# Derivative of Inverse Hyperbolic Secant

## Theorem

Let $S$ denote the open real interval:

$S := \openint 0 1$

Let $x \in S$.

Let $\sech^{-1} x$ denote the inverse hyperbolic secant of $x$.

Then:

$\map {\dfrac \d {\d x} } {\sech^{-1} x} = \dfrac {-1} {x \sqrt{1 - x^2} }$

## Proof

$\sech^{-1} x$ is defined only on the half-open real interval $\hointl 0 1$.

Thus on $\hointl 0 1$:

 $\displaystyle y$ $=$ $\displaystyle \sech^{-1} x$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $=$ $\displaystyle \sech y$ where $y \in \R_{>0}$ $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d x} {\d y}$ $=$ $\displaystyle -\sech y \, \tanh y$ Derivative of Hyperbolic Secant Function $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d y} {\d x}$ $=$ $\displaystyle \dfrac {-1} {\sech y \ \tanh y}$ Derivative of Inverse Function $\displaystyle \leadsto \ \$ $\displaystyle \frac {\d y} {\d x}$ $=$ $\displaystyle \frac {-1} {\sech y \, \sqrt {1 - \sech^2 y} }$ Sum of Squares of Hyperbolic Secant and Tangent $\displaystyle \leadsto \ \$ $\displaystyle \map {\frac \d {\d x} } {\sech^{-1} x}$ $=$ $\displaystyle \frac {-1} {x \sqrt {1 - x^2} }$ Definition of $x$ and $y$

When $x = 1$, however, $\sqrt{1 - x^2} = 0$ and so $\dfrac {-1} {x \sqrt {1 - x^2} }$ is undefined.

Hence $\sech^{-1} x$ can be defined only on $\openint 0 1$.

$\blacksquare$