Hyperbolic Functions: sinh, cosh and tanh

Hyperbolic functions are built from eˣ and e⁻ˣ: $\sinh x = \dfrac{e^x - e^{-x}}{2}$ and $\cosh x = \dfrac{e^x + e^{-x}}{2}$ , with tanh x = sinh x / cosh x. They satisfy cosh²x − sinh²x = 1. Their derivatives are (sinh x)′ = cosh x and (cosh x)′ = sinh x, with no minus sign.

Before this lesson: Derivatives of Exponential Functions, U-Substitution and The Number e

What hyperbolic functions are

Take the average of eˣ and e⁻ˣ and you get cosh x (said “cosh”). Take half their difference and you get sinh x (said “sinch”). These two show up in hanging cables, special relativity and many differential equations.

The names come from geometry. The point (cos t, sin t) lies on the circle x² + y² = 1. The point (cosh t, sinh t) lies on the hyperbola x² − y² = 1. That is why the identities look like trig identities with some signs changed.

Graphs for x from −2.5 to 2.5. The curve y = cosh x is U-shaped with its lowest point at (0, 1). The curve y = sinh x passes through the origin and rises from lower left to upper right. The curve y = tanh x passes through the origin and levels off toward the dashed lines y = 1 and y = −1. — cosh x is even with minimum 1 at x = 0; sinh x and tanh x are odd, and tanh x stays between −1 and 1.

Hyperbolic function definitions and rules

\sinh x = \dfrac{e^x - e^{-x}}{2}, \qquad \cosh x = \dfrac{e^x + e^{-x}}{2}, \qquad \tanh x = \dfrac{\sinh x}{\cosh x}

Rule	Formula
Pythagorean identity	$\cosh^2 x - \sinh^2 x = 1$
Derivative of sinh	$\dfrac{d}{dx}\sinh x = \cosh x$
Derivative of cosh	$\dfrac{d}{dx}\cosh x = \sinh x$
Derivative of tanh	$\dfrac{d}{dx}\tanh x = \operatorname{sech}^2 x$
Integral of sinh	$\int \sinh x\,dx = \cosh x + C$
Integral of cosh	$\int \cosh x\,dx = \sinh x + C$
Integral of tanh	$\int \tanh x\,dx = \ln(\cosh x) + C$

Here sech x = 1/cosh x. The values to remember are sinh 0 = 0 and cosh 0 = 1.

Hyperbolic function examples

Example 1: Exact values

Find sinh(ln 2) and cosh(ln 2). Check that they satisfy the identity.

e^{\ln 2} = 2, \; e^{-\ln 2} = \dfrac{1}{2}

e and ln undo each other (see logarithms).

\sinh(\ln 2) = \dfrac{2 - \tfrac{1}{2}}{2} = \dfrac{3}{4}

Use the definition of sinh.

\cosh(\ln 2) = \dfrac{2 + \tfrac{1}{2}}{2} = \dfrac{5}{4}

Use the definition of cosh.

\left(\dfrac{5}{4}\right)^2 - \left(\dfrac{3}{4}\right)^2 = \dfrac{25}{16} - \dfrac{9}{16} = 1

Check cosh² − sinh² = 1.

Answer: $\sinh(\ln 2) = \dfrac{3}{4}$ and $\cosh(\ln 2) = \dfrac{5}{4}$ .

Example 2: Proving the identity

Show that cosh²x − sinh²x = 1 for every x.

\cosh^2 x = \dfrac{e^{2x} + 2 + e^{-2x}}{4}

Square (eˣ + e⁻ˣ)/2; eˣ · e⁻ˣ = 1 (see exponent rules).

\sinh^2 x = \dfrac{e^{2x} - 2 + e^{-2x}}{4}

Square (eˣ − e⁻ˣ)/2.

\cosh^2 x - \sinh^2 x = \dfrac{2 - (-2)}{4}

Subtract; the e²ˣ and e⁻²ˣ terms cancel.

= 1

4/4 = 1.

Answer: the identity holds for all x. Unlike sin² + cos² = 1, here it is a difference.

Example 3: Derivatives

Find the derivative of $y = x\sinh x - \cosh x$ and of $y = \cosh(3x^2)$ .

