Question

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

Answer 1

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(x)=g(x)*h(x)`
`f'(x)=g'(x)h(x)+g(x)h'(x)`

You get:

`y'=1/(sec(x)tan(x))*[sec(x)tan(x)tan(x)+sec(x)sec^2(x)]=`

`=tan(x)+sec^2(x)/tan(x)`