Derivative of Inverse Hyperbolic Cosecant

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Theorem

Let $x \in \R_{\ne 0}$.

Let $\arcsch x$ denote the inverse hyperbolic cosecant of $x$.


Then:

$\map {\dfrac \d {\d x} } {\arcsch x} = \dfrac {-1} {\size x \sqrt {1 + x^2} }$


Proof

Arcsch.png

Let $x > 1$.

Then we have:

\(\ds y\) \(=\) \(\ds \arcsch x\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \csch y\) where $y \ne 0$
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d y}\) \(=\) \(\ds -\csch y \coth y\) Derivative of Hyperbolic Cosecant
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \dfrac {-1} {\csch y \coth y}\) Derivative of Inverse Function
\(\ds \leadsto \ \ \) \(\ds \frac {\d y} {\d x}\) \(=\) \(\ds \frac {-1} {\csch y \sqrt {1 + \csch^2 y} }\) Difference of Squares of Hyperbolic Cotangent and Cosecant
\(\ds \leadsto \ \ \) \(\ds \map {\frac \d {\d x} } {\arcsch x}\) \(=\) \(\ds \frac {-1} {x \sqrt {1 + x^2} }\) Definition of $x$ and $y$
\(\ds \) \(=\) \(\ds \frac {-1} {\size x \sqrt {1 + x^2} }\) as $x > 0$


We have that Inverse Hyperbolic Cosecant is Odd Function.

Hence from Derivative of Odd Function is Even, $\map {\dfrac \d {\d x} } {\arcsch x}$ is even.

Hence for $x < -1$ we have that:

$\map {\dfrac \d {\d x} } {\arcsch x} = \dfrac {-1} {\paren {-x} \sqrt {1 + x^2} }$

and so for $x < -1$:

$\map {\dfrac \d {\d x} } {\arcsch x} = \dfrac {-1} {\size x \sqrt {1 + x^2} }$

and the result follows.

$\blacksquare$


Sources

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