Answer
$\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x {{\left( {\sqrt x + 1} \right)}^2}}}$
Work Step by Step
$$\eqalign{
& y = \frac{{\sqrt x }}{{\sqrt x + 1}} \cr
& {\text{Differentiating}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{\sqrt x }}{{\sqrt x + 1}}} \right] \cr
& {\text{Differentiate by using the formula }}\left( {\frac{f}{g}} \right)' = \frac{{gf' - fg'}}{{{g^2}}} \cr
& {\text{Let }}f = \sqrt x {\text{ and }}g = \sqrt x + 1,{\text{ then}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x } \right)' - \sqrt x \left( {\sqrt x + 1} \right)'}}{{{{\left( {\sqrt x + 1} \right)}^2}}} \cr
& {\text{Compute derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {\sqrt x + 1} \right)\left( {\frac{1}{{2\sqrt x }}} \right) - \sqrt x \left( {\frac{1}{{2\sqrt x }}} \right)}}{{{{\left( {\sqrt x + 1} \right)}^2}}} \cr
& {\text{Multiply and simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{1}{2} + \frac{1}{{2\sqrt x }} - \frac{1}{2}}}{{{{\left( {\sqrt x + 1} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x {{\left( {\sqrt x + 1} \right)}^2}}} \cr} $$
