## University Calculus: Early Transcendentals (3rd Edition)

$$y'=\frac{1}{2}(\sqrt t)^t(\ln t+1)$$
$$y=(\sqrt t)^t$$ We note that $(\sqrt t)^t=e^{\ln((\sqrt t)^t)}=e^{t\ln \sqrt t}$ Also, $\ln\sqrt t=\ln t^{1/2}=\frac{1}{2}\ln t$ Therefore, $$y=e^{\frac{1}{2}t\ln t}$$ The derivative of $y$ is: $$y'=e^{\frac{1}{2}t\ln t}\Big(\frac{1}{2}t\ln t\Big)'=\frac{1}{2}e^{\frac{1}{2}t\ln t}\Big(\ln t+t\times\frac{1}{t}\Big)$$ $$y'=\frac{1}{2}e^{\frac{1}{2}t\ln t}(\ln t+1)$$ $$y'=\frac{1}{2}(\sqrt t)^t(\ln t+1)$$