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July 29, 2014

July 29, 2014

Posted by **Nurcan** on Saturday, February 25, 2012 at 11:30am.

4x^3 + 3y^3 = 4

=====================================

4x^3 + 3y^3 = 4

12x + 9yy' = 0

24x + 9(2yy')y' + 9y^2(y") = 0

y" = -6[4x-3y(y')^2] answer. (it this right? )

- Calcules -
**Reiny**, Saturday, February 25, 2012 at 1:34pmYour 1st deriv should have been

12x^2 + 9y^2 y' = 0 ----> y' = -12x^2/(9y^2)

2nd derivative:

24x + 9y^2 (y'') + y' (18y) = 0

replacing y'

24x + 9y^2 y'' + (-12x^2)/(9y^2) (18y) = 0

24x + 9y^2 y'' - 24x/y = 0

times y

24xy + 9y^3 - 24x = 0

y'' = (24x - 24xy)/(9y^3) = (8x-8xy)/(3y^3)

I don't think they wanted to see that y' in your answer.

btw, how did you ever get you first derivative ?

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