## Calculus: Early Transcendentals 8th Edition

$f'(x) = 2cotx$
$f(x) = \ln (sin^{2}x)$ 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: $\frac{cosx}{sinx} = cotx$ $f'(x) = 2cotx$