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