y = \operatorname{arcoth} x

The Inverse Hyperbolic Cotangent $y = \operatorname{arcoth} x$

$\operatorname{arcoth} x$ , the inverse hyperbolic cotangent, is the inverse of $\coth x = \dfrac{\cosh x}{\sinh x}$ ¹. Since $\coth$ maps each branch of $x \neq 0$ monotonically onto the region $|y| > 1$ , the domain of its inverse is $|x| > 1$ .

Definition and closed form

$y = \operatorname{arcoth} x$ is the value $y$ with $x = \coth y$ . Written with a logarithm it takes the following form.

\operatorname{arcoth} x = \frac{1}{2}\ln\frac{x+1}{x-1}

The argument $\dfrac{x+1}{x-1}$ is positive exactly when $x > 1$ or $x < -1$ , which matches the domain.

Domain and range

The domain is $x < -1$ or $x > 1$
The range is $y \neq 0$
It decreases monotonically on each branch
It is an odd function

There is no value on $-1 \leq x \leq 1$ . That $0$ is missing from the range is the counterpart of $\coth$ never taking the value $0$ .

Symmetry and asymptotes

Since $\coth$ is odd, so is its inverse, and the graph has point symmetry about the origin.

Approach	$\operatorname{arcoth} x$
$x \to 1^{+}$	$\to +\infty$
$x \to -1^{-}$	$\to -\infty$
$x \to \pm\infty$	$\to 0$

The lines $x = 1$ and $x = -1$ are vertical asymptotes and the $x$ -axis is a horizontal one.

Monotonicity and concavity

The derivative is $\dfrac{d}{dx}\operatorname{arcoth} x = \dfrac{1}{1-x^{2}}$ . On the domain $x^{2} > 1$ , so the denominator is negative and the derivative is always negative: the function decreases on both branches.

The second derivative is $\dfrac{2x}{(1-x^{2})^{2}}$ , carrying the sign of $x$ , so the curve is concave up for $x > 1$ and concave down for $x < -1$ , with no inflection point.

Notable values

$x$	$\operatorname{arcoth} x$
$1.5$	$\dfrac{1}{2}\ln 5 \approx 0.8047$
$2$	$\dfrac{1}{2}\ln 3 \approx 0.5493$
$3$	$\dfrac{1}{2}\ln 2 \approx 0.3466$
$10$	$\approx 0.1003$

Relation to the inverse hyperbolic tangent

The derivative $\dfrac{1}{1-x^{2}}$ is exactly the derivative of $\operatorname{artanh} x$ . Only the domain differs: the two functions take charge of the same formula on complementary ranges.

Item	$\operatorname{artanh} x$	$\operatorname{arcoth} x$
Domain	$\|x\| < 1$	$\|x\| > 1$
Range	all real numbers	$y \neq 0$
Derivative	$\dfrac{1}{1-x^{2}}$	$\dfrac{1}{1-x^{2}}$
Behavior	increasing	decreasing on each branch

For that reason the integral $\int \dfrac{dx}{1-x^{2}}$ has to be written differently on different intervals, though an absolute value lets both cases be combined.

\int \frac{dx}{1-x^{2}} = \frac{1}{2}\ln\left|\frac{1+x}{1-x}\right| + C

The relation through a reciprocal

\operatorname{arcoth} x = \operatorname{artanh}\frac{1}{x}

When $|x| > 1$ we have $\left|\dfrac{1}{x}\right| < 1$ , so the right-hand side falls exactly inside the domain of $\operatorname{artanh}$ . That $\coth$ is the reciprocal of $\tanh$ shows up on the inverse side as a reciprocal of the argument.

Inverse hyperbolic functions, Wikipedia

The Inverse Hyperbolic Cotangent y=arcoth⁡xy = \operatorname{arcoth} xy=arcothx