Q:

How do I find the y double prime by implicit differentiation? 

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? 

A:

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 $y''(0)$.

