The Chain Rule

We define a composite (or compound) function of $x$ as a function that is "composed" of two functions; one function depends on the other function that depends on $x$ , namely $y= f(g(x))$ . The notation you may have seen before is $y=(f \circ g)(x)=f(g(x))$ . Suppose that we have variables $x$ , $y$ , and $u$ , and that $y = f(u)$ and $u = g(x)$ . (Previously we called $u=g(x)$ the inside part of $f(g(x))$ .) Then $f$ is a composite function of $x$ .

DO: Find $f$ and $g$ and write as $f(g(x))$ in the following examples in order to familiarize yourself with the notation. Determine the inside part and outside part of each.

Examples

$y=\sin(x^2)$ is a composite function with $u=x^2$ and $y=\sin(u)$ .

$y=\sin^2(x)$ is a composite function with $u=\sin(x)$ and $y=u^2$ .

$y=e^{3x}$ is a composite function with $u=3x$ and $y=e^u$ .

$y=(x^2 + 4x + 7)^5$ is a composite function with $u=x^2 + 4x + 7$ and $y=u^5$ .

Version 2 of the chain rule says that

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

Note that $\displaystyle \frac{dy}{dx}$ is the same as $\displaystyle{\frac{d}{dx}\Big(f\big(g(x)\big)\Big)}$ , that $\displaystyle{\frac{dy}{du}} = f'(u)=f'(g(x))$ , and that $\displaystyle{\frac{du}{dx}}$ is the same thing as $g'(x)$ . So Version 2 says the exact same thing as Version 1.

Let's see how it applies to our examples above.

DO: After looking at the first example, try to do others before looking at the solutions. You will learn much more by trying and doing than by reading. While you read, write down what you are thinking and what is happening.

Example 1: If

$y=\sin(x^2)$ , then

$u=x^2$ so that

$y=\sin(u)$ .

Then

$\dfrac{du}{dx} =2x$ , and

$\dfrac{dy}{du}=\cos(u)$ . All together, we have

$\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}=\cos(u)\cdot2x = \cos(x^2)\cdot 2x.$

Example 2: If

$y = \sin^2(x)$ , then

$u=\sin(x)$ so that

$y=u^2$ .

Then

$\dfrac{du}{dx}=\cos(x)$ , and

$\dfrac{dy}{du}=2u$ . All together, we have

$\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}=2u\cdot \cos(x) = 2\sin(x)\cos(x).$

Example 3: If

$y=e^{3x}$ , then

$u=3x$ so that

$y=e^u$ .

Then

$\dfrac{du}{dx}=3$ , and

$\dfrac{dy}{du}=e^u$ . All together, we have

$\frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx} =e^u\cdot3 = 3 e^{3x}.$

Example 4: If

$y=(x^2 + 4x + 7)^5$ , then

$u=x^2+4x + 7$ so that

$y=u^5$ .

Then

$\dfrac{du}{dx}=2x+4$ , and

$\dfrac{dy}{du} = 5u^4$ . All together we have

$\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = 5u^4(2x+4)=5(x^2+4x+7)^4(2x+4).$

DO: Try all these examples using version 1,

$(f(g(x)))'=f'(g(x))g'(x)$ , just for practice.

The Six Pillars of Calculus

Trigonometry Review

Exponential Functions

Logarithms and Inverse functions

Intro to Limits

Limit Laws and Computations

Continuity and the Intermediate Value Theorem

Limits at Infinity

Rates of Change

The Derivative Function

Basic Differentiation Rules

Product and Quotient Rules

Derivatives of Trig Functions

The Chain Rule

Implicit Differentiation

Derivatives of Logs

Derivatives in Science

Related Rates

Linear Approximation and Differentials

Differentiation Review

Absolute and Local Extrema

The Mean Value and other Theorems

ff vs. f′f'

Indeterminate Forms and L'Hospital's Rule

Optimization

Newton's Method

Anti-derivatives

The Area under a curve

Definite Integrals

The Fundamental Theorem of Calculus

Chain Rule

Composition of Functions

$f$ vs. $f'$