# How do you differentiate f(x)=ln(sqrt(sin(x^2-3x))) using the chain rule?

##### 1 Answer
Mar 4, 2016

$f ' \left(x\right) = \frac{1}{\sqrt{\sin \left({x}^{2} - 3 x\right)}} \cdot \frac{1}{2 \sqrt{\sin \left({x}^{2} - 3 x\right)}} \cos \left({x}^{2} - 3 x\right) \left(2 x - 3\right)$
$f ' \left(x\right) = \frac{\left(2 x - 3\right) \cos \left({x}^{2} - 3 x\right)}{2 \sin \left({x}^{2} - 3 x\right)} = \frac{1}{2} \left(2 x - 3\right) \cot \left({x}^{2} - 3 x\right)$

#### Explanation:

You have four functions in here. Start with the $\ln$ and keep everything else the same then move to the sqrt, then sine and then ${x}^{2} - 3 x$. Then simplify note that$\cos \frac{{x}^{2} - 3 x}{\sin \left({x}^{2} - 3 x\right)} = \cot \left({x}^{2} - 3 x\right)$