Derivative of Arcsecant Function

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Theorem

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

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


Then:

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


Corollary 1

$\dfrac {\mathrm d \left({\operatorname{arcsec} \left({\frac x a}\right) }\right)} {\mathrm d x} = \dfrac a {\left\vert{x}\right\vert \sqrt {x^2 - a^2} } = \begin{cases} \dfrac a {x \sqrt {x^2 - a^2} } & : 0 < \operatorname{arcsec} \dfrac x a < \dfrac \pi 2 \\ \dfrac {-a} {x \sqrt {x^2 - a^2} } & : \dfrac \pi 2 < \operatorname{arcsec} \dfrac x a < \pi \\ \end{cases}$


Corollary 2

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


Proof

Let $y = \arcsec x$ where $x < -1$ or $x > 1$.

Then:

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

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

Thus:

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


Hence the result.

$\blacksquare$


Also see


Sources