Derivative of Arccosecant Function

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Theorem

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

Let $\arccsc x$ denote the arccosecant of $x$.


Then:

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

\dfrac {+1} {x \sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \ (\text {that is: $x < -1$}) \\ \end{cases}$


Corollary

$\dfrac {\map \d {\arccsc \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 < \arccsc \dfrac x a < \dfrac \pi 2 \\

\dfrac a {x \sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \arccsc \dfrac x a < 0 \\ \end{cases}$


Proof

Arccosecant Function


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

Then:

\(\ds y\) \(=\) \(\ds \arccsc x\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \csc y\) where $y \in \closedint 0 \pi \land y \ne \dfrac pi 2$
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d y}\) \(=\) \(\ds -\csc y \cot y\) Derivative of Cosecant Function
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \frac {-1} {\csc y \cot y}\) Derivative of Inverse Function
\(\ds \leadsto \ \ \) \(\ds \paren {\frac {\d y} {\d x} }^2\) \(=\) \(\ds \frac 1 {\csc^2 y \cot^2 y}\) squaring both sides
\(\ds \) \(=\) \(\ds \frac 1 {\csc^2 y \paren {\csc^2 y - 1} }\) Difference of Squares of Cosecant and Cotangent
\(\ds \) \(=\) \(\ds \frac 1 {x^2 \paren {x^2 - 1} }\) 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 $\dfrac {\d y} {\d x} = \dfrac {-1} {\csc y \cot y}$, the sign of $\dfrac {\d y} {\d x}$ is opposite to the sign of $\csc y \cot y$.

Writing $\csc y \cot y$ as $\dfrac {\cos y} {\sin^2 y}$, it is evident that the sign of $\dfrac {\d y} {\d x}$ is opposite to the sign of $\cos y$.

From Sine and Cosine are Periodic on Reals, $\cos y$ is never negative on its domain ($y \in \closedint {-\dfrac \pi 2} {\dfrac \pi 2} \land y \ne 0$).

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

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

Thus:

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

\dfrac {+1} {x \sqrt {x^2 - 1} } & : -\dfrac \pi 2 < \arccsc x < 0 \ (\text {that is: $x < -1$}) \\ \end{cases}$

$\blacksquare$


Also see


Sources

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