AP Calculus – The Chain Rule and How to Use It

The Chain Rule is a handy tool for dealing with composite functions, making it easier to differentiate more complex functions. For instance, some functions can be differentiated straightforwardly without the Chain Rule, while others are best tackled using it.

If \(y=f\left(u\right)\) is a differentiable function of \(u\) and \(u=g\left(x\right)\) is a differentiable function of \(x\), then \(y=f\left(g\left(x\right)\right)\) is a differentiable function of \(x\) and

\(\displaystyle\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}\)

or

\(\displaystyle\frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]=f’\left(g\left(x\right)\right)\cdot g’\left(x\right)\)

The Chain Rule

Example 1

Find \(\displaystyle F’\left(x\right)\) if \(F\left(x\right)=\sqrt{x^2+1}\).

Solution

Let \(u=x^2+1\) and \(y=\sqrt{u}\):

\(F’\left(x\right)=\displaystyle\frac{dy}{du}\cdot\frac{du}{dx}=\frac{1}{2\sqrt{u}}\cdot 2x= \frac{1}{2\sqrt{x^2+1}}\cdot 2x=\frac{x}{\sqrt{x^2+1}}\)

Example 2

Differentiate \(y=\sin\left(x^2\right)\)

Solution

Let \(u=x^2\) and \(y=\sin u\):

\(\displaystyle\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\cos u\cdot 2x= \cos\left(x^2\right)\cdot 2x=2x\cos\left(x^2\right)\)

Example 3

Differentiate \(y=\sin^2x\)

Solution

Let \(u=\sin x\) and \(y=u^2\):

\(\begin{array}{rcl}\displaystyle\frac{dy}{dx}&=&\displaystyle\frac{dy}{du}\cdot\frac{du}{dx} \\ \displaystyle &=& 2u\cdot \cos x \\ \displaystyle &=&2\sin x\cos x \\ &=&\sin 2x \end{array}\)

Example 4

Differentiate \(y=\left(2x^5+2\right)^{50}\)

Solution

Let \(u=2x^5+2\) and \(y=u^{50}\):

\(\begin{array}{rcl}\displaystyle\frac{dy}{dx}&=&\displaystyle\frac{dy}{du}\cdot\frac{du}{dx} \\ &=& \displaystyle 50u^{49}\cdot 2\cdot 5x^4 \\ &=& \displaystyle 50\left(2x^5+2\right)^{49}\cdot 10x^4 \\ &=& \displaystyle 500x^4\left(2x^5+2\right)^{49}\end{array}\)

Example 5

Differentiate \(y=\left(2x^5+2\right)^{50}\)

Solution

Let \(u=2x^5+2\) and \(y=u^{50}\):

\(\begin{array}{rcl}\displaystyle\frac{dy}{dx}&=&\displaystyle\frac{dy}{du}\cdot\frac{du}{dx} \\ &=& \displaystyle 50u^{49}\cdot 2\cdot 5x^4 \\ &=& \displaystyle 50\left(2x^5+2\right)^{49}\cdot 10x^4 \\ &=& \displaystyle 500x^4\left(2x^5+2\right)^{49}\end{array}\)

Example 6

Find the derivative of the function \(f\left(x\right)=\displaystyle\left(\frac{x-1}{3x+2}\right)^8\)

Solution

Let \(u=\displaystyle\frac{x-1}{3x+2}\) and \(y=u^{8}\):

\(\begin{array}{rcl} \displaystyle\frac{dy}{dx}&=&\displaystyle\frac{dy}{du}\cdot\frac{du}{dx} \\ &=& 8u^7\cdot \displaystyle\frac{\left(x-1\right)’\cdot\left(3x+2\right)-\left(x-1\right)\cdot\left(3x+2\right)’}{\left(3x+2\right)^2} \\ &=& 8u^7\cdot\displaystyle\frac{3x+2-3\left(x-1\right)}{\left(3x+2\right)^2} \\ &=& 8u^7\cdot\displaystyle\frac{5}{\left(3x+2\right)^2} \\ &=& 40\displaystyle\cdot \left(\frac{x-1}{3x+2}\right)^7\cdot \frac{1}{\left(3x+2\right)^2} \\ &=& \displaystyle\frac{40\left(x-1\right)^7}{\left(3x+2\right)^9} \end{array}\)

Example 7

Find the derivative of the function \(f\left(x\right)=\displaystyle e^{\sin x}\).

Solution

Let \(u=\sin x\) and \(y=e^u\):

\(\begin{array}{rcl}\displaystyle\frac{dy}{dx}&=&\displaystyle\frac{dy}{du}\cdot\frac{du}{dx} \\ &=& e^u\cdot \cos x \\ &=& e^{\sin x}\cdot \cos x \end{array}\)

Example 8

Find the derivative of the function \(f\left(x\right)=\displaystyle \sin\left(\cos\left(\tan x\right)\right)\).

Solution

\(\begin{array}{rcl}\displaystyle\frac{dy}{dx}&=& \cos\left(\cos\left(\tan x\right)\right)\cdot \displaystyle\frac{d}{dx}\left(\tan x\right) \\ &=& \cos\left(\cos\left(\tan x\right)\right)\left(-\sin\left(\tan x\right)\right)\cdot \displaystyle\frac{d}{dx}\left(\tan x\right) \\ &=& -\cos\left(\cos\left(\tan x\right)\right)\sin\left(\tan x\right)\sec^2x \end{array}\)

Example 9

Differentiate \(y=\displaystyle e^{\sec 3\theta}\).

Solution

\(\begin{array}{rcl}\displaystyle\frac{dy}{d\theta} &=& e^{\sec 3\theta}\cdot \displaystyle\frac{d}{d\theta}\left(\sec 3\theta\right) \\ &=& e^{\sec 3\theta}\cdot \sec 3\theta\tan 3\theta\cdot\displaystyle\frac{d}{d\theta}\left(3\theta\right) \\ &=& 3e^{\sec 3\theta}\sec 3\theta\tan 3\theta \end{array}\)

More Practice Questions