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