Introduction

Welcome back to AP Calculus AB to the first topic of Unit 3! We will continue to learn more derivative rules and expand on what we’ve learned. A really important differentiation rule you will learn here is called the “Chain Rule”, and it is used to differentiate composite functions. The importance of the Chain Rule cannot be overstated, as it is key to understanding many applications of derivatives and calculus in general. Care should be taken to fully pay attention to the article and gain a comprehensive and thorough understanding of the Chain Rule to ensure your AP Calculus success! As such, this article will contain more content than others to ensure and boost a thorough understanding. Enjoy!

Composite Functions

Before we continue, what is a composite function? Well, the word composite means “made up of various parts or elements”, so a composite function has something to do with being made up of different parts.

Well my friend, it is indeed this! Remember back when you would input numbers into functions like $f(x)$ or $f’(x)$? Well, a composite function is when you input functions into functions.

$f(g(x))$

Notice that we have a very obvious $g(x)$ in the middle, but where is the $f(x)$? Well, the $f(x)$ is actually the outside function! Take a moment and realize that we have simply replaced $x$ with $g(x)$ because $g(x)$ is the input.

$f(x)\to f(g(x))$

There may be some occasions where we have this different notation below, but it means the same exact thing.

$(f\circ g)(x)$

To pronounce this, you say “$f$ of $g$ of $x$”, or “effuf geuf fex”.

Take note that in both cases, $f(x)$ is considered the outer function, and $g(x)$ is considered the inner function. Looking at $f(g(x))$ makes this make sense.

Composite Function Example

Identify the input of f(x) in f(x^2+5).

If we take a quick look at this, notice that $x^2+5$ is the input of the function! It is that simple!

Identify the inner and outer function of (2x+1)^2.

This may be a bit more confusing, but notice how we have what seems to be a whole function being squared, like its inside another function… This means that the input is very likely to be $2x+1$. How do we find the outer function though? Well, if we replace the original function with $g(x)=2x+1$,

$(g(x){)}^2$

The input ($g(x)$) is squared, so this means that $f(x)$ is $x^2$. 

Identify the inner and outer function of cos(2x)+5

This problem seems a bit interesting, but I think we got this. Notice how you can see that $\cos (x)$ has $2x$ inside, so our inner function has got to be $2x$! With the outer function, I think it is easy to see that it is $\cos (x)$. That’s our answer! Time to submit it! 

Well… uhh… we got it wrong. Why? Well, the ENTIRE function is multiplied by $2$ right? And it’s on the outside, right? Also, we just completely ignored the $+5$. This means that our outer function is actually $\cos (x)+5$. 

We can check our answer by plugging back in the inner and outer function into $f(g(x))$. $g(x)$ is the inner function, which we can agree is $2x$, so we have $f(2x)$. Since $f(x)=\cos (x)+5$, $f(2x)=\cos (2x)+5$. Hey! That’s the exact function we got! How nice!

Chain Rule

Now that we are familiar with composite functions, you are ready to learn about the chain rule!

$\frac{d}{dx}f(g(x))=f^{\prime }(g(x))\cdot g^{\prime }(x)$

Notice how this function works similarly to rules we’ve seen before. This simply tells us to identify these functions and find the required info! We will need to identify $f(x)$ and $g(x)$ as well as their derivatives. Hmm, that seems to be the same exact info we needed for the product and quotient rule!

It can also be expressed in a different form, but we aren’t going to analyze this today. This formula isn’t useful to us now, but in the future it will!

$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$

Realizing Composite Functions

First though, it’s important to know when to apply the chain rule. For a function such as $\sin (\ln (x))$ it is quite clear to realize this is a composite function and what the inner and outer functions are. $\sin (x)$ is the outer function and $\ln (x)$ is the inner function. Let’s go over a few examples of this.

Can you find the inner and outer functions for the function $\sqrt{7x^2+2x}$ ?

In this situation the inner function is $7x^2+2x$ and the outer function is $\sqrt{x}$.

How about the function ${\sin }^2(x)$?

In this situation ${\sin }^2(x)$ is equal to $(\sin (x){)}^2$ so the inner function is $\sin (x)$ and the outer function is $x^2$.

Lets go over some examples where there are more than two functions that compose a single function. Like the function $y=e^{\cos (5x+2)}$. To differentiate this function you would need to employ the chain rule multiple times, this could be defined as a composite function. Try finding the individual functions which make up this compose function on your own.

