Solution. Let F(x) = f(g(x)) = \cot^3(x), where f(u) = u^3 and g(x) = \cot(x). To determine the derivative of \cot^3(x), we need to use the chain rule:
\begin{align*}
F'(x) = f'(g(x))g'(x).
\end{align*}\begin{align*}
f'(g(x)) = 3g(x)^2 = 3\cot^2(x) \quad \text{and} \quad g'(x) = -\csc^2(x).
\end{align*}\begin{align*}
F'(x) &= f'(g(x))g'(x) \\
&= 3\cot^2(x)\cdot (-\csc^2(x)) \\
&= -3 \cot^2(x) \csc^2(x).
\end{align*}