Answer
$y'= \dfrac{\sec ^2 x}{\tan x}$
Work Step by Step
In order to derivate this function you have to apply the chain rule
Let's make an «u» substitution to make it easier
$f(u) = \ln(u)$
$u' = \tan x$
Derivate the function:
$f'(u) = \dfrac{u'}{u}$
Now let's find u'
$u' = \sec ^2 x$
Then undo the substitution, simplify and get the answer:
$f'(x) = \dfrac{\sec ^2 x}{\tan x}$
