Question

How do you calculate the derivative of y with respect to x on the equation x^3y+3xy^3=x+y?

Answer 1

$\bf x^3y+3xy^3=x+y\\\\ -----------------------------\\\\ \left[ 3x^2y+x^3\frac{dy}{dx} \right]+3\left[ 1\cdot y^3+x\cdot 3y^2\frac{dy}{dx} \right]=1+\frac{dy}{dx} \\\\\\ 3x^2y+x^3\frac{dy}{dx}+3y^3+9xy^2\frac{dy}{dx}-\frac{dy}{dx}=1\impliedby \textit{common factor} \\\\\\ \cfrac{dy}{dx}(x^3+9xy^2-1)+3x^2y+3y^3=1 \\\\\\ \cfrac{dy}{dx}(x^3+9xy^2-1)=1-3x^2y-3y^3 \\\\\\ \cfrac{dy}{dx}=\cfrac{1-3x^2y-3y^3}{x^3+9xy^2-1}$