Answer
$$\frac{d y}{d t}=\left(\frac{\ln t+2}{2 \sqrt{t}}\right) t^ \sqrt{t}$$
Work Step by Step
Given $$ y=(\sqrt{t})^{t} $$
So, we have
\begin{aligned}
&y=t^{\sqrt{t}}=t^{\left(t^{\frac{1}{2}}\right)}\\
& \Rightarrow \ln y=\ln t^{ t^{\frac{1}{2}}} \\
& \Rightarrow \ln y=\left(t^{1 / 2}\right)(\ln t) \\
&\text{ differentiate with respect to } \ t, \ \\
&\Rightarrow \frac{1}{y} \frac{d y}{d t}
=\left(\frac{1}{2} t^{-1 / 2}\right)(\ln t)+t^{1 / 2}\left(\frac{1}{t}\right)\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\ln t+2}{2 \sqrt{t}} \\
&\Rightarrow \frac{d y}{d t}=\left(\frac{\ln t+2}{2 \sqrt{t}}\right) y\\
&\Rightarrow \frac{d y}{d t}=\left(\frac{\ln t+2}{2 \sqrt{t}}\right) t^ \sqrt{t}
& \ \ \ \ \ \ \ \ \ \ \ \
\end{aligned}
