Derivative of Arccosecant Function

Theorem
Let $x \in \R$ be a real number such that $x < -1$ or $x > 1$.

Let $\operatorname{arccsc} x$ be the arccosecant of $x$.

Then:
 * $\dfrac {\mathrm d \left({\operatorname{arccsc} x}\right)} {\mathrm d x} = \dfrac {-1} {\left|{x}\right| \sqrt {x^2 - 1} } = \begin{cases} \dfrac {-1} {x \sqrt {x^2 - 1} } & : 0 < \operatorname{arccsc} x < \dfrac \pi 2 \\

\dfrac {+1} {x \sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \operatorname{arccsc} x < 0 \\ \end{cases}$

Proof
Let $y = \operatorname{arccsc} x$ where $x < -1$ or $x > 1$.

Then:

Since $\dfrac {\mathrm d y} {\mathrm d x} = \dfrac {-1} {\csc y \cot y}$, the sign of $\dfrac {\mathrm d y} {\mathrm d x}$ is opposite to the sign of $\csc y \cot y$.

Writing $\csc y \cot y$ as $\dfrac {\cos y} {\sin^2 y}$, it is evident that the sign of $\dfrac {\mathrm d y} {\mathrm d x}$ is opposite to the sign of $\cos y$.

From Sine and Cosine are Periodic on Reals, $\cos y$ is never negative on its domain ($y \in \left[{-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right] \land y \ne 0$).

However, by definition of the arccosecant of $x$:
 * $0 < \operatorname{arccsc} x < \dfrac \pi 2 \implies x > 0$
 * $-\dfrac \pi 2 < \operatorname{arccsc} x < 0 \implies x < 0$

Thus:


 * $\dfrac {\mathrm d \left({\operatorname{arccsc} x}\right)} {\mathrm d x} = \dfrac {-1} {\left|{x}\right| \sqrt {x^2 - 1} } = \begin{cases} \dfrac {-1} {x \sqrt {x^2 - 1} } & : 0 < \operatorname{arccsc} x < \dfrac \pi 2 \\

\dfrac {+1} {x \sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \operatorname{arccsc} x < 0 \\ \end{cases}$