step 2\nto find $\frac{dy}{dx}=y$ for $x^{3}+y^{8}=6$, we differentiate both sides of the equation with…

step 2\nto find $\frac{dy}{dx}=y$ for $x^{3}+y^{8}=6$, we differentiate both sides of the equation with respect to $x$.\non the right - hand side, we have the following.\n$\frac{d}{dx}6=square checkmark square$\non 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.\n$\frac{d}{dx}x^{3}=square checkmark square$\nstep 3\nwe now have\n$\frac{d}{dx}x^{3}+y^{8}=3x^{2}+8y^{7}y=\frac{d}{dx}6=0$.\nrearranging this, we get\n$8y^{7}y=square$.\nsolving for $y$, we conclude that the derivative is\n$y=square$.

step 2\nto find $\frac{dy}{dx}=y$ for $x^{3}+y^{8}=6$, we differentiate both sides of the equation with respect to $x$.\non the right - hand side, we have the following.\n$\frac{d}{dx}6=square checkmark square$\non 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.\n$\frac{d}{dx}x^{3}=square checkmark square$\nstep 3\nwe now have\n$\frac{d}{dx}x^{3}+y^{8}=3x^{2}+8y^{7}y=\frac{d}{dx}6=0$.\nrearranging this, we get\n$8y^{7}y=square$.\nsolving for $y$, we conclude that the derivative is\n$y=square$.

Answer

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:

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