Derivative of Inverse Hyperbolic Secant Function

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Theorem

Let $u$ be a differentiable real function of $x$.

Then:

$\map {\dfrac \d {\d x} } {\sech^{-1} u} = \dfrac {-1} {u \sqrt {1 - u^2} } \dfrac {\d u} {\d x}$

where $0 < u < 1$

where $\sech^{-1}$ is the inverse hyperbolic secant.


Proof

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

$\blacksquare$


Also see


Sources