Answer
$$\frac{{dy}}{{dx}} = \frac{{4x + 7 - 4\ln x}}{{x{{\left( {4x + 7} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{\ln x}}{{4x + 7}} \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{\ln x}}{{4x + 7}}} \right] \cr
& {\text{by using quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {4x + 7} \right)\left( {\ln x} \right)' - \left( {\ln x} \right)\left( {4x + 7} \right)'}}{{{{\left( {4x + 7} \right)}^2}}} \cr
& {\text{compute derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {4x + 7} \right)\left( {1/x} \right) - \left( {\ln x} \right)\left( 4 \right)}}{{{{\left( {4x + 7} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{4 + 7/x - 4\ln x}}{{{{\left( {4x + 7} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{4x + 7 - 4\ln x}}{{x{{\left( {4x + 7} \right)}^2}}} \cr} $$
