Solution. Let h(x) = \frac{1}{e^x} = e^{-x}, f(u) = e^u and g(x) = -x. We will use the chain rule, i.e.,
\begin{align*} h'(x) = f'(g(x))g'(x). \end{align*}
\begin{align*} f(u) = e^u \quad \text{and} \quad g(x) = -1. \end{align*}
\begin{align*} h'(x) &= f'(g(x))g'(x) \\ &= e^{-x} \cdot (-1) \\ &= -e^{-x} \\ &= -\frac{1}{e^{-x}} \end{align*}