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.
- Recognise the catenary as cosh at work in a physical curve.
WHAT YOU PROBABLY THINK
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, and the minus sign the opening claim expects never materialises, because the minus in sinh's definition supplies it internally.
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 same swap read backwards integrates the family: ∫cosh x dx = sinh x + c and ∫sinh x dx = cosh x + c, both signs positive.
The inverse integrals
The derivatives of the inverse functions mirror the inverse trig results with one sign changed inside the root:
and ∫1/√(x² − a²) dx = arcosh(x/a) + c. Where a² − x² pointed to arcsin, a 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: 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. The log form turns a hyperbolic answer into an exact one worth full marks.
TRY IT UNSEEN
The arcosh cousin
Evaluate ∫ 1/√(x² − 4) dx from 2 to 4, exactly.
Show the working
a = 2: the integral is arcosh(x/2) from 2 to 4.
arcosh 2 − arcosh 1 = ln(2 + √3) − ln(1 + 0).
= ln(2 + √3), since arcosh 1 = ln 1 = 0. About 1.317 as a check.
THE EXAM BIT
- 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.
- Convert arsinh and arcosh answers to log form when the question says exact.
- For ∫cosh²x dx, use cosh 2x = 2cosh²x − 1, the hyperbolic double angle with no sign flip.
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
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sinh' = cosh and cosh' = sinh: the pair swap with no minus sign, unlike sin and cos.
√(x² + a²) integrals go to arsinh, √(x² − a²) to arcosh; convert to logs for exact answers.
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.
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.
- Recognise the catenary as cosh at work in a physical curve.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.