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Logarithmic, Exponential, and Other Transcendental Functions

5.8 Hyperbolic Functions

Focus on hyperbolic derivatives, hyperbolic integration, the hyperbolic tangent.

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Ideas in this lesson

01Hyperbolic derivativesWorked steps · five questions02Hyperbolic integrationWorked steps · five questions03The hyperbolic tangentWorked steps · five questions
Idea 1 of 3

Hyperbolic derivatives

The derivative of sinh⁡\sinh is cosh⁡\cosh ; the derivative of cosh⁡\cosh is sinh⁡\sinh . Unlike cos⁡,cosh⁡\cos , \cosh introduces no negative sign.

Worked example 1

Find f′(0)f'(0) for f(x)=sinh⁡(2x)f(x) = \sinh (2x).
  1. f′(x)=2cosh⁡(2x)f'(x) = 2 \cosh (2x)
  2. cosh⁡0=1\cosh 0 = 1; sinh⁡0=0\sinh 0 = 0; sech 0=10 = 1
  3. f′(0)=2f'(0) = 2

Answer: 22

Worked example 2

Find f′(0)f'(0) for f(x)=cosh⁡(3x)f(x) = \cosh (3x).
  1. f′(x)=3sinh⁡(3x)f'(x) = 3 \sinh (3x)
  2. cosh⁡0=1\cosh 0 = 1; sinh⁡0=0\sinh 0 = 0; sech 0=10 = 1
  3. f′(0)=0f'(0) = 0

Answer: 00

Idea 2 of 3

Hyperbolic integration

Reverse the derivative formulas and account for any inner chain-rule factor.

Worked example 1

Evaluate ∫012cosh⁡x dx\int_{0}^{1} 2\cosh x\,dx.
  1. F(x)=2sinh⁡xF(x) = 2 \sinh x
  2. F(1)−F(0)=2sinh⁡(1)≈2.35F(1) - F(0) = 2 \sinh (1) \approx 2.35

Answer: 2sinh⁡(1)≈2.352 \sinh (1) \approx 2.35

Worked example 2

Evaluate ∫013sinh⁡x dx\int_{0}^{1} 3\sinh x\,dx.
  1. F(x)=3cosh⁡xF(x) = 3 \cosh x
  2. F(1)−F(0)=3(cosh⁡(1)−1)≈1.63F(1) - F(0) = 3(\cosh (1) - 1) \approx 1.63

Answer: 3(cosh⁡(1)−1)≈1.633(\cosh (1) - 1) \approx 1.63

Idea 3 of 3

The hyperbolic tangent

The derivative of tanh⁡\tanh is sech 2^{2}. Its value at zero follows from cosh⁡0=1\cosh 0 = 1.

Worked example 1

Find f′(0)f'(0) for f(x)=sinh⁡(2x)f(x) = \sinh (2x).
  1. f′(x)=2cosh⁡(2x)f'(x) = 2 \cosh (2x)
  2. cosh⁡0=1\cosh 0 = 1; sinh⁡0=0\sinh 0 = 0; sech 0=10 = 1
  3. f′(0)=2f'(0) = 2

Answer: 22

Worked example 2

Find f′(0)f'(0) for f(x)=cosh⁡(3x)f(x) = \cosh (3x).
  1. f′(x)=3sinh⁡(3x)f'(x) = 3 \sinh (3x)
  2. cosh⁡0=1\cosh 0 = 1; sinh⁡0=0\sinh 0 = 0; sech 0=10 = 1
  3. f′(0)=0f'(0) = 0

Answer: 00

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §5.8,p5.8, p. 383383. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.