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