Find $y^{\text{'}}$ if

$y={\displaystyle \underset{1}{\overset{x^2}{}}}\sqrt[\phantom{2}]{2t+4}dt$

$($A$)2x\sqrt[\phantom{2}]{2x^2+4}$

$($B$)\sqrt[\phantom{2}]{2x^2+4}$

$($C$)\frac{2}{\sqrt[\phantom{2}]{2t+4}}$

$($D$)\frac{1}{3}(\sqrt[\phantom{2}]{2x^2+4}-\sqrt[\phantom{2}]{6})-\frac{1}{3}\sqrt[\phantom{2}]{6}$

$($E$)$ none of the above