Maths › Hyperbolic functions › Calculus with hyperbolic functions
Calculus with hyperbolic functions
sinh and cosh differentiate into each other with no minus sign to remember, and their inverses tame integrals full of x² plus or minus a square.
Builds on Hyperbolic functions and identities and Calculus with inverse trigonometric functions.
IN THIS TOPIC
- Differentiate sinh, cosh and tanh directly from the exponential definitions.
- Integrate 1/√(x² + a²) and 1/√(x² − a²) via arsinh and arcosh.
- Choose a hyperbolic substitution when recognition alone will not do.
COMMON MISCONCEPTION
Just like cos, differentiating cosh must introduce a minus sign.
A derivative pair with no minus
Differentiate the definitions term by term. (ex − e−x)/2 turns into (ex + e−x)/2 and back again, so sinh' = cosh and cosh' = sinh. The pair swap cleanly, with no minus sign anywhere, because the minus already sits inside sinh's definition and supplies itself.
WORKED EXAMPLE
A tangent to sinh
Find the equation of the tangent to y = sinh x at x = 0.
dy/dx = cosh x, and cosh 0 = 1.
sinh 0 = 0, so the tangent is y = x.
Near zero sinh x ≈ x, for exactly the same reason sin x ≈ x. Both have gradient 1 through the origin.
The quotient rule on sinh/cosh then gives tanh' = sech²x, which mirrors tan' = sec²x with no sign change. Read all three backwards and you can integrate the family: ∫cosh x dx = sinh x + c, ∫sinh x dx = cosh x + c, and ∫sech²x dx = tanh x + c.
The inverse integrals
The derivatives of the inverse functions mirror the inverse trig results with one sign changed inside the root:
The booklet's Further Maths integration table carries this one, with the logarithmic form beside it, and it carries ∫1/√(x² − a²) dx = arcosh(x/a) + c for x > a as well. Where a² − x² pointed to arcsin, the sign flip inside the square root points here instead. Reading the quadratic under the root is the whole skill.
WORKED EXAMPLE
An integral that lands on ln 2
Evaluate ∫ 1/√(x² + 16) dx from 0 to 3.
a = 4, so the integral is arsinh(x/4), evaluated from 0 to 3.
arsinh(3/4) = ln(3/4 + √(9/16 + 1)) = ln(3/4 + 5/4).
= ln 2. Converting to log form is what turns a hyperbolic answer into an exact one worth full marks.
INDEPENDENT PRACTICE
The arcosh cousin
Evaluate ∫ 1/√(x² − 4) dx from 2 to 4, exactly.
Show the working
The integrand is unbounded at x = 2, so the integral is improper: write it as the limit as b → 2⁺ of the integral from b to 4.
a = 2 in the standard form, so each piece is arcosh(x/2), evaluated from b to 4, which is arcosh 2 − arcosh(b/2).
As b → 2⁺ this tends to arcosh 2 − arcosh 1 = ln(2 + √3) − ln(1 + 0).
= ln(2 + √3), since arcosh 1 = ln 1 = 0. The limit exists, so the integral converges. About 1.317 as a check.
When recognition runs out
Harder integrands need a substitution, and the identity tells you which one. For anything carrying √(x² + a²) put x = a sinh u, because a² + a²sinh²u = a²cosh²u and the root disappears cleanly. For √(x² − a²) put x = a cosh u instead. Either way dx brings the matching derivative down and the integral collapses into powers of cosh or sinh.
Questions of this kind usually supply the substitution, so the work is in executing it. Change the limits into u values as soon as you substitute, keep the identity you used visible, and convert back to logs at the end if the answer is wanted exactly.
ASSESSMENT FOCUS
- Derive sinh' and cosh' from the definitions when asked to 'show that'. Two lines each.
- Read the sign inside the root first. x² + a² is arsinh, x² − a² is arcosh, a² − x² is arcsin.
- For ∫cosh²x dx, use cosh 2x = 2cosh²x − 1, the hyperbolic double angle with no sign flip.
- Convert arsinh and arcosh answers to log form when the question says exact.
CHECK YOURSELF
Differentiate y = cosh 3x, and state the gradient at x = 0.
Show a hint
Chain rule; cosh' = sinh with no minus.
Show the answer
dy/dx = 3 sinh 3x. At x = 0, sinh 0 = 0, so the gradient is 0. cosh 3x has its minimum there, value 1.
sinh' = cosh and cosh' = sinh, with no minus sign anywhere; tanh' = sech²x.
√(x² + a²) integrals go to arsinh and √(x² − a²) to arcosh.
Substitute x = a sinh u or x = a cosh u when the root will not yield to recognition.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the calculus with hyperbolic functions questions page.
CHECK YOUR PROGRESS
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- Differentiate sinh, cosh and tanh directly from the exponential definitions.
- Integrate 1/√(x² + a²) and 1/√(x² − a²) via arsinh and arcosh.
- Choose a hyperbolic substitution when recognition alone will not do.
Open the full revision checklist to see every objective in the course in one place.