\dfrac{d}{dx}(x\sinh x) = \sinh x + x\cosh x

Product rule (see product rule).

\dfrac{d}{dx}(x\sinh x - \cosh x) = \sinh x + x\cosh x - \sinh x

The derivative of cosh x is sinh x.

= x\cosh x

The sinh terms cancel.

\dfrac{d}{dx}\cosh(3x^2) = \sinh(3x^2)\cdot 6x

Chain rule (see chain rule).

Answer: the first derivative is $x\cosh x$ , and the second is $6x\sinh(3x^2)$ .

Example 4: Integrals

Find $\int x\cosh(x^2)\,dx$ and $\int \tanh x\,dx$ .

u = x^2, \; du = 2x\,dx

For the first integral, let u be the inside.

\int x\cosh(x^2)\,dx = \int \dfrac{1}{2}\cosh u\,du = \dfrac{1}{2}\sinh u + C

Rewrite and integrate.

= \dfrac{1}{2}\sinh(x^2) + C

Substitute back.

\int \tanh x\,dx = \int \dfrac{\sinh x}{\cosh x}\,dx

For the second, write tanh as a quotient.

= \ln(\cosh x) + C

The top is the derivative of the bottom, and cosh x is always positive (see integrals that give natural logarithms).

Answer: $\dfrac{1}{2}\sinh(x^2) + C$ and $\ln(\cosh x) + C$ .

Example 5: A hanging cable

A power line hangs between two poles 60 feet apart. Its shape is $y = 20\cosh\dfrac{x}{20}$ for x from −30 to 30, in feet. How much does it sag below the tops of the poles? How steep is it where it meets the right pole?

y(0) = 20\cosh 0 = 20

The lowest point is in the middle; cosh 0 = 1.

y(30) = 20\cosh 1.5 \approx 47.05

Height at the pole; cosh 1.5 ≈ 2.3524.

47.05 - 20 = 27.05

Sag = pole height minus lowest height.

y'(x) = 20\cdot\dfrac{1}{20}\sinh\dfrac{x}{20} = \sinh\dfrac{x}{20}

Chain rule.

y'(30) = \sinh 1.5 \approx 2.13

Slope at the right pole.

Answer: the cable sags about 27.1 feet, and its slope at the right pole is about 2.13. A hanging cable’s curve, y = a cosh(x/a), is called a catenary.

Common mistakes

Mistake	Example of the error	Fix
Adding a minus sign like trig	$\dfrac{d}{dx}\cosh x = -\sinh x$	There is no minus: $\dfrac{d}{dx}\cosh x = \sinh x$ .
Using the trig identity	$\cosh^2 x + \sinh^2 x = 1$	It’s a difference: $\cosh^2 x - \sinh^2 x = 1$ .
Wrong value at 0	sinh 0 = 1	sinh 0 = (1 − 1)/2 = 0, and cosh 0 = (1 + 1)/2 = 1.
Wrong sign on the integral	$\int \sinh x\,dx = -\cosh x + C$	$\int \sinh x\,dx = \cosh x + C$ .
Reading sinh⁻¹ as a reciprocal	$\sinh^{-1} x = \dfrac{1}{\sinh x}$	sinh⁻¹ is the inverse function. The reciprocal is csch x.

Related methods

Inverse hyperbolic functions

Solving y = sinh x for x gives $\sinh^{-1} x = \ln\left(x + \sqrt{x^2 + 1}\right)$ . Similarly, $\cosh^{-1} x = \ln\left(x + \sqrt{x^2 - 1}\right)$ for x ≥ 1. These are the logs you get from some trigonometric substitution integrals.

Hyperbolic substitution

Because 1 + sinh²u = cosh²u, the substitution x = a sinh u also removes √(a² + x²), just like x = a tan θ.

The other three functions

sech x = 1/cosh x, csch x = 1/sinh x and coth x = cosh x/sinh x, matching the six trig functions. Their derivatives follow from the quotient rule.

Practice problems

Find sinh 0 and cosh 0.
Find cosh(ln 3).
Show that sinh(−x) = −sinh x.
Find the derivative of sinh(5x).
Find the derivative of cosh²x.
Find the derivative of ln(cosh x).
Find $\int \cosh(2x)\,dx$ .
Find $\int \sinh x\cosh x\,dx$ .
Evaluate $\int_0^{\ln 2} \cosh x\,dx$ .
An arch has the shape $y = 40 - 10\cosh\dfrac{x}{10}$ , in meters. How wide is it at ground level?

