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According to my textbook, the second derivative of

\begin{equation*} y^{2}+xy-x^{2}=9 \end{equation*}

is

\begin{equation*} \frac{90}{(2y+x)^{3}}. \end{equation*}

The problem states "Express $\frac{d^{2}y}{dx^{2}}$ in terms of $x$ and $y$." I've tried for two days straight now, and I can't get that answer. I am convinced the book has a typo.

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    muad: "express $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}$ in terms of $x$ and $y$", per the OP. In other words, what is $y^{\prime\prime}(x)$ in terms of $x$ and $y$.2010-09-18

2 Answers 2

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Based on your work as linked in the comments on J.M.'s answer, you've very nearly got it (except for a typo in differentiating *: there's a dy/dx where there should be a d/dx, but the mathematics that follows is correct as if it were d/dx). The numerator you have is $\begin{align} -2 ((2 x - y)^2 &+ (2 x - y) (2 y + x) - (2 y + x)^2) \\\\ &=-10x^2+10xy+10y^2 \\\\ &=10(y^2+xy-x^2) \\\\ &=10\cdot9 \\\\ &=90. \end{align}$

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    I hate basic errors. Thanks Isaac!2010-09-18
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Hint: treat $y$ as a function $y(x)$, so differentiating $xy(x)$ should give something like $x y^{\prime}(x)+y(x)$. Differentiate expressions twice, and solve for $y^{\prime\prime}(x)$

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    M.: the numerator in your comment is $10(y(x+y)-x^2)=10(y^2+xy-x^2)=90$.2010-09-18