Answer
$$\frac{{dy}}{{dt}} = - \frac{{\ln t}}{{{t^2}}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{1 + \ln t}}{t} \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\frac{{1 + \ln t}}{t}} \right] \cr
& {\text{use the quotient rule}} \cr
& \frac{{dy}}{{dt}} = \frac{{t\frac{d}{{dt}}\left[ {1 + \ln t} \right] - \left( {1 + \ln t} \right)\frac{d}{{dt}}\left[ t \right]}}{{{t^2}}} \cr
& {\text{solve the derivatives}} \cr
& \frac{{dy}}{{dt}} = \frac{{t\left( {0 + \frac{1}{t}} \right) - \left( {1 + \ln t} \right)\left( 1 \right)}}{{{t^2}}} \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{dt}} = \frac{{1 - 1 - \ln t}}{{{t^2}}} \cr
& \frac{{dy}}{{dt}} = - \frac{{\ln t}}{{{t^2}}} \cr} $$
