Answer
$$y'=\frac{1}{2y(x+1)^2}$$
Work Step by Step
$$y^2=\frac{x}{x+1}$$
Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$2y\times y'=\frac{x'(x+1)-x(x+1)'}{(x+1)^2}$$
$$2yy'=\frac{x+1-x\times1}{(x+1)^2}$$
$$2yy'=\frac{x+1-x}{(x+1)^2}$$
$$2yy'=\frac{1}{(x+1)^2}$$
Finally, we calculate $y'$: $$y'=\frac{1}{2y(x+1)^2}$$
