \(e^{-x}\) is \(-e^{-x}+ C\), which can be found using \(u\)-substitution.)

Now we can solve for the limit normally. As \(x\) approaches \(\infty\), \(e^{-x}\) approaches 0, so let’s make that substitution:

The entire area under the function to the right of \(x = 0\) is actually finite! The total area is 1.

What we just found is one type of improper integral. This type of improper integral involves one or more bounds that are not finite. In this case, our integral is written like this:

Notice how the upper bound is \(\infty\). This means we’re evaluating an improper integral.

However, this notation is really just shorthand for a limit:

There is also another type of improper integral. Here’s an example of this other type:

Problem: \( \displaystyle \int_0^1 \frac{1}{\sqrt{x}}\dd{x} \)

It might look normal at first, but the value of \(\frac{1}{\sqrt{x}}\) is undefined at the lower bound \(x = 0\) and unbounded as \(x\) approaches 0 from the right.

The value of \(\frac{1}{\sqrt{x}}\) approaches infinity as \(x\) approaches 0 from the right. Can we still find the shaded area?

However, we can still find a finite value for this integral! We just need to use limits again. We’ll try moving the lower bound, \(a\), closer and closer to 0 to see what happens.

The value of the definite integral appears to approach 2. Now let’s find this limit algebraically:

This area is indeed finite, and it’s equal to 2 despite the function going infinitely high around \(x = 0\).

Sometimes when finding an improper integral, the area is unbounded or the limit doesn’t exist. In this case, we say that the improper integral is divergent. If the limit does exist, the improper integral is convergent.

This section isn’t strictly part of the AP Calculus curriculum. It is mainly meant for curious readers wanting to learn more about math!

Recall that the factorial tells you the product of integers from 1 up to a certain integer. For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). But this basic definition of the factorial only works for positive integers.

Looking at the pattern \(3! = \frac{4!}{4}\), \(2! = \frac{3!}{3}\), and \(1! = \frac{2!}{2}\), we can also define \(0! = \frac{1!}{1} = 1\). But can we do better than this? Can we find the factorial of numbers that aren’t integers, such as \(\frac{1}{2}\) factorial or -1.2 factorial?

These expressions might not make sense at first, but there are actually ways to define the factorial for non-integers! The standard way to do this involves the gamma function.

For values of \(x\) greater than 0, the gamma function is defined as this improper integral:

The gamma function evaluated at an integer \(x \ge 1\) is actually equal to the factorial of \(x - 1\). In symbols:

Here are some examples:

• \(\Gamma(1) = 0! = 1\)
• \(\Gamma(2) = 1! = 1\)
• \(\Gamma(3) = 2! = 2\)
• \(\Gamma(4) = 3! = 6\)

Let’s try calculating \(\Gamma(1)\) using the integral definition:

One important property of the gamma function is that \(\Gamma(x+1) = x\cdot\Gamma(x)\) (for example, \(\Gamma(4) = 3\cdot\Gamma(3)\)). This is one reason why the gamma function works as a generalization as the factorial function: the factorial follows the same rule (i.e. \(x! = x\cdot(x-1)!\)).

We can prove this property using integration by parts with \(f(t) = t^x\) and \(g'(t) = e^{-t}\):

Note: If \(x\) is a positive integer, you can show that \(\displaystyle\lim_{b\to\infty}\left[-\frac{b^x}{e^b}\right] = 0\) by using L’Hôpital’s rule \(x\) times. In general, as \(b\) becomes larger, \(b^x\) for any constant \(x\) grows slower than \(e^b\), so the limit equals 0.

