**Proof.**Let F(x) = \sec(x) = \frac{1}{\cos(x)}, f(u) = \frac{1}{u} and g(x) = \cos(x) such that F(x) = f(g(x)). Then we will use the chain rule:

\begin{align*} F'(x) = f'(g(x))g'(x). \end{align*}

\begin{align*} f'(g(x)) = \frac{-1}{g(x)^2} = \frac{-1}{\cos^2(x)} \quad \text{and} \quad g'(x) = -\sin(x). \end{align*}

\begin{align*} F'(x) &= f'(g(x))g'(x) \\ &= \frac{-1}{\cos^2(x)}\cdot(-sin(x)) \\ &= \frac{\sin(x)}{\cos^2(x)} \\ &= \frac{\sin(x)}{\cos(x)}\frac{1}{\cos(x)} \\ &= \tan(x) \frac{1}{\cos(x)} \\ &= \tan(x)\sec(x) \end{align*}