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

Jan 3, 2016

Chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dw}} \frac{\mathrm{dw}}{\mathrm{dx}}$

#### Explanation:

To do so, we'll rename $f \left(x\right) = \ln \left(u\right)$, then $u = \sqrt{v}$, $v = \sin \left(w\right)$ and finally $w = 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{u} \left(\frac{1}{2 \sqrt{v}}\right) \cos \left(w\right) \left(2\right) = \frac{\cancel{2} \cos \left(w\right)}{\cancel{2} u \sqrt{v}}$

Substituting $u$, $v$ and $w$, in this order:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos \left(w\right)}{\sqrt{v} \sqrt{v}} = \frac{\cos \left(w\right)}{v} = \frac{\cos \left(w\right)}{\sin} \left(w\right) = \cos \frac{2 x}{\sin} \left(2 x\right) = \textcolor{g r e e n}{\cot \left(2 x\right)}$