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