Answer
$$\frac{{dy}}{{dt}} = \frac{2}{{t{{\left( {1 - \ln t} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{1 + \ln t}}{{1 - \ln t}} \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\frac{{1 + \ln t}}{{1 - \ln t}}} \right] \cr
& {\text{use the quotient rule}} \cr
& \frac{{dy}}{{dt}} = \frac{{\left( {1 - \ln t} \right)\left( {1 + \ln t} \right)' - \left( {1 + \ln t} \right)\left( {1 - \ln t} \right)'}}{{{{\left( {1 - \ln t} \right)}^2}}} \cr
& {\text{solve the derivatives and simplify}} \cr
& \frac{{dy}}{{dt}} = \frac{{\left( {1 - \ln t} \right)\left( {\frac{1}{t}} \right) - \left( {1 + \ln t} \right)\left( { - \frac{1}{t}} \right)}}{{{{\left( {1 - \ln t} \right)}^2}}} \cr
& \frac{{dy}}{{dt}} = \frac{{\frac{{1 - \ln t}}{t} + \frac{{1 + \ln t}}{t}}}{{{{\left( {1 - \ln t} \right)}^2}}} \cr
& \frac{{dy}}{{dt}} = \frac{{1 - \ln t + 1 + \ln t}}{{t{{\left( {1 - \ln t} \right)}^2}}} \cr
& \frac{{dy}}{{dt}} = \frac{2}{{t{{\left( {1 - \ln t} \right)}^2}}} \cr} $$
