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