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December 21, 2014

December 21, 2014

Posted by **Anonymous** on Friday, October 12, 2012 at 3:32pm.

- Math -
**Steve**, Friday, October 12, 2012 at 3:46pmx^2y^3+4x+2y=12

use the product rule as needed:

2x y^3 + 3x^2y^2 y' + 4 + 2y' = 0

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

y' is thus

-2(xy^3 + 2)

-------------

3x^2 y^2 + 2

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