Derivatives of hyperbolic functions and their inverses

Differentiation, derivatives and Taylor approximations: Differentiating exponential and logarithmic functions

Derivatives of hyperbolic functions and their inverses

The table below lists the derivatives of the hyperbolic functions and their inverses.

$\begin{array}{|c|c|} \hline \mathit{function} & \mathit{derivative} \\ \hline\\ \sinh(x) & \cosh(x) \\ \\ \cosh(x) & \sinh(x) \\ \\ \tanh(x) & \dfrac{1}{\cosh(x)^2}\\ \\ \hline\\ \mathrm{arsinh}(x) & \dfrac{1}{\sqrt{1+x^2}} \\ \\ \mathrm{arcosh}(x) & \dfrac{1}{\sqrt{x-1}\sqrt{x+1}} \\ \\ \mathrm{artanh}(x) & \dfrac{1}{1-x^2}\\ \\ \hline \end{array}$ What is immediately striking is that the derivative of a hyperbolic function is itself a hyperbolic function. It is also striking that the derivative of an inverse hyperbolic function is a function which contains no hyperbolic function occurs anymore.

For math enthusiasts we present the proofs of some formulas.

$\begin{aligned}\frac{\dd}{\dd x}\sinh(x)&= \frac{\dd}{\dd x}\frac{e^x-\e^{-x}}{2}\\[0.25cm] &=\frac{e^x+\e^{-x}}{2}\\[0.25cm] &=\cosh(x)\\[0.25cm] \frac{\dd}{\dd x}\cosh(x)&=\frac{\dd}{\dd x}\frac{e^x+\e^{-x}}{2}\\[0.25cm] &= \frac{e^x-\e^{-x}}{2}\\[0.25cm] &= \sinh(x)\\[0.25cm]\frac{\dd}{\dd x}\mathrm{arsinh}(x) &= \frac{\dd}{\dd x}\ln\bigl(x+\sqrt{x^2+1}\bigr)\\[0.25cm]&=\frac{1+\dfrac{1}{2}\cdot\dfrac{1}{\sqrt{x^2+1}}\cdot 2x}{x+\sqrt{x^2+1}}\\[0.25cm]&=\frac{1+\dfrac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}\\[0.25cm] &=\frac{1}{\sqrt{x^2+1}}\cdot \frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2+1}}\\[0.25cm]&=\frac{1}{\sqrt{x^2+1}}\end{aligned}$