Derivative of cosh

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

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