Derivative of Arccosecant Function

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

Let $\arccsc x$ denote the arccosecant of $x$.

Then:
 * $\dfrac {\map \d {\arccsc x} } {\d x} = \dfrac {-1} {\size x \sqrt {x^2 - 1} } = \begin {cases} \dfrac {-1} {x \sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \ (\text {that is: $x > 1$}) \\

\dfrac {+1} {x \sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \ (\text {that is: $x < -1$}) \\ \end{cases}$

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

Then:

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

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

However, by definition of the arccosecant of $x$:
 * $0 < \arccsc x < \dfrac \pi 2 \implies x > 1$
 * $-\dfrac \pi 2 < \arccsc x < 0 \implies x < -1$

Thus:


 * $\dfrac {\map \d {\arccsc x} } {\d x} = \dfrac {-1} {\size x \sqrt {x^2 - 1} } = \begin {cases} \dfrac {-1} {x \sqrt {x^2 - 1} } & : 0 < \arccsc x < \dfrac \pi 2 \ (\text {that is: $x > 1$}) \\

\dfrac {+1} {x \sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \ (\text {that is: $x < -1$}) \\ \end{cases}$

Also see

 * Derivative of Arcsine Function
 * Derivative of Arccosine Function
 * Derivative of Arctangent Function
 * Derivative of Arccotangent Function
 * Derivative of Arcsecant Function