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December 20, 2014

December 20, 2014

Posted by **Jo** on Tuesday, October 25, 2011 at 12:04am.

- Calculus -
**Steve**, Tuesday, October 25, 2011 at 5:12amUse implicit differentiation:

xy^3 / (1+y^2) = 8/5

(y^3 + 3xy^2 y')(1+y^2) - xy^3 (2yy') = 0

It's all over (1+y^2)^2, but that can be ignored, since it's never 0.

y'(3xy^2 + 3xy^4 - 2xy^4) = -y^3(1 + y^2)

y' = -y^3 (1+y^2)/(3xy^2 + xy^4)

= -y/x * (1+y^2)/(3 + y^2)

Take it from there.

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