$($a$)$ Show that $y^4+x-6=0$ is an implicit solution to $\frac{dy}{dx}=-\frac{1}{4y^3}$ on the interval $(-,6)$

$($b$)$ Show that $x{y}^7-x{y}^{7}sinx=1$ is an implicit solution to the differential equation $\frac{dy}{dx}=\frac{(xcosx+sinx-1)y}{7(x-xsinx)}$ on the interval $(0,\frac{}{2}).$

$($a$)$ To verify the solution and the interval, convert $y^4+x-6=0$ to explicit function$($s$).$

$y=(6-x{)}^{(\frac{1}{4})},-(6-x{)}^{(\frac{1}{4})}($Use a comma to separate answers as needed.$)$

Find the derivative$($s$)$ of the explicit solution$($s$).$

$\frac{dy}{dx}=+-\frac{1}{4}(6-x{)}^{-\frac{3}{4}}($Use a comma to separate answers as needed.$)$

When the explicit solution$($s$)$ for $y$ is$/$are substituted into $\frac{dy}{dx}=-\frac{1}{4y^3},$ the two sides of the equation are equivalent. Explain why the solution is limited to the interval $(-,6).$ Choose the correct answer below.

A$.$ Neither $y$ nor $\frac{dy}{dx}$ are real$-$valued for $x>6,$ and $\frac{dy}{dx}$ is undefined at $x=6.$

B$.$ Neither y nor $\frac{dy}{dx}$ are real$-$valued for $x>6,$ and y is undefined at $x=6.$

C$.$ Neither y nor $\frac{dy}{dx}$ are defined for $x6.$

D$.y$ is undefined for $x6.$