Chain rule with logarithm of a sum

Identify the outer function as $f(u) = \ln(u)$ and the inner function as $u = \sec(x) + \tan(x)$. By the chain rule, $g'(x) = f'(u)\cdot u'(x) = \dfrac{1}{\sec(x)+\tan(x)} \cdot \dfrac{d}{dx}[\sec(x)+\tan(x)]$. Now compute the derivative of the inner function: $\dfrac{d}{dx}[\sec(x)+\tan(x)] = \sec(x)\tan(x) + \sec^2(x)$.

Substitute this back to get $g'(x) = \dfrac{\sec(x)\tan(x) + \sec^2(x)}{\sec(x)+\tan(x)}$. Factoring $\sec(x)$ from the numerator gives $g'(x) = \dfrac{\sec(x)[\tan(x) + \sec(x)]}{\sec(x)+\tan(x)}$, and the common factor $(\sec(x)+\tan(x))$ cancels, leaving the simplified result $g'(x) = \sec(x)$.