# Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 146: 64

$$y'=\frac{-2\sin 2x+4\cos 4x}{2\sqrt{\cos 2x +\sin4x}}.$$

#### 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 $$y'=\frac{1}{2}(\cos 2x +\sin4x)^{-1/2}(-2\sin 2x+4\cos 4x)\\=\frac{-2\sin 2x+4\cos 4x}{2\sqrt{\cos 2x +\sin4x}}.$$

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