Answer
$$\frac{{dy}}{{dx}} = \frac{1}{{x + 1}} + \frac{2}{{x + 2}} - \frac{3}{{x + 3}} - \frac{4}{{x + 4}}$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {\frac{{\left( {x + 1} \right){{\left( {x + 2} \right)}^2}}}{{{{\left( {x + 3} \right)}^3}{{\left( {x + 4} \right)}^4}}}} \right) \cr
& {\text{Using algebraic properties of the natural logarithm function}} \cr
& y = \ln \left( {\left( {x + 1} \right){{\left( {x + 2} \right)}^2}} \right) - \ln \left( {{{\left( {x + 3} \right)}^3}{{\left( {x + 4} \right)}^4}} \right) \cr
& y = \ln \left( {x + 1} \right) + \ln {\left( {x + 2} \right)^2} - \ln {\left( {x + 3} \right)^3} - \ln {\left( {x + 4} \right)^4} \cr
& y = \ln \left( {x + 1} \right) + 2\ln \left( {x + 2} \right) - 3\ln \left( {x + 3} \right) - 4\ln \left( {x + 4} \right) \cr
& {\text{Diffentiate both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{1}{{x + 1}} + \frac{2}{{x + 2}} - \frac{3}{{x + 3}} - \frac{4}{{x + 4}} \cr} $$
