Answer
$g'\left( t \right) = \frac{1}{t} + \frac{{8t}}{{{t^2} + 1}} - \frac{2}{{3\left( {2t - 1} \right)}}$
Work Step by Step
$$\eqalign{
& g\left( t \right) = \ln \frac{{t{{\left( {{t^2} + 1} \right)}^4}}}{{\root 3 \of {2t - 1} }} \cr
& {\text{First we must rewrite the given function using logarithmic}} \cr
& {\text{properties:}} \cr
& {\text{Use the logaritmic property ln}}\left( {\frac{a}{b}} \right) = \ln a - \ln b \cr
& g\left( t \right) = \ln t{\left( {{t^2} + 1} \right)^4} - \ln \root 3 \of {2t - 1} \cr
& {\text{Use the logaritmic property ln}}\left( {ab} \right) = \ln a + \ln b \cr
& g\left( t \right) = \ln t + \ln {\left( {{t^2} + 1} \right)^4} - \ln {\left( {2t - 1} \right)^{1/3}} \cr
& {\text{Use the logaritmic property }}\ln {a^n} = n\ln a \cr
& g\left( t \right) = \ln t + 4\ln \left( {{t^2} + 1} \right) - \frac{1}{3}\ln \left( {2t - 1} \right) \cr
& {\text{Differentiate}} \cr
& g'\left( t \right) = \frac{d}{{dt}}\left[ {\ln t} \right] + 4\frac{d}{{dt}}\left[ {\ln \left( {{t^2} + 1} \right)} \right] - \frac{1}{3}\frac{d}{{dt}}\left[ {\ln \left( {2t - 1} \right)} \right] \cr
& {\text{Use the formula }}\frac{d}{{dt}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dt}},{\text{ }}\left( {{\text{see page 218}}} \right),{\text{ }} \cr
& g'\left( t \right) = \frac{1}{t} + 4\left( {\frac{{2t}}{{{t^2} + 1}}} \right) - \frac{1}{3}\frac{d}{{dt}}\left[ {\ln \left( {2t - 1} \right)} \right] \cr
& {\text{compute the derivative and simplify}} \cr
& g'\left( t \right) = \frac{1}{t} + 4\left( {\frac{{2t}}{{{t^2} + 1}}} \right) - \frac{1}{3}\left( {\frac{2}{{2t - 1}}} \right) \cr
& g'\left( t \right) = \frac{1}{t} + \frac{{8t}}{{{t^2} + 1}} - \frac{2}{{3\left( {2t - 1} \right)}} \cr} $$
