Unit 3 · Topic 3.1 Beta

The Chain Rule

For a composite function f(g(x)) , differentiate the outside function at the unchanged inside, then multiply by the derivative of the inside: d/dx f(g(x)) = f′(g(x))⋅g′(x) .

Practice 1: Implementing Mathematical Processes

Question set for this topic

Part 1 · Hook

Why this matters

Turn the pedals of a bike and the back wheel spins faster than your feet, because the gears multiply the rate. Many functions work the same way: one function feeds its output into another. The chain rule says the rates multiply, and it is the rule you will use most often for the rest of the course.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the derivative of sin⁡x\sin x?

  1. cos⁡x\cos x
  2. −cos⁡x-\cos x
  3. −sin⁡x-\sin x
  4. sec2⁡x\sec^2 x
Show the answer

The derivative of sin⁡x\sin x is cos⁡x\cos x.

  • Correct: cos⁡x\cos x:
  • −cos⁡x-\cos x:
  • −sin⁡x-\sin x:
  • sec2⁡x\sec^2 x:

2. What is the derivative of x5x^5?

  1. 5x45x^4
  2. x4x^4
  3. 5x55x^5
  4. x66\frac{x^6}{6}
Show the answer

The power rule: bring the 5 down and lower the power by one.

  • Correct: 5x45x^4:
  • x4x^4:
  • 5x55x^5:
  • x66\frac{x^6}{6}:

Part 4 · See it

See it first

Graphs of y = sin x (blue, dashed) and y = sin(2x) (crimson) from x = −π to π. Both pass through the origin; the tangent line to sin x there has slope 1 and the tangent line to sin(2x) has slope 2.
The inside function 2x2x squeezes sin⁡x\sin x sideways by a factor of 2, so every slope doubles: at the origin the slope of sin⁡(2x)\sin(2x) is 22, not 11. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A function is built in layers, h(x)=f(g(x))h(x) = f(g(x))xx feeds the inside function gg, and g(x)g(x) feeds the outside function ff
  2. When xx changes a little, gg changes g′(x)g'(x) times as muchand ff changes f′(g(x))f'(g(x)) times as much as its input gg does
  3. The two rates multiply, like gearsh′(x)=f′(g(x))⋅g′(x)h'(x) = f'(g(x)) \cdot g'(x)
  4. In Leibniz form, with u=g(x)u = g(x)dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

Part 6 · Key ideas

Key ideas

  • Spot the layers first: name the outside function and the inside function before you differentiate.
  • Differentiate the outside function and leave the inside alone: f′(g(x))f'(g(x)), not f′(x)f'(x) and not f′(g′(x))f'(g'(x)).
  • Then multiply by the derivative of the inside, g′(x)g'(x). Forgetting this factor is the most common lost point.
  • From a table, read f′f' at the inside value g(a)g(a), then multiply by g′(a)g'(a).

Part 7 · Misconception

A common mistake

The wrong idea: The derivative of sin⁡(x2)\sin(x^2) is cos⁡(x2)\cos(x^2).

What actually happens: That is only the outside derivative. The inside function x2x^2 changes too, at rate 2x2x, so ddxsin⁡(x2)\frac{d}{dx}\sin(x^2)=cos⁡(x2)⋅2x= \cos(x^2) \cdot 2x.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Values of ff, f′f', gg and g′g'

Selected values of the differentiable functions f and g and their derivatives
xxf(x)f(x)f′(x)f'(x)g(x)g(x)g′(x)g'(x)
13−2-225
2463−1-1
31442
42−3-317

1. Let h(x)=f(g(x))h(x) = f(g(x)). What is h′(1)h'(1)?

  1. 3030
  2. −10-10
  3. 66
  4. 1212
Show the answer

The chain rule: h′(x)=f′(g(x))⋅g′(x)h'(x) = f'(g(x)) \cdot g'(x). The outside derivative is read at the inside value g(1)=2g(1) = 2, so h′(1)h'(1)=f′(2)⋅g′(1)= f'(2) \cdot g'(1)=6⋅5= 6 \cdot 5=30= 30.

  • Correct: 3030: h′(1)h'(1)=f′(g(1))⋅g′(1)= f'(g(1)) \cdot g'(1)=f′(2)⋅5= f'(2) \cdot 5=6⋅5= 6 \cdot 5=30= 30.
  • −10-10: This is f′(1)⋅g′(1)f'(1) \cdot g'(1): the outside derivative was evaluated at x=1x = 1 instead of at the inside value g(1)=2g(1) = 2.
  • 66: This is f′(g(1))f'(g(1)) alone: the inside derivative g′(1)=5g'(1) = 5 was dropped.
  • 1212: This multiplies by g(1)=2g(1) = 2, the inside value, instead of by its derivative g′(1)=5g'(1) = 5.

