Answer
$y' = \frac{a}{x} - \ln b$
Work Step by Step
$$\eqalign{
& y = \ln \frac{{{x^a}}}{{{b^x}}} \cr
& {\text{Use the logarithmic property }}\ln \left( {\frac{a}{b}} \right) = \ln a - \ln b \cr
& y = \ln {x^a} - \ln {b^x} \cr
& {\text{Use the logarithmic property }}\ln {u^n} = n\ln u \cr
& y = a\ln x - x\ln b \cr
& {\text{Differentiate both sides}} \cr
& y' = \frac{d}{{dx}}\left[ {a\ln x} \right] - \frac{d}{{dx}}\left[ {x\ln b} \right] \cr
& {\text{Pull out the constants}} \cr
& y' = a\frac{d}{{dx}}\left[ {\ln x} \right] - \ln b\frac{d}{{dx}}\left[ x \right] \cr
& {\text{Compute derivatives}} \cr
& y' = a\left( {\frac{1}{x}} \right) - \ln b\left( 1 \right) \cr
& y' = \frac{a}{x} - \ln b \cr} $$
