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I am to find y double prime at $x = 0$ given $$xy + 7e^y = 7e$$

I am confused as to how to use the information of $x = 0$. The only $x$ term in the equation is $xy$ which after implicit differentiation becomes just $y$. Then there is no more $x$ in $y'$ or $y''$for me to substitute into. Do I substitute it in before beginning implicit differentiation?

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When you take the derivative $$\dfrac{d}{dx}\left[ xy \right] =1 \cdot y + x y'$$ Remember that $y$ is a function of $x$ so the derivative with respect to $x$ is not 1. –  Kris Williams Feb 16 '13 at 3:34

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You have to carry things through before setting $x=0$. Differentiating once: $$y+xy'+7e^y y'=0.$$ A second time, $$y'+y'+xy''+7e^y (y')^2+7e^y y''=0.$$ Solve for $y''$: $$y''=-\frac{7e^y (y')^2+2y'}{x+7e^y}.$$ Now if you plug in $x=0$ to the defining equation and the equation for the first derivative, you see that $$7e^{y(0)}=7e,$$ implying $y(0)=1$, and $$y(0)+7e^{y(0)}y'(0)=0,$$ implying $y'(0)=-\frac{1}{7e}$. Finally, you can evaluate $y''(0)$.

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