Derivative of sinh

Proposition: The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z) = \cosh(z),$$ where $\sinh$ denotes the hyperbolic sine and $\cosh$ denotes the hyperbolic cosine.

Proof: From the definition, $$\sinh(z) = \dfrac{e^z-e^{-z}}{2},$$ and so using the derivative of the exponential function, the linear property of the derivative, the chain rule, and the definition of the hyperbolic cosine, $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z)=\dfrac{e^z + e^{-z}}{2}=\cosh(z),$$ as was to be shown. █