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



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precal
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dw:1333397126713:dw

precal
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dw:1333397156412:dw

precal
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dw:1333397188197:dw

precal
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dw:1333397226975:dw

precal
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ok now take the derivative again. Don't forget to use this statement in again for y'

chance1
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can you help me precal please


precal
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Here is the final solution from wolfram alpha.

anonymous
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\[x^3 + y^3 = 1\]
\[3x^2+3y^2y'=0\]
\[y'=\frac{x^2}{y^2}\]

anonymous
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\[y''=\frac{2xy^22x^2yy'}{y^4}\] by the quotient rule

anonymous
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replace \(y'\) by \(\frac{x^2}{y^2}\) in the above, do some algebra and get the solution

calyne
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got to that but i'm stuck anyway

anonymous
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\[y''=\frac{2xy^22x^2y\times \frac{x^2}{y^2}}{y^4}\]

anonymous
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multiply top and bottom by \(y^2)\] so clear the fraction then cancel some y's

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

anonymous
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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 \(x^36y^3=1\)

anonymous
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rather since \(x^3+y^3=1\)