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.
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.
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 $y' = -2x/y$:
$4 + (2yy'' + 2(-2x/y)^2)=0$
$2yy'' = -4-8x^2/y^2$
$y'' = {-2-4x^2/y^2 \over y}$