Derivada de funciones hiperbólicas

$\parbox{8cm}{ \begin{equation}\quad \frac{d}{dx} \sinh u = \cosh... ...ation}\quad \frac{d}{dx} \tanh u = \sech ^2 u \frac{du}{dx} \end{equation}}$ $\parbox{8cm}{ \begin{equation}\quad \frac{d}{dx} \coth u = - \cs... ...\quad \frac{d}{dx} \csch u = - \csch u\coth u \frac{du}{dx} \end{equation}}$