# Derivative of Arcsecant Function

## Theorem

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

Let $\arcsec x$ be the arcsecant of $x$.

Then:

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

### Corollary 1

$\dfrac {\map \d {\arcsec \frac x a} } {\d x} = \dfrac a {\size x \sqrt {x^2 - a^2} } = \begin {cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \arcsec \dfrac x a < \dfrac \pi 2 \ (\text {that is:$x > a$}) \\ \dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \arcsec \dfrac x a < \pi \ (\text {that is:$x < -a$}) \\ \end{cases}$

### Corollary 2

$\dfrac {\map \d {\arcsec x} } {\d x} = \dfrac 1 {x^2 \sqrt {1 - \frac 1 {x^2} } }$

## Proof

Let $y = \arcsec x$ where $\size x > 1$.

Then:

 $\ds y$ $=$ $\ds \arcsec x$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds \sec y$ where $y \in \closedint 0 \pi \land y \ne \dfrac \pi 2$ $\ds \leadsto \ \$ $\ds \frac {\d x} {\d y}$ $=$ $\ds \sec y \ \tan y$ Derivative of Secant Function $\ds \leadsto \ \$ $\ds \frac {\d y} {\d x}$ $=$ $\ds \dfrac 1 {\sec y \tan y}$ Derivative of Inverse Function $\ds \leadsto \ \$ $\ds \paren {\frac {\d y} {\d x} }^2$ $=$ $\ds \frac 1 {\sec^2 y \ \tan^2 y}$ squaring both sides $\ds$ $=$ $\ds \frac 1 {\sec^2 y \paren {\sec^2 y - 1} }$ Difference of Squares of Secant and Tangent $\ds$ $=$ $\ds \frac 1 {x^2 \paren {x^2 - 1} }$ Definition of $x$ $\ds \leadsto \ \$ $\ds \size {\dfrac {\d y} {\d x} }$ $=$ $\ds \dfrac 1 {\size x \sqrt {x^2 - 1} }$ squaring both sides

Since $\dfrac {\d y} {\d x} = \dfrac 1 {\sec y \tan y}$, the sign of $\dfrac {\d y} {\d x}$ 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 $\dfrac {\d y} {\d x}$ 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 ($y \in \closedint 0 \pi \land y \ne \pi/2$).

However, by definition of the arcsecant of $x$:

$0 < \arcsec x < \dfrac \pi 2 \implies x > 1$
$\dfrac \pi 2 < \arcsec x < \pi \implies x < -1$

Thus:

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

Hence the result.

$\blacksquare$