2. Let k(x)=g(f(x))k(x) = g(f(x)), a composition with ff on the inside. Use the table to find k′(3)k'(3).

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

k′(3)k'(3)=g′(f(3))⋅f′(3)= g'(f(3)) \cdot f'(3)=g′(1)⋅4= g'(1) \cdot 4=5⋅4= 5 \cdot 4=20= 20. Here gg is the outside function, so its derivative is read at f(3)=1f(3) = 1.

  • Answer: 20

3. Let m(x)=[g(x)]3m(x) = [g(x)]^3. What is m′(2)m'(2)?

  1. 2727
  2. −27-27
  3. −9-9
  4. −81-81
Show the answer

The outside function is the cube: m′(x)=3[g(x)]2⋅g′(x)m'(x) = 3[g(x)]^2 \cdot g'(x), so m′(2)=3(3)2(−1)=−27m'(2) = 3(3)^2(-1) = -27.

  • 2727: This is 3[g(2)]23[g(2)]^2 with no inside derivative. The inside function is gg, and g′(2)=−1g'(2) = -1 changes the sign.
  • Correct: −27-27: 3[g(2)]2g′(2)3[g(2)]^2 g'(2)=3⋅9⋅(−1)= 3 \cdot 9 \cdot (-1)=−27= -27.
  • −9-9: This is [g(2)]2g′(2)[g(2)]^2 g'(2): the power rule's factor 3 is missing.
  • −81-81: This is 3[g(2)]3g′(2)3[g(2)]^3 g'(2): the power rule lowers the power by one, so the outside derivative is 3u23u^2, not 3u33u^3.

4. Let p(x)=f(x2)p(x) = f(x^2). What is p′(2)p'(2)?

  1. −3-3
  2. −6-6
  3. −12-12
  4. 2424
Show the answer

p′(x)=f′(x2)⋅2xp'(x) = f'(x^2) \cdot 2x. At x=2x = 2 the inside value is 44: p′(2)p'(2)=f′(4)⋅4= f'(4) \cdot 4=−3⋅4= -3 \cdot 4=−12= -12.

  • −3-3: This is f′(4)f'(4) alone: the derivative of the inside function x2x^2, which is 2x=42x = 4, is missing.
  • −6-6: This multiplies by 2 instead of by 2x=42x = 4. The inside derivative is evaluated at x=2x = 2.
  • Correct: −12-12: f′(4)⋅2(2)=−3⋅4=−12f'(4) \cdot 2(2) = -3 \cdot 4 = -12.
  • 2424: This is f′(2)⋅4f'(2) \cdot 4: the outside derivative was read at x=2x = 2, not at the inside value 22=42^2 = 4.

5. Find f′(x)f'(x) for f(x)=sin⁡(x3)f(x) = \sin(x^3).

Type an expression for f′(x)=f'(x) =.

Show the answer

Outside: sin⁡u\sin u, derivative cos⁡u\cos u, evaluated at the same inside u=x3u = x^3. Inside derivative: 3x23x^2. So f′(x)=cos⁡(x3)⋅3x2f'(x) = \cos(x^3) \cdot 3x^2.

  • Answer: f′(x)=f'(x) = 3x2cos⁡(x3)3x^{2}\,\cos(x^{3})

6. Which of these functions need the chain rule to be differentiated? Select all that apply.

  1. e3xe^{3x}
  2. (x2+1)10(x^2 + 1)^{10}
  3. x2exx^2 e^x
  4. cos⁡xx\frac{\cos x}{x}
Show the answer

The chain rule is needed when a function is applied to another function: e3xe^{3x} and (x2+1)10(x^2 + 1)^{10}.

  • Correct: e3xe^{3x}: The exponent 3x3x is an inside function: the derivative is 3e3x3e^{3x}.
  • Correct: (x2+1)10(x^2 + 1)^{10}: The tenth power acts on the inside function x2+1x^2 + 1.
  • x2exx^2 e^x: This is a product of x2x^2 and exe^x, neither with an inside function: the product rule is enough.
  • cos⁡xx\frac{\cos x}{x}: This is a quotient of two basic functions: the quotient rule is enough.

Part 9 · Summary

Summary

For a composite function f(g(x))f(g(x)), differentiate the outside function at the unchanged inside, then multiply by the derivative of the inside: ddxf(g(x))\frac{d}{dx} f(g(x))=f′(g(x))⋅g′(x)= f'(g(x)) \cdot g'(x). Name the layers before you start, and never drop the inside derivative.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections