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.

$\blacksquare$

Sources

1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Appendix $2$: Table of derivatives and integrals of common functions: Inverse hyperbolic functions
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Appendix: Table $1$: Derivatives
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Appendix: Table $1$: Derivatives