Derivative of Arccosecant Function/Corollary

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Corollary to Derivative of Arccosecant Function

Let $x \in \R$.

Let $\operatorname{arccsc} \left({\dfrac x a}\right)$ be the arccosecant of $\dfrac x a$.


Then:

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


Proof

\(\displaystyle \frac {\mathrm d \left({\operatorname{arccsc} \left({\frac x a}\right)}\right)} {\mathrm d x}\) \(=\) \(\displaystyle \frac 1 a \frac {-1} {\left\vert{\frac x a}\right\vert \sqrt {\left({\frac x a}\right)^2 - 1} }\) Derivative of Arccosecant Function and Derivative of Function of Constant Multiple
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \frac {-1} {\left\vert{\frac x a}\right\vert \frac {\sqrt {x^2 - a^2} } a}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \frac {-a^2} {\left\vert{x}\right\vert {\sqrt {x^2 - a^2} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {-a} {\left\vert{x}\right\vert {\sqrt {x^2 - a^2} } }\)

$\Box$


Similarly:

\(\displaystyle \frac {\mathrm d \left({\operatorname{arccsc} \left({\frac x a}\right)}\right)} {\mathrm d x}\) \(=\) \(\displaystyle \begin{cases} \dfrac 1 a \dfrac {-1} {\frac x a \sqrt {\left({\frac x a}\right)^2 - 1} } & : 0 < \operatorname{arccsc} \dfrac x a < \dfrac \pi 2 \\ \dfrac 1 a \dfrac {+1} {\frac x a \sqrt {\left({\frac x a}\right)^2 - 1} } & : -\dfrac \pi 2 < \operatorname{arccsc} \dfrac x a < 0 \\ \end{cases}\) Derivative of Arccosecant Function
and Derivative of Function of Constant Multiple
\(\displaystyle \) \(=\) \(\displaystyle \begin{cases} \dfrac {-a} {x \sqrt {x^2 - a^2} } & : 0 < \operatorname{arccsc} \dfrac x a < \dfrac \pi 2 \\ \dfrac a {x \sqrt {x^2 - a^2} } & : -\dfrac \pi 2 < \operatorname{arccsc} \dfrac x a < 0 \\ \end{cases}\) simplifying as above

$\blacksquare$


Also see