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Pretty sure I messed up $$ x^3 - 3xy + y^3 = 1 $$ $$ 3x^2-3xy'-3y+3y^2y'=0 $$ $$ 3y^2y'-3xy'=3y-3x^2 $$ $$ (3y^2-3x)y'=3y-3x^2 $$ $$ y' = \dfrac{y-x^2}{y^2-x} $$ $$ y'' = \dfrac{(y^2-x)(y'-2x)-(y-x^2)(2yy'-1)}{(y^2-x)^2} $$ $$ y'' = \dfrac{y^2y'-2xy^2-xy'+2x^2-2y^2y'+y+2x^2yy'-x^2}{(y^2-x)^2} $$ $$ y''=\dfrac{x^2-2xy^2+y+(2x^2y-x-y^2)y'}{(y^2-x)^2} $$ After subbing in $y'$, expanding and simplifying... $$ y'' = \dfrac{x^2-2xy^2+y+3x^2y^2-2x^4y-xy+x^3-y^3}{(y^2-x)^3} $$ The answer should be $$ y'' = -\dfrac{4xy}{(y^2-x)^3} $$

James
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    If you really need to use implicit differentiation, you should consider implicitly differentiating twice. Instead, you are doing it once, then isolating $y'$ and differentiation from there. – 2'5 9'2 Jul 17 '13 at 01:09
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    There may be nothing wrong with your work; keep in mind that there is an implicit relationship between $x$ and $y$. Maybe you can use it to simplify your numerator to the presumed answer's. – 2'5 9'2 Jul 17 '13 at 01:13
  • I'm trying that right now I never thought of doing it twice – James Jul 17 '13 at 01:16
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    OK, I see a problem: when you insert $y'$ at the end, you did not handle the $x^2-2xy^2+y$ part correctly.It should become $(x^2-2xy^2+y)(y^2-x)$ in the next step, which would then multiply out. – 2'5 9'2 Jul 17 '13 at 01:16
  • Yes that's it thanks – James Jul 17 '13 at 01:29

1 Answers1

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As pointed out in the comments, the error occurred when plugging back in the value for $y'$. The part that reads: $$x^2-2xy^2+y$$ Should read: $$(x^2-2xy^2+y)(y^2-x)$$

This multiplies out and helps simplify further.

apnorton
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