# How do you find the derivative of f(x) = sqrt[sin(2x)]?

Jun 10, 2018

$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{\cos \left(2 x\right)}{\setminus \sqrt{\sin \left(2 x\right)}}$

#### Explanation:

Firstly, let $y = \setminus \sqrt{\sin} \left(2 x\right)$

and let $u = \sin \left(2 x\right)$

this means $y = {u}^{\setminus \frac{1}{2}}$

Therefore $\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{\mathrm{dy}}{\mathrm{du}} \cdot \setminus \frac{\mathrm{du}}{\mathrm{dx}}$

$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{1}{2} {u}^{- \setminus \frac{1}{2}} \cdot 2 \cos \left(2 x\right)$

Which goes to

\frac{dy}{dx}=\frac{cos2x}{sqrt{u}

bring back $u = \sin 2 x$

to get

$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{\cos 2 x}{\sqrt{\sin 2 x}}$