**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*}