Derivative of Hyperbolic Tangent/Corollary

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Theorem

$\map {\dfrac \d {\d x} } {\tanh x} = 1 - \tanh^2 x$

where $\tanh$ is the hyperbolic tangent.


Proof

\(\ds \map {\dfrac \d {\d x} } {\tanh x}\) \(=\) \(\ds \sech^2 x\) Derivative of Hyperbolic Tangent
\(\ds \) \(=\) \(\ds 1 - \tanh^2 x\) Sum of Squares of Hyperbolic Secant and Tangent

$\blacksquare$


Sources