How do you find the 2nd derivative of xy+e^y=e?

Question

yttrium · Answer

Derive the equation twice. So far, where are you? @Jmen7164

anonymous · Answer

I got the first derivative as y/(e-xy-x), which I am positive is incorrect.

yttrium · Answer

First derivative (using implicit differentiation)
$$xy+e^y = e$$
$$(y+xy')+e^y y' = 0$$
$$y' = \frac{ -y }{ x+e^y }$$
Second derivative: Derive the first derivative.
$$y'' = \frac{ (x+e^y)(-y')-(-y)(1+e^yy') }{ (x+e^y)^2 }$$
Therefore,
$$y''=\frac{ (x+e^y)(-(\frac{ -y }{ x+e^y })+ y(1+e^y(\frac{ -y }{ x+e^y }) }{ (x+e^y)^2 }$$
And upon simplifying we'll have.
$$y''=\frac{ 2y(x+e^y)-y^2e^y }{ (x+e^y)^2 }$$

yttrium · Answer

@Jmen7164  can you follow?

anonymous · Answer

I guess I am just confused because the back of the book also says the answer is 1/e^2 and I cannot simply an answer to get that.

anonymous · Answer

What class are you in?  And are you sure you're not evaluating it at some point?

anonymous · Answer

I would be evalutating it at 0, but wouldn't that still leave me with -y^2e^y/(e2y)

anonymous · Answer

and is it dx/dy or  dx/dy?

anonymous · Answer

They're wanting you to make it x = (e-e^y)/y then finding the second derivative of that, yah?

yttrium · Answer

Oh my. I just lost my efforts, if that's the case :(

anonymous · Answer

You're definitely more right than me...