anonymous
  • anonymous
Find y'' by implicit differentiation: x^3 + y^3 = 1
Mathematics
schrodinger
  • schrodinger
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precal
  • precal
|dw:1333397126713:dw|
precal
  • precal
|dw:1333397156412:dw|
precal
  • precal
|dw:1333397188197:dw|

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precal
  • precal
|dw:1333397226975:dw|
precal
  • precal
ok now take the derivative again. Don't forget to use this statement in again for y'
anonymous
  • anonymous
can you help me precal please
precal
  • precal
http://www.wolframalpha.com/input/?i=find+the+derivative+of+y%3D%28-x%5E2%29%2Fy%5E2
precal
  • precal
Here is the final solution from wolfram alpha.
anonymous
  • anonymous
\[x^3 + y^3 = 1\] \[3x^2+3y^2y'=0\] \[y'=-\frac{x^2}{y^2}\]
anonymous
  • anonymous
\[y''=\frac{2xy^2-2x^2yy'}{y^4}\] by the quotient rule
anonymous
  • anonymous
replace \(y'\) by \(-\frac{x^2}{y^2}\) in the above, do some algebra and get the solution
anonymous
  • anonymous
got to that but i'm stuck anyway
anonymous
  • anonymous
\[y''=\frac{2xy^2-2x^2y\times \frac{x^2}{y^2}}{y^4}\]
anonymous
  • anonymous
multiply top and bottom by \(y^2)\] so clear the fraction then cancel some y's
anonymous
  • anonymous
typo there i lost the minus sign, should be \[y''=\frac{2xy^2+2x^2y\times \frac{x^2}{y^2}}{y^4}\]
anonymous
  • anonymous
\[y''=\frac{2xy^4+2x^4y}{y^6}\] \[y''=\frac{2xy^3+2x^4}{y^5}\] \[y''=\frac{2x(y^3+x^3)}{y^5}\] \[y''=\frac{2x}{y^5}\] since \(x^36y^3=1\)
anonymous
  • anonymous
rather since \(x^3+y^3=1\)

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