I'm thinking you want something that looks like this [tex]\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}[/tex] or it could be like this [tex]\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)[/tex]
use chain rule dy/dx f(g(x))=f'(g(x))g'(x) and √x=x^(1/2) so I'll do the first one first [tex]\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}[/tex]= [tex]\frac{dy}{dx}(x+\frac{1}{4}sin^2(2x))^{\frac{1}{2}}[/tex]= [tex]\frac{1}{2(x+\frac{1}{4}sin^2(2x))^{\frac{1}{2}}}(1+sin(2x)cos(2x))[/tex]= [tex]\frac{1+sin(2x)cos(2x)}{2\sqrt{x+\frac{1}{4}sin^2(2x}}[/tex]
the 2nd one is a bit simpler [tex]\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)[/tex]= [tex]\frac{dy}{dx}x^{\frac{1}{2}}+\frac{1}{4}sin^2(2x)[/tex]= [tex]\frac{1}{2}x^{\frac{-1}{2}}+\frac{1}{4}sin^2(2x)[/tex]= [tex]\frac{1}{2x^{\frac{1}{2}}}+sin(2x)cos(2x)[/tex]= [tex]\frac{1}{2\sqrt{x}}+sin(2x)cos(2x)[/tex]
in conclusion [tex]\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}[/tex]=[tex]\frac{1+sin(2x)cos(2x)}{2\sqrt{x+\frac{1}{4}sin^2(2x}}[/tex] and [tex]\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)[/tex]=[tex]\frac{1}{2\sqrt{x}}+sin(2x)cos(2x)[/tex] depends which one it was
