Answer
$$y'=\frac{1}{4t\sqrt{\ln\sqrt t}}$$
Work Step by Step
$$y=\sqrt{\ln\sqrt t}$$
We have $$(\sqrt u)'=(u^{1/2})'=\frac{1}{2}u^{-1/2}=\frac{1}{2\sqrt u}$$
Using that, we find the derivative of y: $$y'=\frac{1}{2\sqrt{\ln\sqrt t}}(\ln\sqrt t)'=\frac{1}{2\sqrt{\ln\sqrt t}}\times\frac{1}{\sqrt t}\times(\sqrt t)'$$
$$y'=\frac{1}{2\sqrt{\ln\sqrt t}}\times\frac{1}{\sqrt t}\times\frac{1}{2\sqrt t}$$
$$y'=\frac{1}{4t\sqrt{\ln\sqrt t}}$$
