# How do you find the derivative of ln(sin^2x)?

Nov 10, 2015

$\frac{2 \cos x}{\sin} x$

#### Explanation:

Use the chain rule.

You can break down your function into the logarithm, the square and the sinus function like follows:

$f \left(u\right) = \ln \left(u\right)$
$u \left(v\right) = {v}^{2} \left(x\right)$
$v \left(x\right) = \sin \left(x\right)$

Now, you need to compute the three derivatives of those three functions (and afterwards plug the respective values of $u$ and $v$):

$f ' \left(u\right) = \frac{1}{u} = \frac{1}{v} ^ 2 = \frac{1}{\sin} ^ 2 x$
$u ' \left(v\right) = 2 v = 2 \sin x$
$v ' \left(x\right) = \cos \left(x\right)$

Now, the only thing left to do is multiplying those three derivatives:

$f ' \left(x\right) = f ' \left(u\right) \cdot u ' \left(v\right) \cdot v ' \left(x\right) = \frac{1}{\sin} ^ 2 x \cdot 2 \sin x \cdot \cos x$
$= \frac{2 \cos x}{\sin} x$

Hope that this helped!