Answer
$$y'=\ln\sqrt t+\frac{1}{2}$$
Work Step by Step
$$y=t\ln\sqrt t$$
Recall the following Derivative Rules: $$\frac{d}{dt}(\ln u)=\frac{1}{u}\frac{du}{dt}$$
Therefore, we have $$y'=\Big(t\ln\sqrt t\Big)'=(t)'\ln\sqrt t+t(\ln\sqrt t)'$$
$$y'=\ln\sqrt t+t\times\frac{1}{\sqrt t}\times(\sqrt t)'$$
$$y'=\ln\sqrt t+\sqrt t\times\frac{1}{2\sqrt t}$$
$$y'=\ln\sqrt t+\frac{1}{2}$$