Answer
$$\frac{{dy}}{{dx}} = \frac{{20x - 5}}{{\left( {2\ln 2} \right)\left( {2{x^2} - x} \right)}}$$
Work Step by Step
$$\eqalign{
& y = {\log _2}{\left( {2{x^2} - x} \right)^{5/2}} \cr
& {\text{power property of logarithms}} \cr
& y = \frac{5}{2}{\log _2}\left( {2{x^2} - x} \right) \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\frac{5}{2}{{\log }_2}\left( {2{x^2} - x} \right)} \right) \cr
& \frac{{dy}}{{dx}} = \frac{5}{2}\frac{d}{{dx}}\left( {{{\log }_2}\left( {2{x^2} - x} \right)} \right) \cr
& {\text{use derivative }}{\log _a}x,\,\,\,\,\frac{d}{{dx}}\left[ {{{\log }_a}g\left( x \right)} \right] = \frac{1}{{\ln a}}\left( {\frac{{g'\left( x \right)}}{{g\left( x \right)}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{5}{2}\left[ {\frac{1}{{\ln 2}}\left( {\frac{{\left( {2{x^2} - x} \right)'}}{{2{x^2} - x}}} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \frac{5}{2}\left[ {\frac{1}{{\ln 2}}\left( {\frac{{4x - 1}}{{2{x^2} - x}}} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \frac{{20x - 5}}{{\left( {2\ln 2} \right)\left( {2{x^2} - x} \right)}} \cr} $$