One interesting fact is that \(\Gamma(\frac{1}{2})\) is equal to \(\sqrt{\pi}\). This also means that because \(\Gamma(x+1) = x\Gamma(x)\):

• \(\Gamma(\frac{3}{2}) = \frac{1}{2}\Gamma(\frac{1}{2}) = \frac{1}{2}\sqrt{\pi}\)
• \(\Gamma(\frac{5}{2}) = \frac{3}{2}\Gamma(\frac{3}{2}) = \frac{3}{2} \cdot \frac{1}{2}\sqrt{\pi} = \frac{3}{4}\sqrt{\pi}\)
• \(\Gamma(\frac{7}{2}) = \frac{5}{2}\Gamma(\frac{5}{2}) = \frac{5}{2} \cdot \frac{3}{5}\sqrt{\pi} = \frac{15}{8}\sqrt{\pi}\)
• And so on!

This means that in a way, the factorial of one half is equal to \(\Gamma(\frac{3}{2}) = \frac{\sqrt{\pi}}{2}\).

Even though the integral \(\int_0^\infty t^x e^{-t} \dd{t}\) diverges for negative values of \(x\), we can still evaluate the gamma function at most negative values of \(x\). The key is to use the relation \(\Gamma(x+1) = x\cdot\Gamma(x)\).

For example, \(\Gamma(\frac{1}{2}) = -\frac{1}{2}\Gamma(-\frac{1}{2})\), which means that \(\Gamma(-\frac{1}{2}) = -2\Gamma(\frac{1}{2}) = -2\sqrt{\pi}\). Similarly:

• \(\Gamma(-\frac{1}{2}) = -\frac{3}{2}\Gamma(-\frac{3}{2})\), so \(\Gamma(-\frac{3}{2}) = -\frac{2}{3}(-2\sqrt{\pi}) = \frac{4}{3}\sqrt{\pi}\)
• \(\Gamma(-\frac{3}{2}) = -\frac{5}{2}\Gamma(-\frac{5}{2})\), so \(\Gamma(-\frac{5}{2}) = -\frac{2}{5}(\frac{4}{3}\sqrt{\pi}) = -\frac{8}{15}\sqrt{\pi}\)

However, what happens when we try to evaluate the gamma function at zero? We can try to calculate it using the relation \(\Gamma(1) = 0\Gamma(0)\), but then we end up with \(\Gamma(0) = \frac{\Gamma(1)}{0} = \frac{1}{0}\). Therefore, \(\Gamma(0)\) is undefined. Using a similar argument, we can show that \(\Gamma(-1)\), \(\Gamma(-2)\), \(\Gamma(-3)\), and so on are also undefined.

The gamma function can even be evaluated for complex numbers! For example, \(\Gamma(i) \approx -0.155 - 0.498i\) and \(\Gamma(1+i) \approx 0.498 - 0.155i\), so in a way, we can say that \(i!\) is about \(0.498 - 0.155i\).

Here’s how we define the gamma function evaluated at a complex number \(z\):

• If the real part of \(z\) is greater than 0, then we can use the integral definition \(\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}\dd{t}\). We can use Euler’s formula (note: requires knowledge of infinite series) to define what complex exponents mean!
• If the real part of \(z\) is less than or equal to 0, we can use the relation \(\Gamma(z+1) = z\Gamma(z)\).

This section isn’t strictly part of the AP Calculus curriculum, but is still an important calculus concept.

In this section, we’ll be solving a variety of indefinite integrals involving the sine and cosine functions. We’re going to be using a lot of trig identities, so here are the first three that we’re going to use for now:

These formulas can be derived from the double-angle identities \(\sin(2x) = 2\sin(x)\cos(x)\) and \(\cos(2x) = \cos^2(x) - \sin^2(x)\).

In addition, these integrals, which can be found using \(u\)-substitution, will be very useful:

Let’s start off very simple:

Problem: Evaluate \(\displaystyle\int\sin^2(x)\dd{x}\).

To evaluate this integral, we can simply use the trig identity for \(\sin^2(x)\). After that, a bit of basic integral properties and \(u\)-substitution will help us complete the problem.

Now what happens if we increase the exponent of the sine by 1?

Problem: Evaluate \(\displaystyle\int\sin^3(x)\dd{x}\).

