Respuesta :

[tex]\bf 4x^3+x^2y-xy^3=6\\\\ -------------------------------\\\\ 12x^2+\left( 2xy+x^2\cfrac{dy}{dx} \right)-\left( y^3+x3y^2\cfrac{dy}{dx} \right)=0 \\\\\\ 12x^2+2xy+x^2\cfrac{dy}{dx}-y^3-3xy^2\cfrac{dy}{dx}=0 \\\\\\ x^2\cfrac{dy}{dx}-3xy^2\cfrac{dy}{dx}=y^3-12x^2-2xy \\\\\\ \cfrac{dy}{dx}(x^2-3xy^2)=y^3-12x^2-2xy\implies \cfrac{dy}{dx}=\cfrac{y^3-12x^2-2xy}{x^2-3xy^2}[/tex]
steeleflag19

Answer: dy/dx = y^3 - 12x^2 - 2yx / x(x-3y^2)

Step-by-step explanation: Differentiate each term with respect to x, then solve for y.

I hope this helps you out!

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