Hyperbolic Functions

Two combinations of $e^{x}$ and $e^{-x}$ show up so often that they have names: the hyperbolic sine and cosine. A hanging chain takes the shape of $y = a\cosh\frac{x}{a}$, and these functions also turn up in special relativity and in solutions of differential equations. Everything about them follows from what you know about derivatives of exponentials.

$$\sinh x = \frac{e^{x} - e^{-x}}{2},$$ $$\cosh x = \frac{e^{x} + e^{-x}}{2},$$ $$\tanh x = \frac{\sinh x}{\cosh x}.$$ $\cosh$ is even and always at least 1; $\sinh$ is odd. Every function splits into an even part and an odd part, and for $e^{x}$ those parts are exactly $\cosh x$ and $\sinh x$: $e^{x} = \cosh x + \sinh x$.

$\cosh x$ is the average of the two dashed curves, so it is U-shaped and never below 1. $\sinh x$ is half their difference: it passes through 0 and is odd. For large $x$ both hug $\frac{e^{x}}{2}$.

Why it works

Square and subtract: $$\begin{aligned} &\cosh^{2} x - \sinh^{2} x \\ &= \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} \\ &= \frac{4}{4} \\ &= 1. \end{aligned}$$ So the point $(\cosh t, \sinh t)$ always lies on the hyperbola $X^{2} - Y^{2} = 1$, just as $(\cos t, \sin t)$ lies on the circle $X^{2} + Y^{2} = 1$. That is the reason for the name. Differentiating the definitions term by term: $$\frac{d}{dx}\sinh x = \frac{e^{x} + e^{-x}}{2} = \cosh x,$$ $$\frac{d}{dx}\cosh x = \frac{e^{x} - e^{-x}}{2} = \sinh x.$$

Key idea

$\cosh^{2} x - \sinh^{2} x = 1$; $\;(\sinh x)' = \cosh x$; $\;(\cosh x)' = \sinh x$ (no minus sign); $\;(\tanh x)' = \operatorname{sech}^{2} x = \frac{1}{\cosh^{2} x}$. Integrals reverse these: $\int\cosh x\,dx = \sinh x + C$ and $\int\sinh x\,dx = \cosh x + C$.

	Trig	Hyperbolic
Basic identity	$\cos^{2} x + \sin^{2} x = 1$	$\cosh^{2} x - \sinh^{2} x = 1$
Derivative of cos / cosh	$-\sin x$	$+\sinh x$
Derivative of sin / sinh	$\cos x$	$\cosh x$
Curve traced	circle	right branch of a hyperbola

Worked example: An exact value

Find $\sinh(\ln 3)$.

$e^{\ln 3} = 3$ and $e^{-\ln 3} = \frac{1}{3}$.
$\sinh(\ln 3) = \frac{1}{2}\left(3 - \frac{1}{3}\right) = \frac{1}{2}\cdot\frac{8}{3} = \frac{4}{3}$.

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

Worked example: Using the identity

If $\sinh x = \frac{3}{4}$, find $\cosh x$ and $\tanh x$.

$\cosh^{2} x = 1 + \sinh^{2} x = 1 + \frac{9}{16} = \frac{25}{16}$.
$\cosh x$ is always positive, so $\cosh x = \frac{5}{4}$.
$\tanh x = \frac{3/4}{5/4} = \frac{3}{5}$.

Answer: $\cosh x = \frac{5}{4}$, $\tanh x = \frac{3}{5}$

Worked example: Calculus with them

Find $\frac{d}{dx}\cosh(3x)$ and $\int \sinh x\cosh^{2} x\,dx$.

Chain rule: $\frac{d}{dx}\cosh(3x) = 3\sinh(3x)$.
For the integral let $u = \cosh x$, so $du = \sinh x\,dx$: $\int u^{2}\,du = \frac{u^{3}}{3} + C = \frac{\cosh^{3} x}{3} + C$.

Answer: $3\sinh(3x)$ and $\frac{\cosh^{3} x}{3} + C$

Common mistake

Borrowing the trig identity unchanged: $\cosh^{2} x + \sinh^{2} x = 1$. Testing at $x = 0$ will not catch it (both versions give 1 there). Test at $x = \ln 2$ instead: $\cosh = \frac{5}{4}$ and $\sinh = \frac{3}{4}$, so $\frac{25}{16} - \frac{9}{16} = 1$ but $\frac{25}{16} + \frac{9}{16} \ne 1$. The same goes for derivatives: $\cosh$ differentiates to $+\sinh$.

Hyperbolic functions make the chain curve friendly for arc length: for $y = \cosh x$, $1 + (y')^{2} = 1 + \sinh^{2} x = \cosh^{2} x$, so the length from 0 to $b$ is $\int_0^b \cosh x\,dx = \sinh b$.

Hyperbolic Functions

Practice

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Sources and further reading