One way to solve this problem is to use the trig identity \(\sin^3(x) = \frac{3\sin(x) - \sin(3x)}{4}\), but if you don’t want to remember that, there is another way that doesn’t require you to know this identity.

There is a manipulation we can use in order to apply \(u\)-substitution here. Can you think of it?

Here’s another problem that can be solved in a very similar way.

Problem: Evaluate \(\displaystyle\int\sin^2(x)\cos^5(x)\dd{x}\).

So far we’ve already integrated \(\sin^2(x)\) and \(\sin^3(x)\), so let’s try integrating \(\sin^4(x)\)!

Problem: Evaluate \(\displaystyle\int\sin^4(x)\dd{x}\).

Now for a problem where both the exponents on the sine and cosine are even.

Problem: Evaluate \(\displaystyle\int\sin^2(x)\cos^2(x)\dd{x}\).

Before we get into our next example, here are some more useful identities:

These formulas can be derived from the angle addition and subtraction identities \(\sin(a \pm b) = \sin(a)\cos(b) \pm \cos(a)\sin(b)\) and \(\cos(a \pm b) = \cos(a)\cos(b) \mp \sin(a)\sin(b)\).

Here’s a problem that requires us to make use of these formulas:

Problem: Evaluate \(\displaystyle\int\sin(2x)\cos(3x)\dd{x}\).

This problem simply requires us to use the identity for \(\sin(a)\cos(b)\), with \(a = 2x\) and \(b = 3x\).

As shown by these examples, there is no single way to integrate all of these functions, and you might have to use different strategies depending on the situation.

This section isn’t strictly part of the AP Calculus curriculum, but is still an important calculus concept.

Here are some more integrals involving trig functions that aren’t sine or cosine! As a reminder, here are some important identities and derivatives:

Let’s hop straight into our first problem.

Problem: Evaluate \(\displaystyle\int \sec^3(x)\tan^3(x) \dd{x}\).

Problem: Evaluate \(\displaystyle\int \sec^6(x)\tan^4(x) \dd{x}\).

Problem: Evaluate \(\displaystyle\int \sec(x) \dd{x}\).

Problem: Evaluate \(\displaystyle\int \tan^3(x) \dd{x}\).

Problem: Evaluate \(\displaystyle\int \sec^3(x) \dd{x}\).

The strategies we used will also work for integrals involving cotangent and cosecant, because the associated identities and derivatives are very similar:

Here’s an example of an integral involving cosecant and cotangent.

Problem: Evaluate \(\displaystyle\int \cot^6(x)\csc^6(x) \dd{x}\).

For completeness, here are the integrals of all six trig functions:

This section isn’t strictly part of the AP Calculus curriculum, but is still an important calculus concept.

Before we start doing any integration, let’s talk about the core idea behind trigonometric substitution, which will be useful in the integrals we’re going to do.

Let’s say we have a term like \(\sqrt{1 - 4x^2}\), and we want to use a substitution to rewrite it as a simpler term. How could we do that?

We could use the substitution \(u = 1 - 4x^2\) (which would turn \(\sqrt{1 - 4x^2}\) into \(\sqrt{u}\)), but in the context of integration, that would only be useful if \(\dd{u} = -8x \dd{x}\) appeared somewhere else in the integral. But what if \(-8x\) was nowhere to be seen in the integral?

That’s when we would need to use a different, more clever substitution. Recall this property of square roots:

(This is because if \(x\) is negative, squaring it and then taking the square root will turn \(x\) positive.)

This means that if we could somehow turn \(\sqrt{1 - 4x^2}\) into \(\sqrt{[\text{something}]^2}\), we could then simplify it down to \(|[\text{something}]|\) (the absolute value of \([\text{something}]\)).

What we’re looking for is a way to turn an expression in the form \(\class{red}{a^2} - \class{blue}{b^2x^2}\) (such as \(1 - 4x^2\)) into a single term squared. There is an identity that vaguely resembles this...

