Derivative of coth

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{coth}(z) = -\mathrm{csch}^2(z),$$ where $\mathrm{coth}$ denotes the hyperbolic cotangent and $\mathrm{csch}$ denotes the hyperbolic cosecant.

Proof

By the definition, $$\mathrm{coth}(z) = \dfrac{\mathrm{cosh}(z)}{\mathrm{sinh}(z)}.$$ Using the quotient rule, the derivative of sinh, the derivative of cosh, Pythagorean identity for sinh and cosh, and the definition of $\mathrm{csch}$, we see $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{coth}(z) &= \dfrac{\sinh^2(z)-\cosh^2(z)}{\mathrm{sinh}^2(z)} \\ &= - \dfrac{\mathrm{cosh}^2(z)-\mathrm{sinh}^2(z)}{\mathrm{sinh}^2(z)} \\ &= -\mathrm{csch}^2(z), \end{array}$$ as was to be shown.