we now have \n\frac{d}{dx}x^{4}+y^{3}=4x^{3}+3y^{2}y=\frac{d}{dx}1=0.\nrearranging this, we get \n3y^{2}y=-4…

we now have \n\frac{d}{dx}x^{4}+y^{3}=4x^{3}+3y^{2}y=\frac{d}{dx}1=0.\nrearranging this, we get \n3y^{2}y=-4 \n\nand solving for y, we conclude that the derivative is \ny=

we now have \n\frac{d}{dx}x^{4}+y^{3}=4x^{3}+3y^{2}y=\frac{d}{dx}1=0.\nrearranging this, we get \n3y^{2}y=-4 \n\nand solving for y, we conclude that the derivative is \ny=

Answer

Explanation:

Step1: Rearrange the derivative - equation

Starting with $4x^{3}+3y^{2}y' = 0$, we want to isolate the term with $y'$. Subtract $4x^{3}$ from both sides of the equation. $3y^{2}y'=-4x^{3}$

Step2: Solve for $y'$

Divide both sides of the equation $3y^{2}y'=-4x^{3}$ by $3y^{2}$ (assuming $y\neq0$). $y'=-\frac{4x^{3}}{3y^{2}}$

Answer:

First blank: $x^{3}$ Second blank: $-\frac{4x^{3}}{3y^{2}}$