Answer
$$y'=\frac{(t^{\sqrt t})(\ln t+2)}{2\sqrt t}$$
Work Step by Step
$$y=t^{\sqrt t}$$
We note that $t^{\sqrt t}=e^{\ln(t^{\sqrt t})}=e^{\sqrt t\ln t}$
Therefore, $$y=e^{\sqrt t\ln t}$$
The derivative of $y$ is: $$y'=e^{\sqrt t\ln t}\Big(\sqrt t\ln t\Big)'=e^{\sqrt t\ln t}\Big((\sqrt t)'\ln t+\sqrt t(\ln t)'\Big)$$
$$y'=e^{\sqrt t\ln t}\Big(\frac{\ln t}{2\sqrt t}+\frac{\sqrt t}{t}\Big)$$
$$y'=e^{\sqrt t\ln t}\Big(\frac{t\ln t+2t}{2t\sqrt t}\Big)=e^{\sqrt t\ln t}\Big(\frac{\ln t+2}{2\sqrt t}\Big)$$
$$y'=\frac{(t^{\sqrt t})(\ln t+2)}{2\sqrt t}$$