If we could set up a substitution for \(x\) in terms of another variable \(\theta\) that gets \(1 - 4x^2\) to look like \(1 - \sin^2(\theta)\), we could use this identity to simplify it! But what is this magic substitution?

What if our term was instead \(\sqrt{4-9x^2}\)? What substitution could we use to simplify this?

Notice that in both of these cases we had the square root of a constant minus something involving \(x^2\). But trigonometric substitution works in more cases.

Problem: Simplify \(\sqrt{9x^2 - 16}\) using a trig substitution.

This time we want to turn an expression in the form \(\class{blue}{b^2x^2} - \class{red}{a^2}\) into a single term squared. To do that, we can use this identity that looks similar:

Here’s one more problem that requires us to use a different trig substitution.

Problem: Simplify \(\sqrt{10+25x^2}\) using a trig substitution.

Finally, for expressions like these in the form \(\sqrt{\class{red}{a^2} + \class{blue}{b^2x^2}}\), this identity proves to be the most useful:

Here’s a table that shows the three different trig substitutions we can use depending on the form of the square root term:

Now let’s use trig substitution to actually find an integral!

Problem: Evaluate \(\displaystyle\int \sqrt{1-4x^2} \dd{x}\).

We figured out before that using the substitution \(x = \frac{1}{2}\sin(\theta)\) allows us to turn \(\sqrt{1-4x^2}\) into \(\lvert \cos(\theta) \rvert\).

The problem here is that we are still integrating with respect to \(x\). To fix this, we will differentiate the substitution \(x = \frac{1}{2}\sin(\theta)\) with respect to \(\theta\) to find an expression for \(\dd{x}\) in terms of \(\dd{\theta}\).

Now we have another problem: the absolute value bars around \(\cos(\theta)\). In order to find this integral, we will need to drop the absolute value bars, which essentially means we have to assume \(\cos(\theta)\) is positive. Dropping the absolute value bars means our indefinite integral will only work for a certain range of \(x\)-values, but it does allow us to continue.

Now we have an answer, but it’s in terms of \(\theta\) instead of \(x\). How can we get an answer in terms of \(x\)?

The key is to use our original substitution \(x = \frac{1}{2}\sin(\theta)\). Using this, we can find that \(2x = \sin(\theta)\) and \(\theta = \arcsin(2x)\). To figure out what \(\cos(\theta)\) is in terms of \(x\), we can draw a right triangle:

We know that \(\sin(\theta) = 2x\), so we can draw a right triangle with opposite side \(2x\) and hypotenuse 1. Using the Pythagorean Theorem, we can solve for the adjacent side.

Using this triangle, we can find that \(\cos(\theta) = \sqrt{1-4x^2}\). (Alternatively, you can use the fact that \(\sin(\theta) = 2x\) and the identity \(\sin^2(\theta) + \cos^2(\theta) = 1\).) Now we can rewrite the integral in terms of \(x\)!

Here’s a problem involving a definite integral, which will require a slightly different strategy to solve.

Problem: Evaluate \(\displaystyle\int_0^{3/2} x^3\sqrt{9+4x^2}\dd{x}\).

We have a term of the form \(\sqrt{a^2+b^2x^2}\), so we can use the substitution \(x = \frac{a}{b}\tan(\theta)\). In this case, the substitution is \(x = \frac{3}{2}\tan(\theta)\).

Let’s perform this substitution on \(\sqrt{9+4x^2}\) to see what we can simplify it down to.

Now that we’re doing a definite integral, we can’t just drop the absolute value bars. Instead, we need to check if \(\sec(\theta)\) is positive or negative over the interval we’re integrating over. We can do this by rewriting the bounds of our integral in terms of \(\theta\).

To do this, we plug in the endpoints of the interval we’re integrating over (in this case \(x = 0\) and \(x = \frac{3}{2}\)) into our substitution and solve for \(\theta\).

