Answer
$$h'\left( x \right) = - \frac{1}{{x\left( {2x - 1} \right)}}$$
Work Step by Step
$$\eqalign{
& h\left( x \right) = \ln \left( {\frac{{9x}}{{4x - 2}}} \right) \cr
& {\text{Using logarithmic properties }}\ln \left( {\frac{a}{b}} \right) = \ln \left( a \right) - \ln \left( b \right) \cr
& h\left( x \right) = \ln \left( {9x} \right) - \ln \left( {4x - 2} \right) \cr
& {\text{Differentiate}} \cr
& h'\left( x \right) = \left[ {\ln \left( {9x} \right)} \right]' - \left[ {\ln \left( {4x - 2} \right)} \right]' \cr
& h'\left( x \right) = \frac{9}{{9x}} - \frac{4}{{4x - 2}} \cr
& {\text{Simplifying}} \cr
& h'\left( x \right) = \frac{1}{x} - \frac{4}{{4x - 2}} \cr
& h'\left( x \right) = \frac{{4x - 2 - 4x}}{{x\left( {4x - 2} \right)}} \cr
& h'\left( x \right) = - \frac{2}{{x\left( {4x - 2} \right)}} \cr
& h'\left( x \right) = - \frac{2}{{2x\left( {2x - 1} \right)}} \cr
& h'\left( x \right) = - \frac{1}{{x\left( {2x - 1} \right)}} \cr} $$
