Find dy/dx for x^3-y^3+x^3y=10?

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  • 3 weeks ago
    Favourite answer

    x^3-y^3+x^3y=10

    3x^2 - 3y^2dy/dx + [y * 3x^2 + x^3 * dy/dx] = 0

    3x^2 - 3y^2dy/dx + 3x^2y + x^3dy/dx = 0

     - 3y^2dy/dx + x^3dy/dx = - 3x^2y - 3x^2

    dy/dx[ - 3y^2 + x^3] = - 3x^2y - 3x^2

    ............... - 3x^2y - 3x^2

    dy/dx = ----------------------------

    .................. - 3y^2 + x^3

    ................- 3x^2(y + 1)

    dy/dx = ------------------------

    .................- (3y^2 - x^3)

    ...............3x^2(y + 1)

    dy/dx  =------------------ Answer.

    .................3y^2 - x^3

  • Ash
    Lv 7
    3 weeks ago

    x³ - y³ + x³y = 10

    3x² - 3y² dy/dx + 3x²y + x³ dy/dx = 0

    (x³ - 3y²)dy/dx = -3x² - 3x²y

    dy/dx = -3x²(1 + y)/(x³ - 3y²)

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