Derivative of Hyperbolic Tangent Function

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Theorem

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

Then:

$\map {\dfrac \d {\d x} } {\tanh u} = \sech^2 u \dfrac {\d u} {\d x}$

where $\tanh$ is the hyperbolic tangent and $\sech$ is the hyperbolic secant.


Proof

\(\ds \map {\frac \d {\d x} } {\tanh u}\) \(=\) \(\ds \map {\frac \d {\d u} } {\tanh u} \frac {\d u} {\d x}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \sech^2 u \frac {\d u} {\d x}\) Derivative of Hyperbolic Tangent

$\blacksquare$


Also see


Sources