Answer
$f'(x) = 2cotx$
Work Step by Step
$f(x) = \ln (sin^{2}x)$
Use the chain rule:
$f'(x) = (\frac{1}{sin^{2}x})(2sinx)(cosx)$
$f'(x) = (\frac{2sinx}{sin^{2}x})(cosx)$
Cancel out $sinx$:
$f'(x) = (\frac{2}{sinx})(cosx)$
Multiply:
$f'(x) = (\frac{2cosx}{sinx})$
Given trigonometric function: $\frac{cosx}{sinx} = cotx$
$f'(x) = 2cotx$