Here’s our work for the lower bound \(x = 0\):

And here’s our work for the upper bound \(x = \frac{3}{2}\):

Each value of \(x\) gives us infinitely many corresponding values of \(\theta\). What values of \(\theta\) should we use here? The rule is that if we’re doing a tangent substitution (i.e. a substitution of the form \(x = \frac{a}{b} \tan(\theta)\)), then we choose values of \(\theta\) that are within the range of the inverse tangent function.

The range of the inverse tangent function is \(-\frac{\pi}{2} \lt \theta \lt \frac{\pi}{2}\), so in this case, our bounds in terms of \(\theta\) are \(\theta = 0\) and \(\theta = \frac{\pi}{4}\).

By plugging in these values of \(\theta\) into the secant function, we find that \(\sec(0)\) and \(\sec(\frac{\pi}{4})\) are positive, so we can conclude that \(\sec(\theta)\) is positive over the interval we’re integrating over. This means that we can safely rewrite \(\sqrt{9+4x^2}\) as \(3\sec(\theta)\).

To fully rewrite our integral in terms of \(\theta\), we need to find what \(\dd{x}\) equals in terms of \(\dd{\theta}\). Since \(x = \frac{3}{2}\tan(\theta)\), \(\dd{x} = \frac{3}{2}\sec^2(\theta)\dd{\theta}\). Now we have enough information to write our integral fully in terms of \(\theta\), and we can solve it!

We found the integral of \(\tan^3(\theta)\sec^3(\theta)\) in the previous section, so I’m not going to show the work for it.

• A definite integral tells you the area between a function and the \(x\)-axis over a given interval.
\[ \int_a^b f(x) \dd{x} = \text{area under } f(x) \text{ from } x = a \text{ to } x = b\]
• The definite integral of a function that represents the rate at which a value is changing tells you the total change in that value. For example, the definite integral of a velocity function tells you the total change in position over a certain interval.
• A Riemann sum is a way to approximate the area under a curve using rectangles.
• The trapezoidal rule is a way to approximate the area under a curve using trapezoids.
• The definite integral can be defined as a limit of a Riemann sum as the number of rectangles approaches infinity.
• Properties of definite integrals:
\[ \int_a^a f(x) \dd{x} = 0 \] \[ \int_a^b k \cdot f(x) \dd{x} = k \cdot \int_a^b f(x) \dd{x} \quad (\text{where } k \text{ is a constant}) \] \[ \int_a^b [f(x) + g(x)] \dd{x} = \int_a^b f(x) \dd{x} + \int_a^b g(x) \dd{x} \] \[ \int_a^b [f(x) - g(x)] \dd{x} = \int_a^b f(x) \dd{x} - \int_a^b g(x) \dd{x} \] \[ \int_a^b f(x) \dd{x} = \int_a^c f(x) \dd{x} + \int_c^b f(x) \dd{x} \] \[ \int_b^a f(x) \dd{x} = -\int_a^b f(x) \dd{x} \]
• The fundamental theorem of calculus states that the derivative of a function \(F(x)\) that represents the area under the curve \(f(t)\) from \(t = a\) to \(t = x\) is \(f(x)\).
\[ \dv{x}\int_a^x f(t) \dd{t} = f(x)\]
• An antiderivative or indefinite integral of a function \(f(x)\) is a function whose derivative is \(f(x)\). For example, all of the antiderivatives of \(f(x) = 2x\) can be represented as \(x^2 + C\).
  • The value \(C\) is the constant of integration, which can represent any arbitrary constant. When finding indefinite integrals, you must add \(C\) because any function has infinitely many antiderivatives that differ by a constant.
• The fundamental theorem of calculus also states that the indefinite integral of a function \(f(x)\) from \(x = a\) to \(x = b\) equals \(F(b) - F(a)\), where \(F(x)\) is an antiderivative of \(f(x)\).
\[ \int_a^b f(x) \dd{x} = F(b) - F(a) \]
• Properties of indefinite integrals:
\[ \int [f(x) + g(x)] \dd{x} = \int f(x) \dd{x} + \int g(x) \dd{x} \] \[ \int [f(x) - g(x)] \dd{x} = \int f(x) \dd{x} - \int g(x) \dd{x} \] \[ \int k \cdot f(x) \dd{x} = k \cdot \int f(x) \dd{x} \quad (\text{where } k \text{ is a constant}) \]
• Reverse power rule:
\[ \int x^n \dd{x} = \frac{x^{n+1}}{n+1} + C \quad (n \ne -1) \]
• Indefinite integrals of common functions:
\[ \int \sin(x) \dd{x} = -\cos(x) + C \] \[ \int \cos(x) \dd{x} = \sin(x) + C \] \[ \int e^x \dd{x} = e^x + C \] \[ \int \frac{1}{x} \dd{x} = \int \frac{\dd{x}}{x} = \ln(|x|) + C \]
• \(u\)-substitution is a way to find indefinite integrals of more complicated functions. It is a way to reverse the chain rule for derivatives.
  • To use \(u\)-substitution, you identify an expression and its derivative within the function you are integrating. Define \(u\) as the inner expression, then differentiate it to find \(\dv{u}{x}\). Multiply by \(\dd{x}\) to find an expression for \(\dd{u}\), then substitute \(\dd{u}\) into your integral. Finally, integrate with respect to \(u\), then replace \(u\) with its definition in terms of \(x\).
• Polynomial long division and completing the square can help us find some indefinite integrals.
• Important inverse trig integrals:
\[ \int \frac{1}{\sqrt{1-x^2}} \dd{x} = \arcsin(x) + C \] \[ \int \frac{1}{1+x^2} \dd{x} = \arctan(x) + C \]
• (Calc BC only) Integration by parts is an integration technique that is related to the product rule for derivatives.
\[ \int f(x)g'(x) \dd{x} = f(x)g(x) - \int f'(x)g(x) \dd{x} \]
• (Calc BC only) Partial fraction decomposition is when you take a fraction and express it as a sum of two simpler fractions. This can be used to integrate some functions more easily.
• (Calc BC only) Improper integrals are a special type of integral where either one of the bounds are infinite or the expression inside the integral is undefined at one of the bounds. These integrals can be evaluated by rewriting them as limits.

