Find the integral.$($a$)$ Find dy$,$ and $($b$)$ evaluate dy for the given value of $x$ and $dx.$

$y=x\sqrt[\phantom{2}]{4-2x^2},x=0,dx=-0\mathrm{.}3$

$($a$)$

$($Type an exact answer, using radicals as needed.$)$

$($b$)$ At the given values, $dy=.$

$($Type an integer or a decimal.$)$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(2\sqrt[3]{x^4}-x^{-3})dx$

A$.\frac{6}{7}{x}^{\frac{7}{3}}-2x^{-2}+c$

B$.\frac{4}{3}{x}^{\frac{7}{3}}-3x^{-2}+c$

c$.\frac{6}{7}{x}^{\frac{6}{7}}+\frac{1}{2x^2}+c$

D$.\frac{6}{7}{x}^{\frac{7}{3}}+\frac{1}{2x^2}+C$