Answer
$$\frac{{10}}{9}$$
Work Step by Step
$$\eqalign{
& {\left. {\frac{d}{{dx}}\left( {\frac{{xf\left( x \right)}}{{g\left( x \right)}}} \right)} \right|_{x = 4}} \cr
& {\text{Use the quotient rule}} \cr
& \frac{d}{{dx}}\left( {\frac{{xf\left( x \right)}}{{g\left( x \right)}}} \right) = \frac{{g\left( x \right)\left[ {xf'\left( x \right) + f\left( x \right)} \right] - xf\left( x \right)g'\left( x \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}} \cr
& {\text{Evaluate at }}x = 4 \cr
& {\left. {\frac{d}{{dx}}\left( {\frac{{xf\left( x \right)}}{{g\left( x \right)}}} \right)} \right|_{x = 4}} = \frac{{g\left( 4 \right)\left[ {4f'\left( 4 \right) + f\left( 4 \right)} \right] - 4f\left( 4 \right)g'\left( 4 \right)}}{{{{\left[ {g\left( 4 \right)} \right]}^2}}} \cr
& {\text{From the table we have that }}f'\left( 4 \right) = 1{\text{ and }}f\left( 4 \right) = 2.{\text{ }} \cr
& g'\left( 4 \right) = 1{\text{ and }}g\left( 4 \right) = 3.{\text{ }}\,\,{\text{Then}}{\text{,}} \cr
& {\left. {\frac{d}{{dx}}\left( {\frac{{xf\left( x \right)}}{{g\left( x \right)}}} \right)} \right|_{x = 4}} = \frac{{\left( 3 \right)\left[ {4\left( 1 \right) + 2} \right] - 4\left( 2 \right)\left( 1 \right)}}{{{{\left( 3 \right)}^2}}} \cr
& {\left. {\frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{x + 2}}} \right)} \right|_{x = 4}} = \frac{{10}}{9} \cr} $$
