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