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

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- Engr. RonaldLv 74 weeks agoFavorite 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

- AshLv 74 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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