Khan Academy Link (Calc AB): Differential equations

Khan Academy Link (Calc BC): Differential equations

All topics covered in Khan Academy:
Green underlined topics are topics at least partially covered on my website and red topics are topics not yet covered on my website. Note that even green topics might not be covered in full detail on my page.

1. Modeling situations with differential equations
2. Verifying solutions for differential equations
3. Sketching slope fields
4. Reasoning using slope fields
5. (Calc BC only) Approximating solutions using Euler’s method
6. Finding general solutions using separation of variables
7. Finding particular solutions using initial conditions and separation of variables
8. Exponential models with differential equations
9. (Calc BC only) Logistic models with differential equations

You’ve seen that derivatives and integrals can be used to model the real world, where derivatives describe how something changes and integrals describe how much something has changed by in a specific time frame.

But we have a third tool to describe situations where things are changing: differential equations. A differential equation is an equation involving variables and their derivatives with respect to each other.

Here’s an example of a situation where differential equations come in handy: compound interest. The way interest works is that the more money you have, the more interest you’re earning on it at any given time. So your balance is changing at a rate proportional to the amount of money you have.

If we say that \(B\) is your bank account balance, \(r\) is the interest rate, and \(t\) is the amount of time elapsed, we can model this situation with a differential equation:

(Note: this equation only works if the interest is continuously compounded.)

In words, this means that the balance is changing at a rate that is equal to the interest rate (\(r\)) times the balance (\(B\)). Remember, \(\dv{B}{t}\) just means how fast \(B\) changes with respect to time (\(t\)).

