# Implicit differentiation $xy+e^y = e$

I am trying to find the value of $y''$ when $x= 0$ for $xy+e^y = e$

This seems pretty simple and straightforward to me

$xy' + y + e^y y' = 0$

then derive again

$xy'' + e^y y'' = -e^y y'y' - y'$ which is not correct and I do not know why, infact the book gets some crazy answer with just an e in it which seems impossible to me since e will go away.

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When you differentiate $xy'$ you need both the chain rule and the product rule.

Similarly for $e^y y'$.

Later edit:

When you differentiate $xy'$ you get $xy'' + y'$.

When you differentiate $y$ you get $y'$.

When you differentiate $e^y y'$ you get $e^y y' y' + e^y y''$.

So you've got five terms, and two of them are "like" terms that can get "collected".

And you should notice that when $x=0$ then $y=1$. That implies that when $x=0$ then $y' = -1/e$.

The bottom line I'm getting is that when $x=0$ then $y''=1/e^2$.

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I am pretty sure I did use both, I know I know the correct way to do problems since I have done several other ones correctly. – user138246 May 10 '12 at 15:36
@Jordan: As Micheal pointed, $(xy')'$=$x'y+xy'$=$1+xy'$. Do the same for $e^yy'$. It becomes: $(e^yy')'$=$(e^y)'y'+e^y(y')'$. Now find $(e^y)'$. – Babak S. May 10 '12 at 15:45
I do not know what I am doing wrong, everything appears to be right. – user138246 May 10 '12 at 15:49
@Jordan: Equivalently, if you have an implicit function like $F(x,y)=0$, then $y'=-F_{x}/F_{y}$ . – Babak S. May 10 '12 at 15:51
@Jordan : I don't think Babak is referring to antiderivatives. I think the idea is that "$F(x,y)=0$" is what you've got when you write "$xy+e^y = e$". I.e. $F(x,y)=xy+e^y-e$; it's a function of two variables. And it implicitly defines a function: $y$ is a function of $x$. And $F_x$ and $F_y$ are partial derivatives. But you don't need that point of view to solve the problem that you posed in the initial posting above. – Michael Hardy May 10 '12 at 19:08