Answer
$$y'=\frac{4}{(2\sqrt x+1)^3}$$
Work Step by Step
$$y=\Big(\frac{2\sqrt x}{2\sqrt x+1}\Big)^2$$
According to the Chain Rule, we have $$y'=\frac{d}{dx}\Big(\frac{2\sqrt x}{2\sqrt x+1}\Big)^2$$ $$y'=\frac{d}{d\Big(\frac{2\sqrt x}{2\sqrt x+1}\Big)}\Big(\frac{2\sqrt x}{2\sqrt x+1}\Big)^2\times\frac{d}{dx}\Big(\frac{2\sqrt x}{2\sqrt x+1}\Big)$$ $$y'=2\Big(\frac{2\sqrt x}{2\sqrt x+1}\Big)\times\frac{d}{dx}\Big(\frac{2\sqrt x}{2\sqrt x+1}\Big)$$ $$y'=\frac{4\sqrt x}{2\sqrt x+1}\times\frac{d}{dx}\Big(\frac{2\sqrt x}{2\sqrt x+1}\Big)$$
We calculate $\frac{d}{dx}\Big(\frac{2\sqrt x}{2\sqrt x+1}\Big)$ separately, using the Quotient Rule: $$\frac{d}{dx}\Big(\frac{2\sqrt x}{2\sqrt x+1}\Big)=\Big(\frac{2\sqrt x}{2\sqrt x+1}\Big)'=\frac{(2\sqrt x)'(2\sqrt x+1)-2\sqrt x(2\sqrt x+1)'}{(2\sqrt x+1)^2}$$
$$=\frac{\frac{1}{\sqrt x}(2\sqrt x+1)-2\sqrt x\Big(\frac{1}{\sqrt x}\Big)}{(2\sqrt x+1)^2}=\frac{\frac{1}{\sqrt x}(2\sqrt x+1-2\sqrt x)}{(2\sqrt x+1)^2}$$
$$=\frac{\frac{1}{\sqrt x}\times1}{(2\sqrt x+1)^2}=\frac{1}{\sqrt x(2\sqrt x+1)^2}$$
Therefore, $$y'=\frac{4\sqrt x}{2\sqrt x+1}\times\frac{1}{\sqrt x(2\sqrt x+1)^2}$$ $$y'=\frac{4}{(2\sqrt x+1)^3}$$
