Answer
$$\frac{{dy}}{{dx}} = \frac{2}{{\left( {\ln 7} \right)\left( {4x - 3} \right)}}$$
Work Step by Step
$$\eqalign{
& y = {\log _7}\sqrt {4x - 3} \cr
& {\text{write the radical as }}{\left( {4x - 3} \right)^{1/2}} \cr
& y = {\log _7}{\left( {4x - 3} \right)^{1/2}} \cr
& {\text{power property of logarithms}} \cr
& y = \frac{1}{2}{\log _7}\left( {4x - 3} \right) \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\frac{1}{2}{{\log }_7}\left( {4x - 3} \right)} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\frac{d}{{dx}}\left( {{{\log }_7}\left( {4x - 3} \right)} \right) \cr
& {\text{use the derivative }}{\log _a}x,\,\,\,\,\frac{d}{{dx}}\left[ {{{\log }_a}g\left( x \right)} \right] = \frac{1}{{\ln a}}\left( {\frac{{g'\left( x \right)}}{{g\left( x \right)}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\left[ {\frac{1}{{\ln 7}}\left( {\frac{{\left( {4x - 3} \right)'}}{{4x - 3}}} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\left[ {\frac{1}{{\ln 7}}\left( {\frac{4}{{4x - 3}}} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \frac{2}{{\left( {\ln 7} \right)\left( {4x - 3} \right)}} \cr} $$
