Answer
$$ y'=\frac{-2\sin 2x+4\cos 4x}{2\sqrt{\cos 2x +\sin4x}}.$$
Work Step by Step
Recall that $(\sin x)'=\cos x$.
Recall that $(\cos x)'=-\sin x$.
Recall that $(x^n)'=nx^{n-1}$
Since $ y=\sqrt{\cos 2x +\sin4x}$, by using the chain rule, the derivative $ y'$ is given by
$$ y'=\frac{1}{2}(\cos 2x +\sin4x)^{-1/2}(-2\sin 2x+4\cos 4x)\\=\frac{-2\sin 2x+4\cos 4x}{2\sqrt{\cos 2x +\sin4x}}.$$
