Derivative of Arcsecant Function

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

Let $\operatorname{arcsec} x$ be the arcsecant of $x$.

Then:
 * $\dfrac {\mathrm d \left({\operatorname{arcsec} 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{arcsec} x < \dfrac \pi 2 \\

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

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

Then:

Since $\dfrac {\mathrm d y} {\mathrm d x} = \dfrac 1 {\sec y \tan y}$, the sign of $\dfrac {\mathrm d y} {\mathrm d x}$ is the same as the sign of $\sec y \tan y$.

Writing $\sec y \tan y$ as $\dfrac {\sin y} {\cos^2 y}$, it is evident that the sign of $\dfrac {\mathrm d y} {\mathrm d x}$ is the same as the sign of $\sin y$.

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

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

Thus:


 * $\dfrac {\mathrm d \left({\operatorname{arcsec} 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{arcsec} x < \dfrac \pi 2 \\

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

Hence the result.