Show answers

\sinh 0 = \dfrac{e^0 - e^0}{2} = 0

e⁰ = 1.

\cosh 0 = \dfrac{e^0 + e^0}{2} = 1

(1 + 1)/2.

Answer: sinh 0 = 0 and cosh 0 = 1

\cosh(\ln 3) = \dfrac{3 + \tfrac{1}{3}}{2}

e^(ln 3) = 3 and e^(−ln 3) = 1/3.

= \dfrac{10/3}{2} = \dfrac{5}{3}

Add, then divide by 2.

Answer: $\dfrac{5}{3}$

\sinh(-x) = \dfrac{e^{-x} - e^{x}}{2}

Replace x with −x in the definition.

= -\dfrac{e^x - e^{-x}}{2} = -\sinh x

Factor out −1.

Answer: sinh is an odd function (see even and odd functions).

\dfrac{d}{dx}\sinh(5x) = \cosh(5x)\cdot 5

Chain rule.

Answer: $5\cosh(5x)$

\dfrac{d}{dx}\cosh^2 x = 2\cosh x\cdot\sinh x

Chain rule: outside power, then the derivative of cosh x.

Answer: $2\cosh x\sinh x$ , which equals sinh 2x

\dfrac{d}{dx}\ln(\cosh x) = \dfrac{\sinh x}{\cosh x} = \tanh x

Chain rule with ln.

Answer: $\tanh x$

\int \cosh(2x)\,dx = \dfrac{1}{2}\sinh(2x) + C

Let u = 2x, so dx = du/2.

Answer: $\dfrac{1}{2}\sinh(2x) + C$

\int \sinh x\cosh x\,dx = \int u\,du = \dfrac{u^2}{2} + C

Let u = sinh x, du = cosh x dx.

= \dfrac{\sinh^2 x}{2} + C

Substitute back.

Answer: $\dfrac{\sinh^2 x}{2} + C$

\int_0^{\ln 2} \cosh x\,dx = \Big[\sinh x\Big]_0^{\ln 2}

Integrate.

= \sinh(\ln 2) - \sinh 0 = \dfrac{3}{4} - 0 = \dfrac{3}{4}

Use Example 1.

Answer: $\dfrac{3}{4}$

10.

40 - 10\cosh\dfrac{x}{10} = 0

Ground level is y = 0.

\cosh\dfrac{x}{10} = 4

Add 10 cosh(x/10), then divide by 10.

\dfrac{x}{10} = \ln\left(4 + \sqrt{15}\right) \approx 2.063

Use cosh⁻¹ 4 = ln(4 + √(16 − 1)) for the positive solution.

x \approx 20.63

Multiply by 10.

2(20.63) = 41.26

The arch is symmetric, so double it.

Answer: about 41.3 meters wide

Hyperbolic functions FAQ

What are sinh and cosh?

sinh x = (eˣ − e⁻ˣ)/2 and cosh x = (eˣ + e⁻ˣ)/2. They are called hyperbolic sine and hyperbolic cosine.

What is the derivative of cosh x?

The derivative of cosh x is sinh x, and the derivative of sinh x is cosh x. Neither has a minus sign.

Why are they called hyperbolic functions?

The point (cosh t, sinh t) lies on the hyperbola x² − y² = 1, just as (cos t, sin t) lies on the circle x² + y² = 1.

Where are hyperbolic functions used?

A hanging chain or cable takes the shape y = a cosh(x/a), called a catenary. Hyperbolic functions also appear in physics and in solutions of differential equations.

Summary

sinh x = (eˣ − e⁻ˣ)/2 and cosh x = (eˣ + e⁻ˣ)/2; tanh x = sinh x/cosh x.
cosh²x − sinh²x = 1.
(sinh x)′ = cosh x and (cosh x)′ = sinh x, with no minus sign.
∫sinh x dx = cosh x + C and ∫cosh x dx = sinh x + C.
sinh 0 = 0 and cosh 0 = 1; cosh is even and sinh is odd.

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