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Find the second derivaive of y = x / (x^3 -1)

Mathematics
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have you tried quotient rule or product rule + chain rule?
yes I got the first derivative correctly (verified by Wolfram Alpha) but am messing up when taking the derivative the second time
the first derivative is \[-\frac{ 2x ^{3}+1 }{ (x ^{3}-1)^{2} }\]

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|dw:1352709544762:dw|
I know the quotient rule and chain rule but something is wrong with my execution of the rules
@whatisthequestion let me tell you a good one method. which is much better than this. Do you want to know??????? or want to try with method actually there is just calculation mistakes only:)
I am open to other methods if you have another method let me know
first make it into an equation. can u make.......try :)
\[y = \frac{ x }{ x^3-1 } = \frac{1}{x^2-\frac{1}{x}}=\left[ x^2-x^{-1} \right]^{-1}\] \begin{align} y' &= (-1)\left[ x^2-x^{-1} \right]^{-2} (2x - (-1)x^{-2}) \\ &= -[ x^2-x^{-1} ]^{-2} (2x + x^{-2}) \\ &= \frac{-2x+x^{-2}}{(x^2 - x^{-1})^2}\\ &= \frac{-2x^3+1}{x^2(x^2 - x^{-1})^2}\\ &= \frac{-2x^3+1}{(x(x^2 - x^{-1}))^2}\\ &= \frac{-2x^3+1}{(x^3 - 1)^2}\\ \end{align} \begin{align} y'' &= \frac{d}{dx}[\frac{-2x^3+1}{(x^3 - 1)^2}]\\ &=\frac{d}{dx}[(x^3 - 1)^{-2}(-2x^3+1)]\\ &=\frac{d}{dx}[(x^3 - 1)^{-2}](-2x^3+1) + (x^3 - 1)^{-2} \frac{d}{dx}[(-2x^3+1)]\\ &= -2(3x^2)(x^3 - 1)^{-3}(-2x^3+1) + (x^3 - 1)^{-2} (-6x^2)\\ &= (-6x^2)(-2x^3+1)(x^3 - 1)^{-3} + (-6x^2)(x^3 - 1)^{-2}\\ &= (12x^5-6x^2)(x^3 - 1)^{-3} + (-6x^2)(x^3 - 1)^{-2}\\ &= \frac{12x^5-6x^2}{(x^3 - 1)^3} + \frac{-6x^2}{(x^3 - 1)^2}\\ &= \frac{12x^5-6x^2}{(x^3 - 1)^3} + \frac{(-6x^2)(x^3 - 1)}{(x^3 - 1)^3}\\ &= \frac{12x^5-6x^2 -6x^5 +6x^2}{(x^3 - 1)^3}\\ &= \frac{6x^5}{(x^3 - 1)^3}\\ \end{align}
verification: https://www.wolframalpha.com/input/?i=differentiate+%28%E2%88%922x^3%2B1%29\%28x^3%E2%88%921%29^2+with+respect+to+x

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