Answer
$$\frac{{dy}}{{dx}} = - \frac{{3x + 2}}{{2x\left( {x + 1} \right)}}$$
Work Step by Step
$$\eqalign{
& y = \ln \frac{1}{{x\sqrt {x + 1} }} \cr
& {\text{ write }}\sqrt {x + 1} {\text{ as }}{\left( {x + 1} \right)^{1/2}} \cr
& y = \ln \frac{1}{{x{{\left( {x + 1} \right)}^{1/2}}}} \cr
& {\text{use the quotient property for logarithms}} \cr
& y = \ln 1 - \ln x{\left( {x + 1} \right)^{1/2}} \cr
& y = - \ln x{\left( {x + 1} \right)^{1/2}} \cr
& {\text{use the product property for logarithms}} \cr
& y = - \ln x - \ln {\left( {x + 1} \right)^{1/2}} \cr
& y = - \ln x - \frac{1}{2}\ln \left( {x + 1} \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = - \frac{d}{{dx}}\left[ {\ln x} \right] - \frac{1}{2}\frac{d}{{dx}}\left[ {\ln \left( {x + 1} \right)} \right] \cr
& {\text{solve the derivatives and simplify}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{x} - \frac{1}{2}\left( {\frac{1}{{x + 1}}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{x} - \frac{1}{{2\left( {x + 1} \right)}} \cr
& \frac{{dy}}{{dx}} = \frac{{ - 2x - 2 - x}}{x} \cr
& \frac{{dy}}{{dx}} = - \frac{{3x + 2}}{{2x\left( {x + 1} \right)}} \cr} $$
