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

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Answer 1 · 2018-06-10T17:07:16

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

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

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

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