The inner most function would be $h(x)=5x+2$, the next inner function could be defined as $g(x)=\cos (x)$, and the outer most function would be $f(x)=e^x$. Then our original function would be defined as $y=f(g(h(x)))$.

For a final example, can you find the functions which compose $y=\sqrt[3]{2\ln (x)+4}$?

Here the inner most function can be defined as $h(x)=\ln (x)$, then the next as $g(x)=2x+4$ and the outer function as $f(x)=\sqrt[3]{x}$ or as $f(x)=x^{\frac{1}{3}}$. And again our original function can be defined as $y=f(g(h(x)))$.

Differentiating Using Chain Rule

Example 1

Differentiate $y=(2x+1{)}^2$

Hold on a second… we already identified $f(x)$ and $g(x)$! All we need to do is just figure out the derivatives.

$f(x)=x^2$

$g(x)=2x+1$

$f^{\prime }(x)=2x$

$g^{\prime }(x)=2$

Alright, let’s just plug it into the formula, and please be extremely careful in plugging the quantities in as there are composited functions involved:

$\frac{d}{dx}f(g(x))=f^{\prime }(g(x))\cdot g^{\prime }(x)=2(2x+1)\cdot 2$

So ultimately, $\frac{dy}{dx}=4(2x+1)$

People may tend to omit the last component or perform the formula incorrectly. Erroneous versions could look like:

$\frac{dy}{dx}=2x(2x+1)\cdot 2$

This is incorrect because what is being done here is $[f^{\prime }(x)\cdot g(x)]\cdot g^{\prime }(x)$. It cannot be further emphasized how incorrect and misleading this error is, and is very important to understand this is CONCEPTUALLY INCORRECT. Again, allow me to repeat that what just occurred here is incorrect and is thus a mistake you should NOT be making. $f^{\prime }(g(x))\ne f^{\prime }(x)\cdot g(x)$. So PLEASE do not make this mistake. The way you should think of it is this $f^{\prime }(g(x))⟹f^{\prime }(2x+1)⟹2(2x+1)$.

Example 2

Find $f’(0)$ when $f(x)={\tan }^2(x)$.

Well, I mean, we have a function… yeah. Notice that $\tan (x)$ is squared, almost like the outer function is $x^2$... but that’s just a theory!

If we write the function like this, it is a lot easier to see what the inner and outer functions are.

$f(x)=[\tan (x){]}^2$

Ah, the outer function really was $x^2$, and the inner function is $\tan (x)$. Now, let’s do the same thing we have always done!

$f(x)=x^2$

$g(x)=\tan (x)$

$f’(x)=2x$

$g’(x)={\sec }^2(x)$

Alright, let’s just plug it into the chain rule again!

$f^{\prime }(g(x))\cdot g^{\prime }(x)=2x(\tan (x))\cdot {\sec }^2(x)$

Anddddd how fantastic! We made the SAME mistake again. It’s a function INSIDE a function, not multiplied! Can we like, stop doing that please? The only multiplication I see is between $f’(g(x))$ and $g’(x)$, but nowhere else. 

Let’s just, ignore $f’(x)$ again.

$f^{\prime }(g(x))\cdot g^{\prime }(x)=f’(\tan (x))\cdot {\sec }^2(x)$

Ah, it looks cleaner. What is $f’(\tan (x))$? Well if $f’(x)=2x$, then $f’(\tan (x))=2\tan (x)$. Replaced the $x$ and now we flex!

$f’(\tan (x))\cdot {\sec }^2(x)=2\tan (x)\cdot {\sec }^2(x)$

Let’s try one more example. WAITTTTTTTTTTTTT!!!! We forgot that we were trying to find $f’(0)$! Absolutely crucial to remember! Maybe the MCQ answers will remind you about it, but the FRQs won’t! Let’s just plug it in and complete this final step.

$f’(0)=2\tan (0)\cdot {\sec }^2(0)=2(0)\cdot (1{)}^2=0$

Nice. 

Example 3

Find the derivative of $f(x)$ when $f(x)=\ln (4^{\cos (x)}+3\cos (x))$

Oh for goodness sake! This function looks like somebody grabbed a bunch of Legos and blindly threw them in hopes that it would create something! Let’s just take a moment to analyze it.

For starters, I think we can all see that the outer function is $\ln (x)$, yeah? The inner function has gotta be $4^{\cos (x)}+3\cos (x)$, yeah? Yeah, that looks correct. Normal process, let’s just do it!

