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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.

  • 3
    $2(1+x)=y^2+2y+1=(y+1)^2$.2012-05-27

1 Answers 1

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

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

$2yy'+2y'=2$

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

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

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

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

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 $2(1+x)$:

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

  • 0
    Awesome. Thanks. :)2012-05-27