Introduction

Welcome to the first topic of Unit 3 in FiveHive's AP Calculus BC course. The primary focus of this unit is to expand on your knowledge of derivatives from Unit 2. One of the most important rules for differentiation that you have yet to learn is the chain rule. It is used to differentiate composite functions but you will see its use in contexts that are less obvious later on.

Composite Functions

Just as a review by composite functions, I mean a function within another function. Something along the lines of this:

$f(g(x))$

Or this:

$(f\circ g)(x)$

In both of these situations $f$ is the outer function and $g$ is the inner function. So this is the function: $f$ of $g$ of $x$.

Differentiating Composite Functions

The method we will use for differentiating composite functions is the chain rule. The chain rule states that the derivative of a composite function is equal to the derivative of the outer function with respect to the inner function, multiplied by the derivative of the inner function. In mathematical notion this looks like:

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

It can also be expressed as

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

Both are valid ways of expressing the chain rule but you may find the second one more useful in future applications.

Let's go through an example of differentiating a function using the chain rule. The function we will differentiate is:

$y=(2x+1{)}^2$

First we need to break it down into an inner and an outer function. In this situation the inner function which we will call $g(x)$ is $g(x)=2x+1$, while the outer function which we will call $f(x)$ is $f(x)=x^2$. Since we have defined an $f(x)$ and a $g(x)$ we can now express this composite function as: $f(g(x))$

Now to differentiate it we can apply the aforementioned formula which is:

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

Since the inner and outer functions have been identified the derivatives, $f’(x)$ and $g’(x)$ can be found to be $f’(x)=2x$ and $g’(x)=2$ by applying our basic rules for differentiation.

Then we can plug the derivates into the formula mentioned above:

$2[g(x)]\cdot 2$

Then plug in $2x+1$ for $g(x)$ and solve for our answer:

$2[2x+1]\cdot 2$

$4[2x+1]$

$\frac{d}{dx}[(2x+1{)}^2]=8x+4$

That’s the basic process for differentiating a composite function. The function I gave as an example could have been differentiated much easier by just multiplying out the polynomial and then applying the power rule. I wanted to go over an easy example first though. I will go over a more difficult function in a minute.

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 $y=f(g(h(x)))$.

A More Difficult Example

The composite function we will differentiate this time is:

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

This situation is a bit more difficult, I encourage you to attempt to differentiate this function on your own before going over the example. Notice that this is a composite function, it is much easier to see when you rewrite it as:

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

First we need to define the inner and outer functions. This time I will define the inner function $\tan (x)$ as $u=\tan (x)$, and the outer function $x^2$ as $v=x^2$. This time we will use the expression: $f’(x)=\frac{dv}{du}\cdot \frac{du}{dx}$

Note that this is the same as: $f’(x)=v’(u(x))\cdot u’(x)$

And can be solved the same way I solved the previous problem. I am just choosing to solve it this way to show a different method. First I will plug in for $v$ and $u$: $\frac{d(\tan (x){)}^2}{d(\tan (x))}\cdot \frac{d(\tan (x))}{dx}$

For the first term here we take the derivative of $(\tan (x){)}^2$ with respect to $\tan (x)$. So we treat the whole expression $\tan (x)$ as if we would treat $x$ when normally differentiating. So we get: $\frac{d(\tan (x){)}^2}{d(\tan (x))}=2\tan (x)$

Then we are left with:

$2\tan (x)\cdot \frac{d}{dx}(\tan (x))$

So to simplify the second term we are just taking the derivative of $\tan (x)$ with respect to $x$. Which you may remember from Unit 2 is ${\sec }^2(x)$. So we end up with: $2\tan (x)\cdot {\sec }^2(x)$

And our final answer for the derivative is: $\frac{d}{dx}[{\tan }^2(x)]=2\tan (x){\sec }^2(x)$

That example was a bit more complicated but once you understand the chain rule it becomes quite simple. Let's quickly go over a final challenging example of differentiating composite functions. I again would encourage you to attempt to differentiate this function before I go over a quick explanation of how to do it (Note that this example will require you to use the chain rule multiple times):

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

This function can be seen as composed of 3 functions. The first and the innermost function is $h(x)=\cos (x)$, the next function is $g(x)=4^x+3x$ and the outermost function is $f(x)=\ln (x)$. And can therefore be composed as $y=f(g(h(x)))$ and when the differential is applied can be written as $\frac{dy}{dx}=f’(g(h(x)))\cdot g’(h(x))\cdot h’(x)$. This is the formula you get when applying the chain rule to a doubly composed function. I will now quickly go over the steps of solving it.

