Answer
$$\frac{7}{8}$$
Work Step by Step
$$\eqalign{ & {\left. {\frac{d}{{dx}}\left( {\frac{{xf\left( x \right)}}{{g\left( x \right)}}} \right)} \right|_{x = 4}} \cr & {\text{Use the quotient rule}} \cr & \frac{d}{{dx}}\left( {\frac{{f\left( x \right)g\left( x \right)}}{x}} \right) = \frac{{x\left[ {g\left( x \right)f'\left( x \right) + g'\left( x \right)f\left( x \right)} \right] - f\left( x \right)g\left( x \right)}}{{{x^2}}} \cr & {\text{Evaluate at }}x = 4 \cr & {\left. {\frac{d}{{dx}}\left( {\frac{{xf\left( x \right)}}{{g\left( x \right)}}} \right)} \right|_{x = 4}} = \frac{{4\left[ {g\left( 4 \right)f'\left( 4 \right) + g'\left( 4 \right)f\left( 4 \right)} \right] - f\left( 4 \right)g\left( 4 \right)}}{{{4^2}}} \cr & {\text{From the table we have that }}f'\left( 4 \right) = 1{\text{ and }}f\left( 4 \right) = 2.{\text{ }} \cr & g'\left( 4 \right) = 1{\text{ and }}g\left( 4 \right) = 3.{\text{ }}\,\,{\text{Then}}{\text{,}} \cr & {\left. {\frac{d}{{dx}}\left( {\frac{{xf\left( x \right)}}{{g\left( x \right)}}} \right)} \right|_{x = 4}} = \frac{{4\left[ {\left( 3 \right)\left( 1 \right) + \left( 1 \right)\left( 2 \right)} \right] - \left( 2 \right)\left( 3 \right)}}{{16}} \cr & {\left. {\frac{d}{{dx}}\left( {\frac{{xf\left( x \right)}}{{g\left( x \right)}}} \right)} \right|_{x = 4}} = \frac{7}{8} \cr} $$
