Question

step 2
to find $\frac{dy}{dx}=y$ for $x^{3}+y^{8}=6$, we differentiate both sides of the equation with respect to $x$.
on the right - hand side, we have the following.
$\frac{d}{dx}6=square checkmark square$
on the left - hand side, we have $\frac{d}{dx}x^{3}+y^{8}$. we already know that $\frac{d}{dx}y^{8}=8y^{7}y$. for the term $x^{3}$, we have the following.
$\frac{d}{dx}x^{3}=square checkmark square$
step 3
we now have
$\frac{d}{dx}x^{3}+y^{8}=3x^{2}+8y^{7}y=\frac{d}{dx}6=0$.
rearranging this, we get
$8y^{7}y=square$.
solving for $y$, we conclude that the derivative is
$y=square$.

Explanation:

Step1: Rearrange the derivative - equation

We start with $3x^{2}+8y^{7}y' = 0$. To isolate the term with $y'$, we subtract $3x^{2}$ from both sides.
$8y^{7}y'=- 3x^{2}$

Step2: Solve for $y'$

Divide both sides of the equation $8y^{7}y'=-3x^{2}$ by $8y^{7}$ to get the value of $y'$.
$y'=-\frac{3x^{2}}{8y^{7}}$

Answer:

$8y^{7}y'=-3x^{2}$; $y'=-\frac{3x^{2}}{8y^{7}}$