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calyne

  • 4 years ago

Find y'' by implicit differentiation: x^3 + y^3 = 1

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  1. precal
    • 4 years ago
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    |dw:1333397126713:dw|

  2. precal
    • 4 years ago
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    |dw:1333397156412:dw|

  3. precal
    • 4 years ago
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    |dw:1333397188197:dw|

  4. precal
    • 4 years ago
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    |dw:1333397226975:dw|

  5. precal
    • 4 years ago
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    ok now take the derivative again. Don't forget to use this statement in again for y'

  6. chance1
    • 4 years ago
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    can you help me precal please

  7. precal
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=find+the+derivative+of+y%3D%28-x%5E2%29%2Fy%5E2

  8. precal
    • 4 years ago
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    Here is the final solution from wolfram alpha.

  9. anonymous
    • 4 years ago
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    \[x^3 + y^3 = 1\] \[3x^2+3y^2y'=0\] \[y'=-\frac{x^2}{y^2}\]

  10. anonymous
    • 4 years ago
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    \[y''=\frac{2xy^2-2x^2yy'}{y^4}\] by the quotient rule

  11. anonymous
    • 4 years ago
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    replace \(y'\) by \(-\frac{x^2}{y^2}\) in the above, do some algebra and get the solution

  12. calyne
    • 4 years ago
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    got to that but i'm stuck anyway

  13. anonymous
    • 4 years ago
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    \[y''=\frac{2xy^2-2x^2y\times \frac{x^2}{y^2}}{y^4}\]

  14. anonymous
    • 4 years ago
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    multiply top and bottom by \(y^2)\] so clear the fraction then cancel some y's

  15. anonymous
    • 4 years ago
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    typo there i lost the minus sign, should be \[y''=\frac{2xy^2+2x^2y\times \frac{x^2}{y^2}}{y^4}\]

  16. anonymous
    • 4 years ago
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    \[y''=\frac{2xy^4+2x^4y}{y^6}\] \[y''=\frac{2xy^3+2x^4}{y^5}\] \[y''=\frac{2x(y^3+x^3)}{y^5}\] \[y''=\frac{2x}{y^5}\] since \(x^36y^3=1\)

  17. anonymous
    • 4 years ago
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    rather since \(x^3+y^3=1\)

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