Question $2$

$2$ pts

Given $g(x)={\displaystyle \underset{0}{\overset{x}{}}}\sqrt[\phantom{2}]{t+t^3}dt,$ then

$\frac{d}{dx}g(x)=\sqrt[\phantom{2}]{t+t^2}$

$\frac{d}{dx}g(x)=\sqrt[\phantom{2}]{x+x^2}$

$\frac{d}{dx}g(x)=\frac{1+2x}{2\sqrt[\phantom{2}]{x+x^2}}$

$g(x)=\sqrt[\phantom{2}]{x+x^2}$