The Chain Rule

Derivatives of functions inside functions. Rates multiply.

0 / 10 problems mastered

Chain Rule

\big(f(g(x))\big)'=f'(g(x))\cdot g'(x)

Derivative of the outside (leave the inside alone) × derivative of the inside.

Leibniz form

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

If u changes 3× as fast as x, and y changes 2× as fast as u, y changes 6× as fast as x.

Power + chain

\frac{d}{dx}[g(x)]^n=n[g(x)]^{n-1}g'(x)

Exponential + chain

\frac{d}{dx}e^{g(x)}=e^{g(x)}g'(x)

\frac{d}{dx}a^x=a^x\ln a

A nudge travels down the chain

Lab 6 · y = sin(x²)

x Δx

Δu / Δx ≈ du/dx = 2x–

Δy / Δu ≈ dy/du = cos u–

product of the two–

exact dy/dx = 2x·cos(x²)–

The inner function u = x² stretches the nudge by a factor of 2x. The outer function y = sin u stretches it again by cos u. Total stretch = the product. That is the chain rule.

How to spot inside vs. outside

Ask: if I plugged in a number, what would I compute last? That's the outer function.

$\sqrt{x^3+1}$ : last step is the square root. Outer $\sqrt u$ , inner $u=x^3+1$ .
$\sin^2x=(\sin x)^2$ : last step is squaring. Outer $u^2$ , inner $\sin x$ .
$\sin(x^2)$ : last step is sin. Outer $\sin u$ , inner $x^2$ .
Long chains peel like an onion: $\big(f(g(h(x)))\big)'=f'(g(h(x)))\cdot g'(h(x))\cdot h'(x)$ .

Worked examples

Example · three layers

Differentiate $y=e^{\sec 3\theta}$ .

Layers: $e^{(\ )}$ → $\sec(\ )$ → $3\theta$ .
Peel: $e^{\sec3\theta}\cdot\sec3\theta\tan3\theta\cdot3$ .
$y'=3e^{\sec3\theta}\sec3\theta\tan3\theta$

Example · chain inside a product (§3.4 #21)

Differentiate $F(x)=(4x+5)^3(x^2-2x+5)^4$ .

Product Rule, and each factor needs the Chain Rule:
$F'=3(4x+5)^2(4)\cdot(x^2-2x+5)^4+(4x+5)^3\cdot4(x^2-2x+5)^3(2x-2)$
Factor out the lowest powers: $4(4x+5)^2(x^2-2x+5)^3\big[3(x^2-2x+5)+(4x+5)(2x-2)\big]$
Bracket: $3x^2-6x+15+8x^2+2x-10=11x^2-4x+5$ .
$F'(x)=4(4x+5)^2(x^2-2x+5)^3(11x^2-4x+5)$

Practice

§3.4 #1

Write

y=(5-x^4)^3

f(g(x))

and find

dy/dx

§3.4 #3

Differentiate

y=\sin(\cos x)

§3.4 #13 & #14

Differentiate

f(\theta)=\cos(\theta^2)

and

g(\theta)=\cos^2\theta

. Why are they different?

§3.4 #17

Differentiate

y=x^2e^{-3x}

§3.4 #19

Differentiate

f(t)=e^{at}\sin bt

§3.4 #23

Differentiate

y=\sqrt{\dfrac{x}{x+1}}

§3.4 #29

Differentiate

r(t)=10^{2\sqrt t}

§3.4 #41

Differentiate

y=\sin^2(x^2+1)

§3.4 #58

Find the tangent line to

y=\sqrt{1+x^3}

(2,3)

§3.4 · table style

h(x)=f(g(x))

. Given

g(1)=3

g'(1)=4

f(3)=7

f'(3)=-2

f'(1)=5

. Find

h'(1)

Traps

Forgetting the inside derivative

$(\sin3x)'=3\cos3x$ , not $\cos3x$ . Every layer contributes a factor.

Changing the inside

$(\cos(x^2))'$ is $-\sin(x^2)\cdot2x$ . The inside stays $x^2$ ; don't write $-\sin(2x)$ .

f′(g(x)) vs f′(x)

In table problems, evaluate the outer derivative at the inner value.