Answer
$y'=\dfrac{1}{2\sqrt{x}}+2x\cos{(2x)}^2.$
Work Step by Step
$y=f(x)+g(x)\rightarrow f(x)=\sqrt{x}$ ; $g(x)=\frac{1}{4}\sin{(2x)}^2$
$f'(x)=\dfrac{1}{2\sqrt{x}}.$
$g'(x)$ is found using the Chain Rule:
$u=(2x)^2$; $\dfrac{du}{dx}=8x$
$g(u)=\frac{1}{4}\sin{u};\dfrac{d}{du}g(u)=\frac{1}{4}\cos{u}$
$\dfrac{d}{dx}g(x)=\dfrac{d}{du}g(u)\times\dfrac{du}{dx}=(8x)(\frac{1}{4}\cos{(2x)}^2)=2x\cos{(2x)}^2.$
$y'=f'(x)+g'(x)=\dfrac{1}{2\sqrt{x}}+2x\cos{(2x)}^2.$
