Answer
$$Q'\left( t \right) = {t^{1/t}}\left( {\frac{{1 - \ln t}}{{{t^2}}}} \right)$$
Work Step by Step
$$\eqalign{
& Q\left( t \right) = {t^{1/t}} \cr
& {\text{we use the inverse relationship }}{e^{\ln x}} = x \cr
& Q\left( t \right) = {e^{\ln {t^{1/t}}}} \cr
& Q\left( t \right) = {e^{\frac{1}{t}\ln t}} \cr
& {\text{evaluate the derivative}} \cr
& Q'\left( t \right) = \frac{d}{{dt}}\left( {{e^{\frac{1}{t}\ln t}}} \right) \cr
& {\text{by }}\frac{d}{{dx}}\left( {{e^{u\left( x \right)}}} \right) = {e^{u\left( x \right)}}u'\left( x \right) \cr
& Q'\left( t \right) = {e^{\frac{1}{t}\ln t}}\frac{d}{{dt}}\left( {\frac{1}{t}\ln t} \right) \cr
& {\text{product rule}} \cr
& Q'\left( t \right) = {e^{\frac{1}{t}\ln t}}\left( {\frac{1}{t}\left( {\frac{1}{t}} \right) - \frac{{\ln t}}{{{t^2}}}} \right) \cr
& {\text{simplify}} \cr
& Q'\left( t \right) = {t^{1/t}}\left( {\frac{{1 - \ln t}}{{{t^2}}}} \right) \cr} $$
