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*}