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

1 Answer
Jun 10, 2018

Answer:

#\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}}#