Answer
$$y'=\frac{1}{2\sqrt{\frac{x}{x+1}}(x+1)^2}$$
Work Step by Step
$$y'=\left(\sqrt{\frac{x}{x+1}}\right)'=\frac{1}{2\sqrt{\frac{x}{x+1}}}\left(\frac{x}{x+1}\right)'=\frac{1}{2\sqrt{\frac{x}{x+1}}}\frac{x'(x+1)-x(x+1)'}{(x+1)^2}=\frac{1}{2\sqrt{\frac{x}{x+1}}}\frac{1\cdot(x+1)-x\cdot1}{(x+1)^2}=\frac{1}{2\sqrt{\frac{x}{x+1}}(x+1)^2}$$
