Answer
$F^{'}(x)={\sqrt[4] x }$
Work Step by Step
$F(x)=\int_{1}^{x}{\sqrt[4] t }{dt}$
$F^{'}(x)=\frac{d}{dx}\int_{1}^{x}{\sqrt[4] t }{dt}$
$={\sqrt[4] x }$
This is valid since the function, $f(t)={\sqrt[4] t }$ is a continuous function over the implied domain, $t\geq0$
