Solution. Let F(x) = \sec^2(x), f(u) = u^2 and g(x) = \sec(x). Then we will apply the chain rule:
\begin{align*}
F'(x) = f'(g(x))g'(x).
\end{align*}\begin{align*}
f'(u) = 2u \quad \text{and} \quad g'(x) = \tan(x)\sec(x).
\end{align*}\begin{align*}
F'(x) &= f'(g(x))g'(x) \\
&= 2\sec(x)\tan(x)\sec(x) \\
&= 2\tan(x)\sec^2(x).
\end{align*}