Answer
$$\frac{{dy}}{{dx}} = \frac{{6x + 14}}{{2x - 1}} + 3\ln \left( {2x - 1} \right)$$
Work Step by Step
$$\eqalign{
& y = \left( {3x + 7} \right)\ln \left( {2x - 1} \right) \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\left( {3x + 7} \right)\ln \left( {2x - 1} \right)} \right] \cr
& {\text{by using product rule}} \cr
& \frac{{dy}}{{dx}} = \left( {3x + 7} \right)\frac{d}{{dx}}\left[ {\ln \left( {2x - 1} \right)} \right] + \ln \left( {2x - 1} \right)\frac{d}{{dx}}\left[ {\left( {3x + 7} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \left( {3x + 7} \right)\left( {\frac{2}{{2x - 1}}} \right) + \ln \left( {2x - 1} \right)\left( 3 \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{6x + 14}}{{2x - 1}} + 3\ln \left( {2x - 1} \right) \cr} $$
