Answer
$$f'\left( x \right) = \frac{2}{{2x - 1}} + \frac{3}{{x + 2}} + \frac{8}{{1 - 4x}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln \frac{{\left( {2x - 1} \right){{\left( {x + 2} \right)}^3}}}{{{{\left( {1 - 4x} \right)}^2}}} \cr
& {\text{use the quotient property for logarithms}} \cr
& f\left( x \right) = \ln \left( {2x - 1} \right){\left( {x + 2} \right)^3} - \ln {\left( {1 - 4x} \right)^2} \cr
& {\text{use the product property for logarithms}} \cr
& f\left( x \right) = \ln \left( {2x - 1} \right) + \ln {\left( {x + 2} \right)^3} - \ln {\left( {1 - 4x} \right)^2} \cr
& {\text{use the power property for logarithms}} \cr
& f\left( x \right) = \ln \left( {2x - 1} \right) + 3\ln \left( {x + 2} \right) - 2\ln \left( {1 - 4x} \right) \cr
& {\text{differentiate both sides}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {2x - 1} \right)} \right] + \frac{d}{{dx}}\left[ {3\ln \left( {x + 2} \right)} \right] - \frac{d}{{dx}}\left[ {2\ln \left( {1 - 4x} \right)} \right] \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {2x - 1} \right)} \right] + 3\frac{d}{{dx}}\left[ {\ln \left( {x + 2} \right)} \right] - 2\frac{d}{{dx}}\left[ {\ln \left( {1 - 4x} \right)} \right] \cr
& {\text{computing derivatives}} \cr
& f'\left( x \right) = \frac{2}{{2x - 1}} + 3\left( {\frac{1}{{x + 2}}} \right) - 2\left( {\frac{{ - 4}}{{1 - 4x}}} \right) \cr
& {\text{simplifying}} \cr
& f'\left( x \right) = \frac{2}{{2x - 1}} + \frac{3}{{x + 2}} + \frac{8}{{1 - 4x}} \cr} $$
