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

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

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

Then $x = \sec y$ where $y \in [0..\pi] \land y \ne \pi/2$.

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

From Derivative of an 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 the corollary to Sum of Squares of Sine and Cosine:
 * $1 + \tan^2 y = \sec^2 y \implies \tan^2 y = \sec^2 y - 1$

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 $\sec y$ so:


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

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} }$

This is precisely the desired result.

Proof of Corollary
Since for all $x \in \R$, we have $\left|{x}\right| = \sqrt{x^2}$, we can write:


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

Multiplying the denominator by $1 = \dfrac {\sqrt{x^2}} {\sqrt{x^2}}$ yields:


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