Answer
$$\frac{{dy}}{{dx}} = \frac{{10x}}{{{x^2} + 1}} + \frac{1}{{2\left( {1 - x} \right)}}$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {\frac{{{{\left( {{x^2} + 1} \right)}^5}}}{{\sqrt {1 - x} }}} \right) \cr
& {\text{write the radical }}\sqrt {1 - x} {\text{ as }}{\left( {1 - x} \right)^{1/2}} \cr
& y = \ln \left( {\frac{{{{\left( {{x^2} + 1} \right)}^5}}}{{{{\left( {1 - x} \right)}^{1/2}}}}} \right) \cr
& {\text{use the quotient property for logarithms}} \cr
& y = \ln {\left( {{x^2} + 1} \right)^5} - \ln {\left( {1 - x} \right)^{1/2}} \cr
& {\text{use the power property for logarithms}} \cr
& y = 5\ln \left( {{x^2} + 1} \right) - \frac{1}{2}\ln \left( {1 - x} \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = 5\frac{d}{{dx}}\left[ {\ln \left( {{x^2} + 1} \right)} \right] - \frac{1}{2}\frac{d}{{dx}}\left[ {\ln \left( {1 - x} \right)} \right] \cr
& {\text{solve the derivatives using }}\frac{d}{{dx}}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = 5\left( {\frac{{2x}}{{{x^2} + 1}}} \right) - \frac{1}{2}\left( {\frac{{ - 1}}{{1 - x}}} \right) \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{{10x}}{{{x^2} + 1}} + \frac{1}{{2\left( {1 - x} \right)}} \cr} $$
