Find $y''$ by implicit differentiation.

Question

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.

Answer 1

Given:

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

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 $y' = -2x/y$:

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

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

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