Answer
$$dy = \frac{8}{{{{\left( {4 - x} \right)}^2}}}dx$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{4 + x}}{{4 - x}} \cr
& {\text{Differentiate by using the quotient rule}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{4 + x}}{{4 - x}}} \right] \cr
& f'\left( x \right) = \frac{{\left( {4 - x} \right)\left( 1 \right) - \left( {4 + x} \right)\left( { - 1} \right)}}{{{{\left( {4 - x} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{4 - x + 4 + x}}{{{{\left( {4 - x} \right)}^2}}} \cr
& f'\left( x \right) = \frac{8}{{{{\left( {4 - x} \right)}^2}}} \cr
& {\text{Write in the form }}dy = f'\left( x \right)dx \cr
& dy = \frac{8}{{{{\left( {4 - x} \right)}^2}}}dx \cr} $$
