Answer
$4e^t\left(\sqrt{t}+\frac{1}{2\sqrt{t}}\right)$
Work Step by Step
We are trying to find $\frac{d}{dt}4e^t\sqrt{t}$.
Use the Product Rule, $\frac{d}{dt}u(t)v(t)=u'(t)v(t)+u(t)v'(t)$.
In this case, $u(t)=4e^t$, so $u'(t)=4e^t$. Also, $v(t)=\sqrt{t}$, so $v'(w)=\frac{1}{2\sqrt{t}}$.
Plugging these in, we get:
$\frac{d}{dt}4e^t\sqrt{t}$
$=4e^t\sqrt{t}+4e^t*\frac{1}{2\sqrt{t}}$
$4e^t\left(\sqrt{t}+\frac{1}{2\sqrt{t}}\right)$
