Answer
$$y'=\frac{\ln \left(t^4\right)+4}{2\sqrt{t\ln \left(t^4\right)}}$$
Work Step by Step
Given $$y=\sqrt{t\ln t^4}$$
Then
\begin{align*}
y'&=\frac{1}{2\sqrt{t\ln \left(t^4\right)}}\left(\frac{d}{dt}\left(t\right)\ln \left(t^4\right)+\frac{d}{dt}\left(\ln \left(t^4\right)\right)t\right)\\
&=\frac{\ln \left(t^4\right)+4}{2\sqrt{t\ln \left(t^4\right)}}
\end{align*}