📂Functions

Derivative of the Inverse Hyperbolic Functions

Formulas¹

The derivatives of the inverse hyperbolic functions are as follows:

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

Description

The closed forms of the inverse hyperbolic functions are 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

$(\sinh^{-1} x)^{\prime}$

Using the Chain Rule

Assume $y = \sinh^{-1} x$. Then, since $\sinh y = x$, differentiate both sides with respect to $x$ using the chain rule,

$$ \dfrac{d}{dx} \sinh y = (x)^{\prime} \implies \dfrac{d}{dy} \sinh y \dfrac{dy}{dx} = 1 \implies \cosh y \dfrac{dy}{dx} = 1 $$

Therefore, the following is obtained:

$$ \dfrac{dy}{dx} = \dfrac{1}{\cosh y} $$

However, since $\cosh^{2} y - \sinh^{2} y = 1$ and $\sinh y = x$,

$$ \dfrac{dy}{dx} = \dfrac{1}{\sqrt{\sinh^{2} y + 1}} = \dfrac{1}{\sqrt{x^{2} + 1}} $$

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Using the Logarithmic Differentiation

By the differentiation of logarithmic functions,

$$ \begin{align*} \dfrac{d}{dx} (\sinh^{-1} x) &= \dfrac{d}{dx} \ln \left( x + \sqrt{x^{2} + 1} \right) = \dfrac{(x + \sqrt{x^{2} + 1})^{\prime}}{x + \sqrt{x^{2} + 1}} \\[1em] &= \dfrac{1 + \dfrac{2x}{2\sqrt{x^{2} + 1}}}{x + \sqrt{x^{2} + 1}} = \dfrac{\dfrac{x + \sqrt{x^{2} + 1}}{\sqrt{x^{2} + 1}}}{x + \sqrt{x^{2} + 1}} \\[1em] &= \dfrac{x + \sqrt{x^{2} + 1}}{(x + \sqrt{x^{2} + 1})\sqrt{x^{2} + 1}} = \dfrac{1}{\sqrt{x^{2} + 1}} \\[1em] \end{align*} $$

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$(\cosh^{-1} x)^{\prime}$

The same method yields the following, so only the process is briefly listed:

$$ \begin{align*} && & y = \cosh^{-1} x \implies \cosh y = x \implies \dfrac{d}{dx} \cosh y = 1 \\ \implies && & \sinh y \dfrac{dy}{dx} = 1 \implies \dfrac{dy}{dx} = \dfrac{1}{\sinh y} \\ \implies && & \dfrac{dy}{dx} = \dfrac{1}{\sqrt{\cosh^{2} y + 1}} = \dfrac{1}{\sqrt{x^{2} + 1}} \end{align*} $$

$(\tanh^{-1} x)^{\prime}$

The same method yields the following, so only the process is briefly listed:

$$ \begin{align*} && & y = \tanh^{-1} x \implies \tanh y = x \implies \dfrac{d}{dx} \tanh y = 1 \\ \implies && & \sech^{2} y \dfrac{dy}{dx} = 1 \implies \dfrac{dy}{dx} = \dfrac{1}{\sech^{2} y} \\ \implies && & \dfrac{dy}{dx} = \dfrac{1}{1 - \tanh^{2} y} = \dfrac{1}{1 - x^{2}} \end{align*} $$