Answer
$\frac{{dy}}{{dx}} = \frac{{\sqrt {x + 1} {{\left( {2 - x} \right)}^5}}}{{{{\left( {x + 3} \right)}^7}}}\left[ {\frac{1}{{2\left( {x + 1} \right)}} + \frac{5}{{x - 2}} - \frac{7}{{x + 3}}} \right]$
Work Step by Step
$$\eqalign{
& y = \frac{{\sqrt {x + 1} {{\left( {2 - x} \right)}^5}}}{{{{\left( {x + 3} \right)}^7}}} \cr
& {\text{Use radical properties}} \cr
& y = \frac{{{{\left( {x + 1} \right)}^{1/2}}{{\left( {2 - x} \right)}^5}}}{{{{\left( {x + 3} \right)}^7}}} \cr
& {\text{Differentiate by using the Logarithmic Differentiation }} \cr
& {\text{First}}{\text{, take logarithms of both sides of the equation}} \cr
& \ln y = \ln \left( {\frac{{{{\left( {x + 1} \right)}^{1/2}}{{\left( {2 - x} \right)}^5}}}{{{{\left( {x + 3} \right)}^7}}}} \right) \cr
& {\text{Use the Laws of Logarithms to simplify}} \cr
& \ln y = \ln {\left( {x + 1} \right)^{1/2}}{\left( {2 - x} \right)^5} - \ln {\left( {x + 3} \right)^7} \cr
& \ln y = \ln {\left( {x + 1} \right)^{1/2}} + \ln {\left( {2 - x} \right)^5} - \ln {\left( {x + 3} \right)^7} \cr
& \ln y = \frac{1}{2}\ln \left( {x + 1} \right) + 5\ln \left( {2 - x} \right) - 7\ln \left( {x + 3} \right) \cr
& {\text{Taking the derivative on both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {\ln y} \right] = \frac{d}{{dx}}\left[ {\frac{1}{2}\ln \left( {x + 1} \right)} \right] + \frac{d}{{dx}}\left[ {5\ln \left( {2 - x} \right)} \right] - \frac{d}{{dx}}\left[ {7\ln \left( {x + 3} \right)} \right] \cr
& {\text{Compute the derivatives}} \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{2}\left( {\frac{1}{{x + 1}}} \right) + 5\left( {\frac{{ - 1}}{{2 - x}}} \right) - 7\left( {\frac{1}{{x + 3}}} \right) \cr
& \frac{1}{y}\frac{{dy}}{{dx}} = \frac{1}{{2\left( {x + 1} \right)}} + \frac{5}{{x - 2}} - \frac{7}{{x + 3}} \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = y\left[ {\frac{1}{{2\left( {x + 1} \right)}} + \frac{5}{{x - 2}} - \frac{7}{{x + 3}}} \right] \cr
& {\text{Substitute }}\frac{{\sqrt {x + 1} {{\left( {2 - x} \right)}^5}}}{{{{\left( {x + 3} \right)}^7}}}{\text{ for }}y \cr
& \frac{{dy}}{{dx}} = \frac{{\sqrt {x + 1} {{\left( {2 - x} \right)}^5}}}{{{{\left( {x + 3} \right)}^7}}}\left[ {\frac{1}{{2\left( {x + 1} \right)}} + \frac{5}{{x - 2}} - \frac{7}{{x + 3}}} \right] \cr} $$
