Solution. Let h(x) = \ln(\cos(x)), f(u) = \ln(u) and g(x) = \cos(x). Then we apply the chain rule, i.e.,
\begin{align*}
h'(x) = f'(g(x))g'(x).
\end{align*}\begin{align*}
f'(u) = \frac{1}{u} \quad \text{and} \quad g'(x) = -\sin(x).
\end{align*}\begin{align*}
h'(x) &= f'(g(x))g'(x) \\
&= \frac{1}{\cos(x)}\cdot (-\sin(x)) \\
&= \frac{-\sin(x)}{\cos(x)} \\
&= -\tan(x).
\end{align*}