# Simple $d^2 y/dx^2$ - don't know what I'm missing

I'm studying some Calculus on my own. I get what appears to be the wrong answer for this one implicit differentiation exercise, and I don't know why.

I have to find $\frac {d^2 y}{dx^2}$ of $y^2-2x = 1-2y$. I calculate

$\frac{dy}{dx} = \frac{1}{1+y}$

and from there I get

$\frac {d^2 y}{dx^2} = \frac{-y'}{(1+y)^2} = -\frac{1}{(1+y)^3}$

Now, Wolfram Alpha tells me I got $f'$ right but $f''$ wrong. It says

$\frac{\delta^2y(x)}{\delta x^2} = - \frac{1}{2(1+x)(1+y)}$

WA hasn't steered me wrong so far. Pointers to where I'm going wrong would be appreciated.

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$2(1+x)=y^2+2y+1=(y+1)^2$. – Did May 27 '12 at 10:52

$$y^2-2x = 1-2y\tag{1}$$

$$2yy'-2=-2y'$$

$$2yy'+2y'=2$$

$$2y'(y+1)=2$$

$$y'=\frac{1}{y+1}$$

$$y'' = \frac{(1)'(y+1)-(y+1)'(1)}{(y-1)^2}=\frac{-y'}{(y+1)^2}$$

$$y''=\frac{-\frac{1}{y+1}}{(y+1)^2}=-\frac{1}{(y+1)^3}$$

As Didier pointed out in his comment, from $(1)$ we have:

$$y^2+2y\color{red}{+1} = 1+2x\color{red}{+1}$$

$$(y+1)^2 = 2(1+x)$$

So WA just substituted $(y+1)^2$ with $2(1+x)$:

$$y''=-\frac{1}{2(x+1)(y+1)}$$

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Awesome. Thanks. :) – Apteris May 27 '12 at 12:33