Answer
$$\frac{{dy}}{{dx}} = 1 + \frac{2}{{{{\left( {x + 1} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = x + 1 + 2\left( {\frac{x}{{x + 1}}} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {x + 1 + 2\left( {\frac{x}{{x + 1}}} \right)} \right] \cr
& {\text{Use sum rule}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ x \right] + \frac{d}{{dx}}\left[ 1 \right] + \frac{d}{{dx}}\left[ {2\left( {\frac{x}{{x + 1}}} \right)} \right] \cr
& {\text{Use power rule and quotient rule}} \cr
& \frac{{dy}}{{dx}} = 1 + 0 + 2\left( {\frac{{\left( {x + 1} \right)\left( 1 \right) - x\left( 1 \right)}}{{{{\left( {x + 1} \right)}^2}}}} \right) \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = 1 + 2\left( {\frac{{x + 1 - x}}{{{{\left( {x + 1} \right)}^2}}}} \right) \cr
& \frac{{dy}}{{dx}} = 1 + \frac{2}{{{{\left( {x + 1} \right)}^2}}} \cr} $$
