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