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

Corollary
$ \dfrac {d(\operatorname{arcsec} x)} {dx} = \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 $\dfrac {dx} {dy} = \sec y \tan y$ from Derivative of Secant Function.

From Derivative of an Inverse Function it follows that $\dfrac {dy} {dx} = \frac{1}{\sec y \tan y}$.

Squaring both sides we have:


 * $\left({\dfrac {dy} {dx}}\right)^2 = \frac{1}{\sec^2 y \tan^2 y}$

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 {dy} {dx}}\right)^2 = \dfrac{1}{(\sec^2 y) \sec^2 y - 1} $

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


 * $\left({\dfrac {dy} {dx}}\right)^2 = \dfrac{1}{(x^2) \ x^2 - 1} $

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


 * $\left|{\dfrac {dy} {dx}}\right| = \dfrac{1}{|x|\sqrt {x^2 - 1}}$

Since $\dfrac {dy} {dx} = \dfrac{1}{\sec y \tan y}$ the sign of $\dfrac {dy} {dx}$ is the same as the sign of $\sec y \tan y$.

Writing $\sec y \tan y$ as $\dfrac{\sin y}{\cos^2y}$ it is evident that the sign of $\dfrac {dy} {dx}$ 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 ($y \in [0..\pi] \land y \ne \pi/2$). Thus our absolute value is unnecessary and we have:


 * $ \dfrac {dy} {dx} = \dfrac{1}{|x|\sqrt {x^2 - 1}}$

which is what we desired to prove.

Proof of Corollary
Since $|x| = \sqrt{x^2}$ we can write:


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

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


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