Respuesta :

Answer:

[tex]\frac{dy}{dx} =\frac{4x^3+3x^2y-5y^2}{10xy-3y^2-x^3}[/tex]

Step-by-step explanation:

solving for dy/dx

multiply the equation out to remove parentheses

[tex]x^4 + x^3y = 5xy^2 -y^3[/tex]

now differentiating in terms of x ([tex]\frac{dy}{dx}[/tex])

[tex]4x^3 +3x^2y + x^3(\frac{dy}{dx} ) = 5y^2 + 10xy(\frac{dy}{dx} )-3y^2(\frac{dy}{dx} )[/tex]

isolating dy/dx to one side

[tex]4x^3 +3x^2y - 5y^2= 10xy(\frac{dy}{dx} )-3y^2(\frac{dy}{dx} )-x^3(\frac{dy}{dx} )[/tex]

[tex]4x^3 +3x^2y - 5y^2= \frac{dy}{dx}(10xy-3y^2-x^3)[/tex]

[tex]\frac{dy}{dx} =\frac{4x^3+3x^2y-5y^2}{10xy-3y^2-x^3}[/tex]

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