Answer
$$g'\left( x \right) = - \frac{1}{{\ln 3\left( {2 - 2x} \right)}}$$
Work Step by Step
$$\eqalign{
& g\left( x \right) = {\log _3}\sqrt {1 - x} \cr
& {\text{Use the logarithmic properties}} \cr
& g\left( x \right) = \frac{{\ln \sqrt {1 - x} }}{{\ln 3}} \cr
& {\text{Differentiate}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{\ln \sqrt {1 - x} }}{{\ln 3}}} \right] \cr
& g'\left( x \right) = \frac{1}{{\ln 3}}\frac{d}{{dx}}\left[ {\ln \sqrt {1 - x} } \right] \cr
& g'\left( x \right) = \frac{1}{{\ln 3}}\left( {\frac{1}{{\sqrt {1 - x} }}} \right)\frac{d}{{dx}}\left[ {\sqrt {1 - x} } \right] \cr
& g'\left( x \right) = \frac{1}{{\ln 3}}\left( {\frac{1}{{\sqrt {1 - x} }}} \right)\left( {\frac{{ - 1}}{{2\sqrt {1 - x} }}} \right) \cr
& {\text{Simplifying}} \cr
& g'\left( x \right) = - \frac{1}{{2\ln 3\left( {1 - x} \right)}} \cr
& g'\left( x \right) = - \frac{1}{{\ln 3\left( {2 - 2x} \right)}} \cr} $$
