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