📂Functions

Inverse Hyperbolic Functions

Definition¹

The inverse functions of hyperbolic functions are called inverse hyperbolic functions.

$$ \begin{align*} y = \sinh^{-1} x &\iff \sinh y = x \\ y = \cosh^{-1} x &\iff \cosh y = x \\ y = \tanh^{-1} x &\iff \tanh y = x \\ \end{align*} $$

Closed Form

The values of the inverse hyperbolic functions are concretely as follows.

$$ \begin{align*} \sinh^{-1} x &= \ln \left( x + \sqrt{x^{2} + 1} \right) & x \in \mathbb{R} \\ \cosh^{-1} x &= \ln \left( x + \sqrt{x^{2} - 1} \right) & x \le 1 \\ \tanh^{-1} x &= \dfrac{1}{2} \ln \left( \dfrac{1 + x}{1 - x} \right) & -1 \lt x \lt 1 \\ \end{align*} $$

Proof

$\sinh^{-1} x$

Method 1

Let’s say $y = \sinh^{-1} x$. Then, since $\sinh y = x$,

$$ \dfrac{e^{y} - e^{-y}}{2} = x \implies e^{y} - e^{-y} - 2x = 0 $$

Multiply by $e^{y}$ and rearrange as a quadratic equation with respect to $e^{y}$.

$$ (e^{y})^{2} - 2x(e^{y}) - 1 = 0 $$

Using the quadratic formula to find the roots,

$$ e^{y} = x \pm \sqrt{x^{2} + 1} $$

In this case, $x \le \sqrt{x^{2}} < \sqrt{x^{2} + 1}$ and since $e^{y} > 0$, the possible cases are,

$$ e^{y} = x + \sqrt{x^{2} + 1} $$

Taking logarithms on both sides,

$$ y = \ln (x + \sqrt{x^{2} + 1}) $$

Method 2

Let’s say $y = \sinh^{-1} x$. Then, since $\sinh y = x$ and $\cosh x + \sinh x = e^{x}$, add $\cosh y$ to both sides,

$$ e^{y} = x + \cosh y $$

Also, using the identity $\cosh^{2}x - \sinh^{2}x = 1$, we obtain the following.

$$ e^{y} = x + \sqrt{\sinh^{2} y + 1} = x + \sqrt{x^{2} + 1} $$

Taking logarithms on both sides,

$$ y = \ln (x + \sqrt{x^{2} + 1}) $$

---

$\cosh^{-1} x$

The method to find $\sinh^{-1} x$ is the same, so to summarize simply,

$$ \begin{align*} && y &= \cosh^{-1} x \\ \implies && \cosh y &= x \\ \implies && \cosh y + \sinh y &= x + \sinh y \\ \implies && e^{y} &= x + \sqrt{\cosh^{2}y - 1} \\ \implies && e^{y} &= x + \sqrt{x^{2} - 1} \\ \implies && y &= \ln (x + \sqrt{x^{2} - 1}) \end{align*} $$

---

$\tanh^{-1} x$

Simply stating the process without explanation is as follows.

$$ \begin{align*} && y &= \tanh^{-1} x \\ \implies && \tanh y &= x \\ \implies && \dfrac{e^{y} - e^{-y}}{e^{y} + e^{-y}} &= x \\ \implies && \dfrac{e^{2y} - 1}{e^{2y} + 1} &= x \\ \implies && e^{2y} - 1 &= x (e^{2y} + 1) \\ \implies && e^{2y} - xe^{2y} &= x + 1 \\ \implies && (1 - x)e^{2y} &= x + 1 \\ \implies && e^{2y} &= \dfrac{1 + x}{1 - x} \\ \implies && y &= \dfrac{1}{2} \ln \left( \dfrac{1 + x}{1 - x} \right) \\ \end{align*} $$

Domain and Range

$$ \begin{align*} \sinh^{-1} &: \mathbb{R} \to \mathbb{R} \\ \cosh^{-1} &: [1, \infty) \to [0, \infty) \\ \tanh^{-1} &: (-1, 1) \to \mathbb{R} \end{align*} $$

Derivatives

• Derivation process

$$ \begin{align*} \dfrac{d}{dx} (\sinh^{-1} x) &= \dfrac{1}{\sqrt{x^{2} + 1}} \qquad & \dfrac{d}{dx} (\csch^{-1} x) &= - \dfrac{1}{|x|\sqrt{x^{2} + 1}} \\ \dfrac{d}{dx} (\cosh^{-1} x) &= \dfrac{1}{\sqrt{x^{2} - 1}} \qquad & \dfrac{d}{dx} (\sech^{-1} x) &= - \dfrac{1}{x\sqrt{1 - x^{2}}} \\ \dfrac{d}{dx} (\tanh^{-1} x) &= \dfrac{1}{1 - x^{2}} \qquad & \dfrac{d}{dx} (\coth^{-1} x) &= \dfrac{1}{1 - x^{2}} \end{align*} $$