Answer
$$g'\left( x \right) = - \frac{2}{{\left( {\ln 5} \right)\left( x \right)}} + \frac{1}{{2\ln 5\left( {1 - x} \right)}}$$
Work Step by Step
$$\eqalign{
& g\left( x \right) = {\log _5}\frac{4}{{{x^2}\sqrt {1 - x} }} \cr
& {\text{Using logarithmic properties}} \cr
& g\left( x \right) = {\log _5}4 - {\log _5}{x^2}\sqrt {1 - x} \cr
& g\left( x \right) = {\log _5}4 - {\log _5}{x^2} - {\log _5}\sqrt {1 - x} \cr
& {\text{Differentiate both sides with respect to }}x \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {{{\log }_5}4 - {{\log }_5}{x^2} - {{\log }_5}\sqrt {1 - x} } \right] \cr
& {\text{Apply }}\frac{d}{{dx}}\left[ {{{\log }_a}u} \right] = \frac{1}{{\left( {\ln a} \right)u}}\frac{{du}}{{dx}} \cr
& g'\left( x \right) = 0 - \frac{1}{{\left( {\ln 5} \right)\left( {{x^2}} \right)}}\left( {2x} \right) - \frac{1}{{\ln 5\left( {\sqrt {1 - x} } \right)}}\left( { - \frac{1}{{2\sqrt {1 - x} }}} \right) \cr
& {\text{Simplifying}} \cr
& g'\left( x \right) = - \frac{2}{{\left( {\ln 5} \right)\left( x \right)}} + \frac{1}{{2\ln 5\left( {1 - x} \right)}} \cr} $$
