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precalBest ResponseYou've already chosen the best response.4
ok now take the derivative again. Don't forget to use this statement in again for y'
 2 years ago

chance1Best ResponseYou've already chosen the best response.0
can you help me precal please
 2 years ago

precalBest ResponseYou've already chosen the best response.4
http://www.wolframalpha.com/input/?i=find+the+derivative+of+y%3D%28x%5E2%29%2Fy%5E2
 2 years ago

precalBest ResponseYou've already chosen the best response.4
Here is the final solution from wolfram alpha.
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
\[x^3 + y^3 = 1\] \[3x^2+3y^2y'=0\] \[y'=\frac{x^2}{y^2}\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
\[y''=\frac{2xy^22x^2yy'}{y^4}\] by the quotient rule
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
replace \(y'\) by \(\frac{x^2}{y^2}\) in the above, do some algebra and get the solution
 2 years ago

calyneBest ResponseYou've already chosen the best response.0
got to that but i'm stuck anyway
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
\[y''=\frac{2xy^22x^2y\times \frac{x^2}{y^2}}{y^4}\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
multiply top and bottom by \(y^2)\] so clear the fraction then cancel some y's
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
typo there i lost the minus sign, should be \[y''=\frac{2xy^2+2x^2y\times \frac{x^2}{y^2}}{y^4}\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
\[y''=\frac{2xy^4+2x^4y}{y^6}\] \[y''=\frac{2xy^3+2x^4}{y^5}\] \[y''=\frac{2x(y^3+x^3)}{y^5}\] \[y''=\frac{2x}{y^5}\] since \(x^36y^3=1\)
 2 years ago

satellite73Best ResponseYou've already chosen the best response.2
rather since \(x^3+y^3=1\)
 2 years ago
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