Answer
$$g'\left( x \right) = - \frac{1}{{x\left( {x - 1} \right)\ln 5}}$$
Work Step by Step
$$\eqalign{
& h\left( x \right) = {\log _5}\frac{x}{{x - 1}} \cr
& {\text{Use the logarithmic properties}} \cr
& g\left( x \right) = \frac{{\ln \frac{x}{{x - 1}}}}{{\ln 5}} \cr
& {\text{Differentiate}} \cr
& g'\left( x \right) = \frac{1}{{\ln 5}}\frac{d}{{dx}}\left[ {\ln \frac{x}{{x - 1}}} \right] \cr
& g'\left( x \right) = \frac{1}{{\ln 5}}\left( {\frac{{x - 1}}{x}} \right)\frac{d}{{dx}}\left[ {\frac{x}{{x - 1}}} \right] \cr
& {\text{Use quotient rule}} \cr
& g'\left( x \right) = \frac{1}{{\ln 5}}\left( {\frac{{x - 1}}{x}} \right)\left( {\frac{{x - 1 - x}}{{{{\left( {x - 1} \right)}^2}}}} \right) \cr
& {\text{Simplifying}} \cr
& g'\left( x \right) = - \frac{1}{{\ln 5}}\left( {\frac{1}{x}} \right)\left( {\frac{1}{{x - 1}}} \right) \cr
& g'\left( x \right) = - \frac{1}{{x\left( {x - 1} \right)\ln 5}} \cr} $$
