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Logarithmic, Exponential, and Other Transcendental Functions

5.1 The Natural Logarithmic Function: Differentiation

Focus on logarithm properties and derivatives, derivative of a logarithm, logarithmic differentiation, logarithm with a composite input.

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Ideas in this lesson

01Logarithm properties and derivativesWorked steps · five questions02Derivative of a logarithmWorked steps · five questions03Logarithmic differentiationWorked steps · five questions04Logarithm with a composite inputWorked steps · five questions
Idea 1 of 4

Logarithm properties and derivatives

For positive inputs, ln⁡(xn)=nln⁡x\ln (x^{n}) = n \ln x. Simplify the logarithm before differentiating.

Worked example 1

For x>0x > 0, simplify ln⁡(x2)+ln⁡(e2)\ln (x^{2}) + \ln (e^{2}). Find its derivative at x=1x = 1.
  1. ln⁡(x2)+ln⁡(e2)=2ln⁡x+2\ln (x^{2}) + \ln (e^{2}) = 2 \ln x + 2
  2. f′(x)=2xf'(x) = \dfrac{2}{x}
  3. f′(1)=2f'(1) = 2

Answer: 22

Worked example 2

For x>0x > 0, simplify ln⁡(x3)+ln⁡(e3)\ln (x^{3}) + \ln (e^{3}). Find its derivative at x=1x = 1.
  1. ln⁡(x3)+ln⁡(e3)=3ln⁡x+3\ln (x^{3}) + \ln (e^{3}) = 3 \ln x + 3
  2. f′(x)=3xf'(x) = \dfrac{3}{x}
  3. f′(1)=3f'(1) = 3

Answer: 33

Idea 2 of 4

Derivative of a logarithm

For positive uu, differentiate ln⁡u\ln u as u′u\frac{u'}{u}. For ln⁡∣u∣\ln |u|, the same rule holds where uu is nonzero.

Worked example 1

Find f′(1)f'(1) for f(x)=ln⁡xf(x) = \ln x.
  1. u=xu = x; u′=1u' = 1
  2. f′(x)=1xf'(x) = \dfrac{1}{x}
  3. f′(1)=1f'(1) = 1

Answer: 11

Worked example 2

Find f′(2)f'(2) for f(x)=ln⁡(3x+1)f(x) = \ln (3x + 1).
  1. u=3x+1u = 3x + 1; u′=3u' = 3
  2. f′(x)=33x+1f'(x) = \dfrac{3}{3x + 1}
  3. f′(2)=37≈0.43f'(2) = \dfrac{3}{7} \approx 0.43

Answer: 37≈0.43\dfrac{3}{7} \approx 0.43

Idea 3 of 4

Logarithmic differentiation

Take logs to simplify products, quotients, and variable powers. Differentiate, then multiply by the original yy.

Worked example 1

Use logarithmic differentiation to find y′y' at x=2x = 2 for y=xxy = x^{x}.
  1. ln⁡y=xln⁡x\ln y = x \ln x
  2. y′y=ln⁡x+1\dfrac{y'}{y} = \ln x + 1
  3. y′=xx(ln⁡x+1)y' = x^{x}(\ln x + 1)
  4. y′(2)=4(ln⁡2+1)≈6.77y'(2) = 4(\ln 2 + 1) \approx 6.77

Answer: 4(ln⁡2+1)≈6.774(\ln 2 + 1) \approx 6.77

Worked example 2

Use logarithmic differentiation to find y′y' at x=3x = 3 for y=(x2+1)3y = (x^{2} + 1)^{3}.
  1. ln⁡y=3ln⁡(x2+1)\ln y = 3 \ln (x^{2} + 1)
  2. y′y=6xx2+1\dfrac{y'}{y} = \dfrac{6x}{x^{2} + 1}
  3. y′=6x(x2+1)2y' = 6x(x^{2} + 1)^{2}
  4. y′(3)=1800y'(3) = 1800

Answer: 18001800

Idea 4 of 4

Logarithm with a composite input

A logarithm does not cancel its inner derivative. Apply the chain rule before evaluating.

Worked example 1

Find f′(1)f'(1) for f(x)=ln⁡xf(x) = \ln x.
  1. u=xu = x; u′=1u' = 1
  2. f′(x)=1xf'(x) = \dfrac{1}{x}
  3. f′(1)=1f'(1) = 1

Answer: 11

Worked example 2

Find f′(2)f'(2) for f(x)=ln⁡(3x+1)f(x) = \ln (3x + 1).
  1. u=3x+1u = 3x + 1; u′=3u' = 3
  2. f′(x)=33x+1f'(x) = \dfrac{3}{3x + 1}
  3. f′(2)=37≈0.43f'(2) = \dfrac{3}{7} \approx 0.43

Answer: 37≈0.43\dfrac{3}{7} \approx 0.43

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §5.1,p5.1, p. 318318. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.