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:

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.