$f(x)=\ln (x)$

$g(x)=4^{\cos (x)}+3\cos (x)$

$f’(x)=\frac{1}{x}$

$g’(x)=$

Wait, hold up, isn’t… $4^{\cos (x)}$ a composite function? A function inside a function? Wait… you’re telling me the input has a function inside a function??? A triple function chain? A mega super composite function? Well, we need to differentiate this composite function to find $g’(x)$, so… let’s do it.

I think we can all agree that the outer function is $4^x$, and the inner function is $\cos (x)$. We don’t need to add the $3\cos (x)$ to the outer function this time around, because uhh… it has a variable and it’ll lead to a completely different answer. Let’s go!

$f(x)=4^x$

$g(x)=\cos (x)$

$f’(x)=4^x(\ln 4)$

$g’(x)=-\sin (x)$

Use the chain rule, and let’s ignore $f’(x)$ again cause I think we’ve learned our lesson from last time.

$f’(g(x))\cdot g’(x)=f’(\cos (x))\cdot -\sin (x)$

If $f’(x)=4^x(\ln 4)$, then $f’(\cos (x))=4^{\cos (x)}(\ln 4)$, so we have

$f’(\cos (x))\cdot -\sin (x)=4^{\cos (x)}(\ln 4)\cdot -\sin (x)$

Now, we can bring the other chain rule function list down! We’ve completed this chain rule!

$f(x)=\ln (x)$

$g(x)=4^{\cos (x)}+3\cos (x)$

$f’(x)=\frac{1}{x}$

$g’(x)=4^{\cos (x)}(\ln 4)\cdot -\sin (x)$

Hold on a second, this is missing something. Oh! It’s missing the $3\cos (x)$ from before! I mean, that is part of the input of the original function. Remember to differentiate this guy to $-3\sin (x)$, and we get:

$f(x)=\ln (x)$

$g(x)=4^{\cos (x)}+3\cos (x)$

$f’(x)=\frac{1}{x}$

$g’(x)=4^{\cos (x)}(\ln 4)\cdot -\sin (x)-3\sin (x)$

Plugging this all into the chain rule finally, we get:

$f’(g(x))\cdot g’(x)=f’(4^{\cos (x)}+3\cos (x))\cdot (4^{\cos (x)}(\ln 4)\cdot -\sin (x)-3\sin (x))$

Finally, if $f’(x)=\frac{1}{x}$, $f’(4^{\cos (x)}+3\cos (x))=\frac{1}{4^{\cos (x)}+3\cos (x)}$. We can finally finish this!

$f’(4^{\cos (x)}+3\cos (x))\cdot (4^{\cos (x)}(\ln 4)\cdot -\sin (x)-3\sin (x))=\frac{1}{4^{\cos (x)}+3\cos (x)}\cdot (4^{\cos (x)}(\ln 4)\cdot -\sin (x)-3\sin (x))$

What an absolutely insane derivative. You won’t be expected to differentiate derivatives of such complexity and tediousness, so rest assured. But hopefully, in following this example, you understand how the Chain Rule process can be extremely layered, in that you can have a function and break it down into pieces for the formula. Yet, said pieces within the formula need to be further Chain Ruled for a complete differentiation. Let this example help clarify and ensure that Chain Rule problems can be very layered in nature.

Example 4:

Calculate the derivative of $x^2\sin (e^x)$ with respect to $x$

First we will do the product rule as this is differentiating two multiplied functions.

$\frac{d}{dx}[x^2\sin (e^x)]=\frac{d}{dx}[x^2\cdot \sin (e^x)]+x^2\frac{d}{dx}[\sin (e^x)]$

Now to differentiate we need to recognize the outer and inner functions.

Outer function is $\sin x$ and inner function is $e^x$

Step 1: Derivative of the outside function with the inside function inputted

$\frac{d}{dx}[\sin x]=\cos x$

$\cos (e^x)$

Step 2: Multiply by the derivative of the inside function

$\frac{d}{dx}[e^x]=e^x$

$\frac{d}{dx}[\sin (e^x)]=e^x\cos (e^x)$

Now going back to the original problem we can input this result. 

$\frac{d}{dx}[x^2\sin (e^x)]=\frac{d}{dx}[(x^2)\cdot \sin (e^x)]+x^2\frac{d}{dx}[\sin (e^x)]$

$\frac{d}{dx}[x^2\sin (e^x)]=2x\cdot \sin (e^x)+x^2\cdot e^x\cos (e^x)=2x\sin (e^x)+x^2{e}^x\cos (e^x)$

If you ever have three or more functions, just keep the chain going.

Notes for the Chain Rule

Practice Questions