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