Please someone please find the second differential for this equation
(x^{3}) + (y^{3}) = 1
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well for te first differential i got
3x^{2} + 3y^{2}(dy/dx)=0
dy/dx = (-x^{2} )/(y^{2})
Then seoncd differntial using quotient rule
[(y^{2})(-2x)-(-x^{2} )(2y(dy/dx))]/(y^{4})
then finally the second differntial came down to for me
[-2xy^{3} + 2x^{4} ]/y^{5}
Please help
You are correct up to where you have used the Quotient Rule to find the second derivative. So you have
$\displaystyle \displaystyle \begin{align*} \frac{d^2y}{dx^2} &= \frac{-2xy^2 + 2x^2y\,\frac{dy}{dx}}{y^4} \\ &= \frac{-2xy + 2x^2\frac{dy}{dx}}{y^3} \\ &= \frac{-2xy + 2x^2\left(-\frac{x^2}{y^2}\right)}{y^3} \\ &= \frac{-2xy - \frac{2x^4}{y^2}}{y^3} \\ &= \frac{-2xy^3 - 2x^4}{y^5} \end{align*}$
Your final answer has a sign error but is otherwise correct.