# What is the derivative of y = ln(sin^2(theta))?

Oct 9, 2016

$C o {t}^{2} \theta$

#### Explanation:

Using the chain rule we have:

$F \left(x\right) ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

$y ' = \ln ' \left({\sin}^{2} \left(\theta\right)\right) \cdot \left({\cos}^{2} \left(\theta\right)\right)$
$\frac{1}{\sin} ^ 2 \theta \cdot C o {s}^{2} \theta$

$C o {t}^{2} \theta$