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Given $2x^2 + y^2 = 4$, find $y''$. I found the first derivative which is $y' =$ ${-2x}\over y$.

I then got to $2(xy' - y)\over y^2$ and I don't know where to go from there.

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Plug in the first derivative you've found. – Andrew Oct 19 '12 at 22:56
so $2({{-2x^2}\over y} - y)\over y^2$? – dsta Oct 19 '12 at 22:58
Yep, although you may want to simplify the expression. – Andrew Oct 19 '12 at 22:59
The same way you'd do with any fraction. Just be careful so you don't multiply or divide by $0$. – Arthur Oct 19 '12 at 23:01
When you simplify by bringing your top to the common denominator $y$, you may want to use the fact that $2x^2+y^2=4$. – André Nicolas Oct 19 '12 at 23:11


$2x^2 + y^2 = 4$

Take the derivative of both sides with respect to x:

$4x + 2yy' = 0$

Now take the derivative again:

$4 + (2yy'' + 2y'^2)=0$

Now substitute $y' = -2x/y$:

$4 + (2yy'' + 2(-2x/y)^2)=0$

$2yy'' = -4-8x^2/y^2$

$y'' = {-2-4x^2/y^2 \over y}$

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