Answer
$$\frac{{dy}}{{dx}} = \frac{1}{2}\sqrt {x\left( {x + 1} \right)} \left( {\frac{{2x + 1}}{x(x+1)}} \right)$$
Work Step by Step
$$\eqalign{
& y = \sqrt {x\left( {x + 1} \right)} \cr
& y = {x^{1/2}}{\left( {x + 1} \right)^{1/2}} \cr
& {\text{Take the natural log of both sides:}} \cr
& {\text{Use the properties of logarithms:}} \cr
& \ln y = \ln {x^{1/2}}{\left( {x + 1} \right)^{1/2}} \cr
& {\text{product rule:}} \cr
& \ln y = \ln {x^{1/2}} + \ln {\left( {x + 1} \right)^{1/2}} \cr
& {\text{power rule:}} \cr
& \ln y = \frac{1}{2}\ln x + \frac{1}{2}\ln \left( {x + 1} \right) \cr
& {\text{Take derivatives of both sides with respect to }}x \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{2}\left( {\frac{1}{x}} \right) + \frac{1}{2}\left( {\frac{1}{{x + 1}}} \right) \cr
& {\text{solve for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}y\left( {\frac{1}{x} + \frac{1}{{x + 1}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}y\left( {\frac{{2x + 1}}{x(x+1)}} \right) \cr
& {\text{substitute }}\sqrt {x\left( {x + 1} \right)} {\text{ for }}y{\text{ }} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\sqrt {x\left( {x + 1} \right)} \left( {\frac{{2x + 1}}{x(x+1)}} \right) \cr} $$
