# How do I differentiate y=ln(sec(x) tan(x))?

You use the Chain Rule deriving first $\ln$ as it is and then multiplying by the derivative of the argument and also the Product Rule where:
if $f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$
$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$
$y ' = \frac{1}{\sec \left(x\right) \tan \left(x\right)} \cdot \left[\sec \left(x\right) \tan \left(x\right) \tan \left(x\right) + \sec \left(x\right) {\sec}^{2} \left(x\right)\right] =$
$= \tan \left(x\right) + {\sec}^{2} \frac{x}{\tan} \left(x\right)$