Respuesta :

[tex]\bf x^3y^2-x^4y=4xy^3=0\\\\ -------------------------------\\\\ \left( 3x^2y^2+x^32y\frac{dy}{dx} \right)-\left( 4x^3y+x^4\frac{dy}{dx} \right)+4\left( 1\cdot y^3+x3y^2\frac{dy}{dx} \right)=0 \\\\\\ 2x^3y\frac{dy}{dx}-x^4\frac{dy}{dx}+12xy^2\frac{dy}{dx}=4x^3y-3x^2y^2-4y^3 \\\\\\ \cfrac{dy}{dx}\left( 2x^3y-x^4+12xy^2 \right)=4x^3y-3x^2y^2-4y^3 \\\\\\ \cfrac{dy}{dx}=\cfrac{4x^3y-3x^2y^2-4y^3}{2x^3y-x^4+12xy^2}[/tex]

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