Question

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

Answer 1

`\frac{dy}{dx}=\frac{cos(2x)}{\sqrt{sin(2x)}}`

Explanation:

Chain rule:

Firstly, let `y=\sqrtsin(2x)`

and let `u=sin(2x)`

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

Therefore `\frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}`

`\frac{dy}{dx}=\frac{1}{2}u^{-\frac{1}{2}} * 2cos(2x)`

Which goes to

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

bring back `u=sin2x`

to get

`\frac{dy}{dx}=\frac{cos2x}{sqrt{sin2x}}`