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Logarithmic, Exponential, and Other Transcendental Functions

5.4 Exponential Functions: Differentiation and Integration

Focus on exponential derivatives, exponential antiderivatives, continuous growth.

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Ideas in this lesson

01Exponential derivativesWorked steps · five questions02Exponential antiderivativesWorked steps · five questions03Continuous growthWorked steps · five questions
Idea 1 of 3

Exponential derivatives

The exponential eue^{u} reproduces itself when differentiated, multiplied by u′u'.

Worked example 1

Find f′(0)f'(0) for f(x)=e2xf(x) = e^{2x}.
  1. f′(x)=2e2xf'(x) = 2e^{2x}
  2. f′(0)=2f'(0) = 2

Answer: 22

Worked example 2

Find f′(0)f'(0) for f(x)=xexf(x) = xe^{x}.
  1. f′(x)=ex+xexf'(x) = e^{x} + xe^{x}
  2. f′(0)=1f'(0) = 1

Answer: 11

Idea 2 of 3

Exponential antiderivatives

Match the derivative of the exponent through substitution. A linear exponent contributes a reciprocal scale factor.

Worked example 1

Evaluate ∫01e2x dx\int_{0}^{1} e^{2x}\,dx.
  1. u=2xu = 2x; du=2dxdu = 2 dx
  2. F(x)=e2x2F(x) = \dfrac{e^{2x}}{2}
  3. F(1)−F(0)=e2−12≈3.19F(1) - F(0) = \dfrac{e^{2} - 1}{2} \approx 3.19

Answer: e2−12≈3.19\dfrac{e^{2} - 1}{2} \approx 3.19

Worked example 2

Evaluate ∫012xex2 dx\int_{0}^{1} 2xe^{x^{2}}\,dx.
  1. u=x2u = x^{2}; du=2xdxdu = 2x dx
  2. F(x)=ex2F(x) = e^{x^{2}}
  3. F(1)−F(0)=e−1≈1.72F(1) - F(0) = e - 1 \approx 1.72

Answer: e−1≈1.72e - 1 \approx 1.72

Idea 3 of 3

Continuous growth

In Q=Q0ektQ = Q_{0}e^{kt}, the constant kk is the relative growth rate, and Q′=Q' = kQ.

Worked example 1

A quantity follows Q(t)=2e0.1tQ(t) = 2e^{0.1t}. Find Q(2)Q(2).
  1. Q(t)=2e0.1tQ(t) = 2e^{0.1t}; Q′(t)=0.1Q(t)Q'(t) = 0.1Q(t)
  2. Q(2)=2e0.2=2e0.2≈2.44Q(2) = 2e^{0.2} = 2e^{0.2} \approx 2.44

Answer: 2e0.2≈2.442e^{0.2} \approx 2.44

Worked example 2

A quantity follows Q(t)=3e0.2tQ(t) = 3e^{0.2t}. Find its doubling time.
  1. Q(t)=3e0.2tQ(t) = 3e^{0.2t}; Q′(t)=0.2Q(t)Q'(t) = 0.2Q(t)
  2. Doubling: e0.2t=2  ⟹  t=ln⁡20.2=ln⁡20.2≈3.47e^{0.2t} = 2 \implies t = \dfrac{\ln 2}{0.2} = \dfrac{\ln 2}{0.2} \approx 3.47

Answer: ln⁡20.2≈3.47\dfrac{\ln 2}{0.2} \approx 3.47

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §5.4,p5.4, p. 346346. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.