Here are some more examples of differential equations:

Unlike the solutions to a traditional algebraic equation (which are values), the solutions to a differential equation are functions. For example, a solution to the differential equation \(\dv{y}{x} = 2xy\) is \(y = e^{x^2}\). (Note: when you see an “exponent tower” like \(e^{x^2}\), it is evaluated from right to left. So \(e^{x^2} = e^{(x^2)}\).) We can verify this solution by differentiating the function:

Then, we plug these values into the differential equation to make sure the equality is always true:

The last line is always true no matter what value of \(x\) we put in, so \(y = e^{x^2}\) is indeed a solution to this differential equation. If the last line wasn’t true for all values of \(x\), then the function wouldn’t be a solution.

Differential equations can have multiple solutions: in fact, this differential equation has infinitely many! Any function of the form \(y = e^{x^2+C}\) where \(C\) is a constant is a solution to this differental equation.

One important skill is being able to take a real-world situation and model it with a differential equation. For example, let’s say that an object is moving at a speed that is inversely proportional to how far it has traveled, \(x\).

First, we need to identify what is changing. In this case, the object is moving, meaning that the total distance it has traveled is constantly changing. The rate at which the distance is changing is its speed, or the derivative of the distance with respect to time (\(\dv{x}{t}\)).

We know that the speed is inversely proportional to the distance. This means that the speed is proportional to the reciprocal of the distance, or \(\frac{1}{x}\). However, because all we know is that the speed is proportional to \(\frac{1}{x}\), we need to multiply the speed by \(k\), a constant of proportionality. This \(k\) could be any number since we don’t know the exact factor.

So our speed is equal to \(\frac{k}{x}\) and it’s also equal to \(\dv{x}{t}\) by definition. Setting these equal to each other, we get our differential equation:

One way to visualize a differential equation is to use a slope field. Here’s how it works.

Let’s say we have a differential equation like \(\dv{y}{x} = y\). To generate a slope field, we draw the slope (\(\dv{y}{x}\)) of the differential equation at every point on the coordinate grid.

For example, at the point \((1, 1)\), our slope \(\dv{y}{x}\) is 1 (since \(y = 1\)), so we would draw a line segment with a slope of 1. At \((2, -3)\), \(\dv{y}{x} = -3\), so we would draw a line with a slope of -3.

If we do this for many points on the coordinate plane, we get a slope field that looks like this:

The slopes of the short lines show the slope of this differential equation at different points. They show the value of \(\dv{y}{x}\) at different values of \(x\) and \(y\).

Here’s another example. This time, our differential equation is \(\dv{y}{x} = x - y\). To get the slope at a particular point \((x, y)\), we simply subtract the \(y\)-value from the \(x\)-value. Here are a few examples of the slope at specific points:

We can use slope fields to visualize solutions to differential equations. For example, let’s go back to our first slope field for \(\dv{y}{x} = y\). What is a solution to that differential equation that goes through the point \((0, 1)\)?

What is a solution to the differential equation \(\dv{y}{x} = y\) that goes through \(\class{red}{(0, 1)}\)?

Our slope field tells us that at \((0, 1)\), the slope of our solution is 1, and as our function goes higher and higher, the slope continues to increase. If we follow the lines of the slope field, we can sketch a solution that looks like this:

This is a solution to the differential equation \(\dv{y}{x} = y\). Notice how the slope at any point on the solution is given by the slope field. (The actual function for this solution is \(y = e^x\).)

We can find more solutions using this slope field. For example, here’s the solution that goes through \((0, -1)\):

The actual function for this solution is \(y = -e^x\).

Slope fields might not always be able to give us the exact solutions to a differential equation, but they’re a useful tool for approximating them. (I know the exact solutions because I used a technique to solve differential equations known as “separation of variables”. You’ll learn about this soon!)

Interestingly, the differential equation \(\dv{y}{x} = y\) can be written as \(f'(x) = f(x)\). This equation is essentially asking, “what functions are their own derivatives?” \(y = e^x\) and \(y = -e^x\), two of the solutions, are indeed examples of functions equal to their own derivatives!

