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Differentiation

2.3 Product and Quotient Rules and Higher-Order Derivatives

Focus on product rule, quotient rule, higher-order derivatives.

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Ideas in this lesson

01Product RuleWorked steps · five questions02Quotient RuleWorked steps · five questions03Higher-order derivativesWorked steps · five questions
Idea 1 of 3

Product Rule

Differentiate one factor at a time: (uv)′=u′v+uv′(uv)' = u'v + uv'. Multiplying the two derivatives is incorrect.

Worked example 1

Find the derivative at x=1x = 1 of f(x)=x(x+2)f(x) = x(x + 2).
  1. u=x,u′=x0u = x, u' = x^{0}; v=x+2,v′=1v = x + 2, v' = 1
  2. f′(x)=x0(x+2)+xf'(x) = x^{0}(x + 2) + x
  3. f′(1)=1(1)0(1+2)+1=4f'(1) = 1(1)^{0}(1 + 2) + 1 = 4

Answer: 44

Worked example 2

Find the derivative at x=2x = 2 of f(x)=x2(x+3)f(x) = x^{2}(x + 3).
  1. u=x2,u′=2xu = x^{2}, u' = 2x; v=x+3,v′=1v = x + 3, v' = 1
  2. f′(x)=2x(x+3)+x2f'(x) = 2x(x + 3) + x^{2}
  3. f′(2)=2(2)(2+3)+22=24f'(2) = 2(2)(2 + 3) + 2^{2} = 24

Answer: 2424

Idea 2 of 3

Quotient Rule

Use u′v−uv′v2\frac{u'v - uv'}{v^{2}}. Keep the subtraction order and exclude denominator zeros.

Worked example 1

Find f′(1)f'(1) for f(x)=xx+2f(x) = \dfrac{x}{x + 2}.
  1. u=x,v=x+2u = x, v = x + 2
  2. f′(x)=x0(x+2)−x(x+2)2f'(x) = \dfrac{x^{0}(x + 2) - x}{(x + 2)^{2}}
  3. f′(1)=29≈0.22f'(1) = \dfrac{2}{9} \approx 0.22

Answer: 29≈0.22\dfrac{2}{9} \approx 0.22

Worked example 2

Find f′(2)f'(2) for f(x)=x2x+3f(x) = \dfrac{x^{2}}{x + 3}.
  1. u=x2,v=x+3u = x^{2}, v = x + 3
  2. f′(x)=2x(x+3)−x2(x+3)2f'(x) = \dfrac{2x(x + 3) - x^{2}}{(x + 3)^{2}}
  3. f′(2)=1625≈0.64f'(2) = \dfrac{16}{25} \approx 0.64

Answer: 1625≈0.64\dfrac{16}{25} \approx 0.64

Idea 3 of 3

Higher-order derivatives

Differentiate the derivative again for f′′f''. Interpret higher derivatives in the relevant physical or graphical context.

Worked example 1

For f(x)=2x2−xf(x) = 2x^{2} - x, find f′(1)f'(1).
  1. f′(x)=4x−1f'(x) = 4x - 1
  2. f′(1)=3f'(1) = 3

Answer: 33

Worked example 2

For f(x)=3x3−xf(x) = 3x^{3} - x, find f′(2)f'(2).
  1. f′(x)=9x2−1f'(x) = 9x^{2} - 1
  2. f′(2)=35f'(2) = 35

Answer: 3535

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §2.3,p2.3, p. 118118. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.