# Derivative of Inverse Hyperbolic Cosecant

## Theorem

Let $x \in \R_{\ne 0}$.

Let $\arcsch x$ denote the inverse hyperbolic cosecant of $x$.

Then:

$\map {\dfrac \d {\d x} } {\arcsch x} = \dfrac {-1} {\size x \sqrt {1 + x^2} }$

## Proof

Let $x > 1$.

Then we have:

 $\ds y$ $=$ $\ds \arcsch x$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds \csch y$ where $y \ne 0$ $\ds \leadsto \ \$ $\ds \frac {\d x} {\d y}$ $=$ $\ds -\csch y \coth y$ Derivative of Hyperbolic Cosecant $\ds \leadsto \ \$ $\ds \frac {\d y} {\d x}$ $=$ $\ds \dfrac {-1} {\csch y \coth y}$ Derivative of Inverse Function $\ds \leadsto \ \$ $\ds \frac {\d y} {\d x}$ $=$ $\ds \frac {-1} {\csch y \sqrt {1 + \csch^2 y} }$ Difference of Squares of Hyperbolic Cotangent and Cosecant $\ds \leadsto \ \$ $\ds \map {\frac \d {\d x} } {\arcsch x}$ $=$ $\ds \frac {-1} {x \sqrt {1 + x^2} }$ Definition of $x$ and $y$ $\ds$ $=$ $\ds \frac {-1} {\size x \sqrt {1 + x^2} }$ as $x > 0$

We have that Inverse Hyperbolic Cosecant is Odd Function.

Hence from Derivative of Odd Function is Even, $\map {\dfrac \d {\d x} } {\arcsch x}$ is even.

Hence for $x < -1$ we have that:

$\map {\dfrac \d {\d x} } {\arcsch x} = \dfrac {-1} {\paren {-x} \sqrt {1 + x^2} }$

and so for $x < -1$:

$\map {\dfrac \d {\d x} } {\arcsch x} = \dfrac {-1} {\size x \sqrt {1 + x^2} }$

and the result follows.

$\blacksquare$