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*}