Derivative of tanh

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \tanh(z)=\mathrm{sech}^2(z),$$ where $\tanh$ denotes the hyperbolic tangent and $\mathrm{sech}$ denotes the hyperbolic secant.

Proof

From the definition, $$\tanh(z) = \dfrac{\sinh(z)}{\cosh(z)},$$ and so using the derivative of sinh, the derivative of cosh, the quotient rule, the Pythagorean identity for sinh and cosh, and the definition of the hyperbolic secant, $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \tanh(z) &= \dfrac{\mathrm{d}}{\mathrm{d}z}\left[ \dfrac{\sinh(z)}{\cosh(z)} \right] \\ &= \dfrac{\cosh^2(z)-\sinh^2(z)}{\cosh^2(z)} \\ &= \dfrac{1}{\cosh^2(z)} \\ &= \mathrm{sech}^2(z), \end{array}$$ as was to be shown. █