Derivative of a product of trigonometric functions

Apply the product rule: $\frac{d}{dx}[u v] = u'v + uv'$ with $u = \tan(x)$ and $v = \sec(x)$. Since $u' = \sec^2(x)$ and $v' = \sec(x)\tan(x)$, substituting gives $f'(x) = \sec^2(x)\sec(x) + \tan(x)\sec(x)\tan(x)$.

Simplify each term: $\sec^2(x)\sec(x) = \sec^3(x)$ and $\tan(x)\sec(x)\tan(x) = \sec(x)\tan^2(x)$. Therefore $f'(x) = \sec^3(x) + \sec(x)\tan^2(x)$, which can also be factored as $\sec(x)\left(\sec^2(x) + \tan^2(x)\right)$.