Hyperbolic calculus
Objective. Differentiate and integrate hyperbolic functions.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
Hyperbolic functions differentiate and integrate much like trigonometric ones, but without the alternating signs: d/dx(sinh x) = cosh x, and d/dx(cosh x) = sinh x, both without a negative sign, unlike d/dx(cos x) = −sin x. Correspondingly, ∫sinh x dx = cosh x + c and ∫cosh x dx = sinh x + c. These rules extend using the chain rule for composite functions, such as d/dx(cosh(3x)) = 3 sinh(3x). The absence of sign changes is a direct consequence of the exponential definitions, where differentiating eˣ and e⁻ˣ never introduces unexpected signs beyond the chain rule.
Worked example
Find the derivative of f(x) = sinh(2x) and evaluate ∫₀^1 cosh(2x) dx.
- 1By the chain rule, f'(x) = 2 cosh(2x)
- 2For the integral, ∫cosh(2x)dx = (1/2)sinh(2x) + c
- 3Evaluate between 0 and 1: (1/2)sinh(2) − (1/2)sinh(0)
- 4Since sinh(0)=0, the definite integral equals (1/2)sinh(2)
Answer: f'(x) = 2cosh(2x); integral = (1/2)sinh(2)
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Start practisingCommon mistake. Students differentiate cosh x as −sinh x, copying the trigonometric rule for cos x. Remember hyperbolic differentiation introduces no extra minus sign: d/dx(cosh x) = +sinh x.
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