# Derivative of Hyperbolic Tangent Function

## 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

$\blacksquare$