Sometimes, slope fields can guide us towards the general solution of a differential equation. For example, here’s the slope field for \(\dv{y}{x} = -\frac{x}{y}\):

If you try to use this slope field to draw a solution, it will always be a circle (and the circle can be any radius). This means the general solution to this differential equation is \(x^2 + y^2 = C\), since this equation corresponds to a circle of any radius.

This makes sense because when we use implicit differentiation to differentiate the equation for a circle, we get \(\dv{y}{x} = -\frac{x}{y}\).

This content is covered in AP^® Calculus BC but not in AP^® Calculus AB.

Slope fields are one way we can estimate a solution to a differential equation without having to solve it. There is another more precise way to numerically estimate a solution, and it’s called Euler’s method.

Let’s take a look at the differential equation \(\dv{y}{x} = y\) once again.

Problem: How can we use Euler’s method to estimate a solution of the differential equation \(\displaystyle\dv{y}{x} = y\)?

In order to come up with an estimation of a solution using Euler’s method, we need to start with an initial condition (i.e. a point that lies on the solution curve). Let’s say that our initial condition is \((0, 1)\), so that when \(x = 0\), \(y = 1\).

You will need to click the “Previous Step” and “Next Step” buttons below to move through the steps of Euler’s method.

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To use Euler’s method, we need to create a table of points along with the derivative \(\dv{y}{x}\) at each point. Our first point is \((0, 1)\) and \(\dv{y}{x} = 1\) at that point (because \(\dv{y}{x} = y\) and the point has a \(y\)-value of 1).

\(x\)	\(y\)	\(\dv{y}{x}\)
0	1	1

This is a visual representation of the first point.

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Even though \(\dv{y}{x}\) constantly changes as we change \(x\) and \(y\), to make our lives easier, we will assume that the derivative \(\dv{y}{x} = 1\) at the point \((0, 1)\) stays constant for a while as we increase \(x\). We’ll call the length of that “while” \(\Delta x\). \(\Delta x\) determines the accuracy of our estimation; a lower \(\Delta x\) means a more accurate estimate of our solution.

Let’s say that our \(\Delta x\) is 1. This means that we will assume that \(\dv{y}{x}\) stays constant all the way from \(x = 0\) (which is the \(x\)-value of our point \((0, 1))\) to \(x = 1\), a width of \(\Delta x = 1\) unit.

If our derivative or slope stays constant at 1, this means that if we increase \(x\) by 1, then \(y\) will also increase by 1. This allows us to find a second point simply by taking our first point \((0, 1)\) and increasing both the \(x\)- and \(y\)-values by 1.

Doing this gives us a new point \((1, 2)\). Note that because \(\dv{y}{x} = y\) for any point, the derivative at this point is different from the first point (it is 2 instead of 1). Let’s add this point to our table.

\(x\)	\(y\)	\(\dv{y}{x}\)
0	1	1
1	2	2

The graph with the new point added. The blue line has a slope of 1.

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We can do the same thing for our second point \((1, 2)\) to get a third point. The slope at \((1, 2)\) is 2 and we will assume that the slope stays the same all the way up until \(x = 2\).

Because the slope is 2, when we increase \(x\) by 1 again, \(y\) will increase by 2. Adding 1 to the \(x\)-value of our point and adding 2 to the \(y\)-value of our point, we get another point \((2, 4)\).

Notice how to get the next point, we need to add \(\Delta x\) to the \(x\)-coordinate and \(\dv{y}{x} \cdot \Delta x\) to the \(y\)-coordinate. In this case, we are increasing the \(x\)-coordinate by \(\Delta x = 1\) and increasing the \(y\)-coordinate by \(\dv{y}{x} \cdot \Delta x = 2\).

We find that \(\dv{y}{x} = 4\) at the new point \((2, 4)\). Let’s add this new point to our table.

\(x\)	\(y\)	\(\dv{y}{x}\)
0	1	1
1	2	2
2	4	4

The graph with the new point added. The green line has a slope of 2.