Solution. Let f(x) = \sinh(x). We know that
\begin{align*} \sinh(x) = \frac{e^x - e^{-x}}{2} \end{align*}
\begin{align*} f'(x) &= \frac{d}{dx} \sinh(x) \\ &= \frac{d}{dx} \frac{e^x - e^{-x}}{2} \\ &= \frac{d}{dx} \frac{e^x}{2} - \frac{d}{dx} \frac{e^{-x}}{2} \\ &= \frac{e^x}{2} - \frac{-e^{-x}}{2} \\ &= \frac{e^x}{2} + \frac{e^{-x}}{2} \\ &= \frac{e^x + e^{-x}}{2} \\ &= \cosh(x). \end{align*}