Respuesta :

alessandroarenas

[tex]3x^2y +x^3\frac{dy}{dx} -4x + 4y\frac{dy}{dx} = 0\\\frac{dy}{dx} (x^3 +4y) = 4x -3x^2y\\\frac{dy}{dx} = \frac{4x-3x^2y}{x^3+4y}[/tex]

wegnerkolmp2741o

x^3y-2x^2+y^4=8

take the derivative of each part, using the product rule on the first term

x^3 dy +y*3x^2dx -4xdx+4y^3 dy = 0

gather the dy and dx terms together

dy (x^3+4y^3) -dx(-3x^2y+4x) =0

dy terms on the left, dx terms on the right

dy (x^3+4y^3) =dx(-3x^2y+4x)

dy divided by dx

dy/dx (x^3+4y^3) =(-3x^2y+4x)

dy/dx  =(-3x^2y+4x) / (x^3+4y^3)

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