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:
 * $\dfrac {\mathrm d \left({\operatorname{arcsec} x}\right)} {\mathrm d x} = \dfrac 1 {\left|{x}\right| \sqrt {x^2 - 1} }$

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

Then $x = \sec y$ where $y \in \left[{0 \,.\,.\, \pi}\right] \land y \ne \pi/2$.

Then $\dfrac {\mathrm d x} {\mathrm d y} = \sec y \tan y$ from Derivative of Secant Function.

From Derivative of Inverse Function it follows that $\dfrac {\mathrm d y} {\mathrm d x} = \dfrac 1 {\sec y \tan y}$.

Squaring both sides we have:


 * $\left({\dfrac {\mathrm d y} {\mathrm d x} }\right)^2 = \dfrac 1 {\sec^2 y \ \tan^2 y}$

From Sum of Squares of Sine and Cosine: Corollary 1:
 * $1 + \tan^2 y = \sec^2 y \implies \tan^2 y = \sec^2 y - 1$

Using this identity we can write:


 * $\left({\dfrac {\mathrm d y} {\mathrm d x} }\right)^2 = \dfrac 1 {\sec^2 y \left({\sec^2 y - 1}\right)}$

$x$ was defined as $\sec y$:


 * $\left({\dfrac {\mathrm d y} {\mathrm d x} }\right)^2 = \dfrac 1 {x^2 \left({x^2 - 1}\right)}$

Taking the square root of each side of this equation yields:


 * $\left|{\dfrac {\mathrm d y} {\mathrm d x} }\right| = \dfrac 1 {\left|{x}\right| \sqrt {x^2 - 1} }$

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 ($y \in \left[{0 .. \pi}\right] \land y \ne \pi/2$).

Thus the absolute value is redundant, and we have, for all $x$ considered:


 * $ \dfrac {\mathrm d y} {\mathrm d x} = \dfrac 1 { \left|{x}\right| \sqrt {x^2 - 1} }$

Hence the result.