Answer
$$\frac{{dy}}{{dx}} = \frac{{x + 2}}{{{{\left( {x + 1} \right)}^2}}} + \frac{x}{{x + 1}}$$
Work Step by Step
$$\eqalign{
& y = \left( {x + 2} \right)\left( {\frac{x}{{x + 1}}} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\left( {x + 2} \right)\left( {\frac{x}{{x + 1}}} \right)} \right] \cr
& {\text{Use the product rule}} \cr
& \frac{{dy}}{{dx}} = \left( {x + 2} \right)\frac{d}{{dx}}\left[ {\frac{x}{{x + 1}}} \right] + \frac{x}{{x + 1}}\frac{d}{{dx}}\left[ {x + 2} \right] \cr
& {\text{Use quotient rule}} \cr
& \frac{{dy}}{{dx}} = \left( {x + 2} \right)\left( {\frac{{\left( {x + 1} \right)\left( 1 \right) - x\left( 1 \right)}}{{{{\left( {x + 1} \right)}^2}}}} \right) + \frac{x}{{x + 1}}\left( 1 \right) \cr
& \frac{{dy}}{{dx}} = \left( {x + 2} \right)\left( {\frac{{x + 1 - x}}{{{{\left( {x + 1} \right)}^2}}}} \right) + \frac{x}{{x + 1}} \cr
& \frac{{dy}}{{dx}} = \left( {x + 2} \right)\left( {\frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right) + \frac{x}{{x + 1}} \cr
& \frac{{dy}}{{dx}} = \frac{{x + 2}}{{{{\left( {x + 1} \right)}^2}}} + \frac{x}{{x + 1}} \cr} $$
