y = \operatorname{arcosh} x

Inverse Hyperbolic Cosine $y = \operatorname{arcosh} x$

$\operatorname{arcosh} x$ , the inverse hyperbolic cosine or area hyperbolic cosine, is the inverse of $\cosh x = \dfrac{e^x + e^{-x}}{2}$ ¹. Since $\cosh$ is even and takes the same value at a point and at its negative, it has no inverse as it stands, so the inverse of the branch with $x \geq 0$ , the principal branch, is what is called $\operatorname{arcosh}$ .

Definition and closed form

$y = \operatorname{arcosh} x$ is the value $y$ with $x = \cosh y$ and $y \geq 0$ . Written with a logarithm it has the following closed form.

\operatorname{arcosh} x = \ln\left(x + \sqrt{x^2 - 1}\right)

For the quantity under the root to be non-negative we need $x \geq 1$ , and for $x < 1$ the expression takes no real value.

Domain and range

The domain is $x \geq 1$
The range is $y \geq 0$
It increases monotonically
It is neither even nor odd

The range of $\cosh$ , which was $y \geq 1$ , has passed straight over to become the domain of the inverse.

Monotonicity

The derivative is as follows.

\frac{d}{dx}\operatorname{arcosh} x = \frac{1}{\sqrt{x^2 - 1}} \quad (x > 1)

It is always positive, so the function increases monotonically over its whole domain.

The tangent at the endpoint

The graph begins at the point $(1, 0)$ . There the denominator of the derivative tends to $0$ , so the slope grows without bound and the tangent is vertical. The horizontal tangent that $\cosh$ has at its minimum, $x = 0$ , appears as a vertical tangent on the inverse. A small increase of $x$ beyond $1$ lifts $y$ abruptly.

Notable values

$x$	$\operatorname{arcosh} x$
$1$	$0$
$2$	$\ln(2 + \sqrt{3}) \approx 1.3170$
$3$	$\approx 1.7627$
$10$	$\approx 2.9932$

For large $x$ we have $\sqrt{x^2-1} \approx x$ , so $\operatorname{arcosh} x \approx \ln(2x)$ and the growth is logarithmic and slow. There is no horizontal asymptote.

Comparison with the inverse hyperbolic sine

Item	$\operatorname{arsinh} x$	$\operatorname{arcosh} x$
Domain	all real numbers	$x \geq 1$
Range	all real numbers	$y \geq 0$
Closed form	$\ln(x + \sqrt{x^2+1})$	$\ln(x + \sqrt{x^2-1})$
Derivative	$\dfrac{1}{\sqrt{x^2+1}}$	$\dfrac{1}{\sqrt{x^2-1}}$
Tangent at the end	none	vertical at $(1, 0)$

A single sign under the root is the whole difference, and yet it decides whether the domain is the entire line or a half-line.

Applications

It appears as the result of an integral.

\int \frac{dx}{\sqrt{x^2 - 1}} = \operatorname{arcosh} x + C \quad (x > 1)

It is used in the geometry of hyperbolas and in physical calculations such as electromagnetism and heat conduction.

Inverse hyperbolic functions, Wikipedia

Inverse Hyperbolic Cosine y=arcosh⁡xy = \operatorname{arcosh} xy=arcoshx