Derivative of Inverse Hyperbolic Cosecant Function

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Theorem

Let $u$ be a differentiable real function of $x$ such that $u \ne 0$.

Then:

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

where $\arcsch$ is the real inverse hyperbolic cosecant.


Proof

\(\ds \map {\frac \d {\d x} } {\arcsch u}\) \(=\) \(\ds \map {\frac \d {\d u} } {\arcsch u} \frac {\d u} {\d x}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \dfrac {-1} {\size u \sqrt {1 + u^2} } \frac {\d u} {\d x}\) Derivative of Inverse Hyperbolic Cosecant

$\blacksquare$


Also see


Sources