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Differentiation

2.4 The Chain Rule

Focus on chain rule, radicals and reciprocal powers, trigonometric compositions, exponential compositions.

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Ideas in this lesson

01Chain RuleWorked steps · five questions02Radicals and reciprocal powersWorked steps · five questions03Trigonometric compositionsWorked steps · five questions04Exponential compositionsWorked steps · five questions
Idea 1 of 4

Chain Rule

Differentiate the outer function at the inner expression, then multiply by the inner derivative.

Worked example 1

Find f′(1)f'(1) for f(x)=(2x+2)2f(x) = (2x + 2)^{2}.
  1. u=2x+2,u′=2u = 2x + 2, u' = 2
  2. f′(x)=2(2x+2)⋅2=4(2x+2)f'(x) = 2(2x + 2) \cdot 2 = 4(2x + 2)
  3. f′(1)=4(4)=16f'(1) = 4(4) = 16

Answer: 1616

Worked example 2

Find f′(2)f'(2) for f(x)=(2x+3)3f(x) = (2x + 3)^{3}.
  1. u=2x+3,u′=2u = 2x + 3, u' = 2
  2. f′(x)=3(2x+3)2⋅2=6(2x+3)2f'(x) = 3(2x + 3)^{2} \cdot 2 = 6(2x + 3)^{2}
  3. f′(2)=6(7)2=294f'(2) = 6(7)^{2} = 294

Answer: 294294

Idea 2 of 4

Radicals and reciprocal powers

Rewrite a radical or reciprocal using a power before differentiating, while preserving domain restrictions.

Worked example 1

Find f′(4)f'(4) for f(x)=xf(x) = \sqrt{x}.
  1. f(x)=x12f(x) = x^{\dfrac{1}{2}}
  2. f′(x)=12 xf'(x) = \dfrac{1}{2\,\sqrt{x}}
  3. f′(4)=14≈0.25f'(4) = \dfrac{1}{4} \approx 0.25

Answer: 14≈0.25\dfrac{1}{4} \approx 0.25

Worked example 2

Find f′(1)f'(1) for f(x)=3x+1f(x) = \sqrt{3x + 1}.
  1. f(x)=(3x+1)12f(x) = (3x + 1)^{\dfrac{1}{2}}
  2. f′(x)=32 3x+1f'(x) = \dfrac{3}{2\,\sqrt{3x + 1}}
  3. f′(1)=32 4≈0.75f'(1) = \dfrac{3}{2\,\sqrt{4}} \approx 0.75

Answer: 32 4≈0.75\dfrac{3}{2\,\sqrt{4}} \approx 0.75

Idea 3 of 4

Trigonometric compositions

For f(x)=sin⁡u(x),f′(x)=cos⁡u(x)⋅u′(x)f(x) = \sin u(x), f'(x) = \cos u(x) \cdot u'(x).

Worked example 1

Let f(x)=2sin⁡(x)f(x) = 2 \sin (x). Find f′(0)f'(0).
  1. u=xu = x; u′=1u' = 1
  2. f′(x)=2cos⁡(x)⋅1f'(x) = 2 \cos (x) \cdot 1
  3. f′(0)=2cos⁡0=2f'(0) = 2 \cos 0 = 2

Answer: 22

Worked example 2

Let f(x)=3sin⁡(2x)f(x) = 3 \sin (2x). Find f′(0)f'(0).
  1. u=2xu = 2x; u′=2u' = 2
  2. f′(x)=3cos⁡(2x)⋅2f'(x) = 3 \cos (2x) \cdot 2
  3. f′(0)=6cos⁡0=6f'(0) = 6 \cos 0 = 6

Answer: 66

Idea 4 of 4

Exponential compositions

For f(x)=eu(x),f′(x)=eu(x)⋅u′(x)f(x) = e^{u}(x), f'(x) = e^{u}(x) \cdot u'(x). Keep every inner factor.

Worked example 1

Let f(x)=e2x2f(x) = e^{2x^{2}}. Find f′(1)f'(1).
  1. u=2x2u = 2x^{2}; u′=4xu' = 4x
  2. f′(x)=e2x2⋅4xf'(x) = e^{2x^{2}} \cdot 4x
  3. f′(1)=4e2f'(1) = 4e^{2}

Answer: 4e2≈29.564e^{2} \approx 29.56

Worked example 2

Let f(x)=e3x3f(x) = e^{3x^{3}}. Find f′(1)f'(1).
  1. u=3x3u = 3x^{3}; u′=9x2u' = 9x^{2}
  2. f′(x)=e3x3⋅9x2f'(x) = e^{3x^{3}} \cdot 9x^{2}
  3. f′(1)=9e3f'(1) = 9e^{3}

Answer: 9e3≈180.779e^{3} \approx 180.77

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §2.4,p2.4, p. 129129. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.