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April 18, 2014

April 18, 2014

Posted by **Aria** on Tuesday, June 5, 2012 at 12:12am.

- Calculus -
**Steve**, Tuesday, June 5, 2012 at 12:16am2xy^3 + 3x^2y^2y' + 3x^2y^4 + 4x^3y^3y' = 0

y' = -(2xy^3 + 3x^2y^4)/(3x^2y^2 + 4x^3y^3)

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

assuming x,y not zero

- Calculus -
**Aria**, Tuesday, June 5, 2012 at 12:22amThanks a lot, it makes so much more sense now! :)

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