Solution. Let F'(x) = \tan^3(x) = f(g(x)), where f(u) = u^3 and g(x) = \tan(x). We will use the chain rule to determine the derivative of \tan^3(x):
\begin{align*}
F'(x) = f'(g(x))g'(x).
\end{align*}\begin{align*}
f'(g(x)) = f'(\tan(x)) = 3\tan^2(x) \quad \text{and} \quad g'(x) = \frac{1}{\cos^2(x)}.
\end{align*}\begin{align*}
F'(x) &= f'(g(x))g'(x) \\
&= 3\tan^2(x) \frac{1}{\cos^2(x)} \\
&= 3\tan^2(x)\sec^2(x),
\end{align*}