$\frac{dy}{dx}=\frac{1}{g(h(x))}\cdot (4^{h(x)}\cdot \ln (4)+3)\cdot -\sin (x)$

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

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

$\frac{dy}{dx}=-\frac{\sin (x)(4^{\cos (x)}\ln (4)+3)}{4^{\cos (x)}+3\cos (x)}$

As you can see I just plugged in the values for the functions and their derivatives and then simplified it down into the final form. That example is a bit lengthier than something you might usually see but as long as you understand the basic idea of the chain rule you should be able to differentiate something like that (even if it takes a while).

Future topics in this unit will expand upon this rule but for now lets go over some notes and then practice problems.

Even More Examples!

Example 1:

Calculate the derivative of $\ln (2x)$ with respect to $x$. 

Before we knew the chain rule we would split $\ln (2x)=\ln (2)+\ln (x)$ and then differentiate term by term. Now this is important to recognize, but we can also use the chain rule.

Step 1: Derivative of the outside function and then input the inside function. 

The outside function is $\ln x$

Recall $\frac{d}{dx}\ln x=\frac{1}{x}$

The inside function is $2x$

So the first step is $\frac{1}{2x}$

But wait! There’s more!

We know from properties of logarithms and differentiation that $\frac{d}{dx}\ln (2x)=\frac{1}{x}$

This is the chain rule, never forget the chain!

Step 2: Multiply by the derivative of the inside.

The inside function’s derivative is $2$.

$\frac{1}{2x}\cdot 2=\frac{1}{x}$

So we calculated the derivative of $\ln (2x)$ with respect to x using the chain rule.

Let’s do one where we can’t use logarithmic properties though.

Example 1b:

Calculate the derivative of $\ln (2x+1)$ with respect to $x$. 

Let’s identify the outside and inside functions. Once you do enough problems, this will be done in your head. For now this process helps with carrying out the rule since you just learned it.

Outside function: $\ln (x)$

Inside function: $2x+1$

Step 1: Derivative of the outside function and then input the inside function.

$\frac{d}{dx}\ln x=\frac{1}{x}$

We have $\frac{1}{2x+1}$

Step 2: Multiply by the derivative of the inside.

The inside function’s derivative is 2.

$\frac{1}{2x+1}\cdot 2=\frac{2}{2x+1}$

$\frac{d}{dx}\ln (2x+1)=\frac{2}{2x+1}$

Like everything else, this rule doesn’t exist in a vacuum. 

Example 2:

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.

Example 3:

Calculate the derivative of ${\sin }^3(4t+2)$ with respect to $t$.

Identify the composition:

Outer function: $t^3$

Middle function: $\sin t$

Inner function: $4t+2$

Now differentiate and insert the inner functions

Step 1: 

$\frac{d}{dt}{t}^3=3t^2$

Insert the rest of the composition (middle and inner)

$3(\sin (4t+2){)}^2$

Step 2:

$\frac{d}{dt}\sin t=\cos t$

$\cos (4t+2)$

Step 3:

$\frac{d}{dt}4t+2=4$

Now multiply (or chain)  it together:

$\frac{d}{dt}{\sin }^3(4t+2)=3(\sin (4t+2){)}^2\cdot \cos (4t+2)\cdot 4=12(\sin (4t+2){)}^2\cos (4t+2)$

Future topics in this unit will expand upon this rule but for now lets go over some notes and then practice problems.

Notes for the Chain Rule

Practice Questions