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Suppose $2x^2y - 3y^2 = 4$.  Find $y''$

1.)  First find $y'$:

Easiest method is to use implicit differentiation.  Take derivative (with 
respect to x) of both sides.
\v
\centerline{${d \over dx}(2x^2y - 3y^2) = {d \over dx}(4)$}
\v
\centerline{$4xy + 2x^2y' - 6yy' = 0$}

Solve for $y'$ (note this step is easy as one can factor $y'$ from some 
terms):


\centerline{$y'(2x^2 - 6y) = -4xy$}
\v

\centerline{$y'= {-4xy \over 2x^2 - 6y} ={-2(2xy) \over -2(3y - x^2)} = 
{2xy \over 3y - x^2}$}

Hence $y'= {2xy \over 3y - x^2}$

2.)  To find $y''$, take derivative of $y'$:


$y''= \big({2xy \over 3y - x^2}\big)'$ 
$= {(2xy' + 2y)(3y - x^2) - 2xy(3yy' - 2x)\over (3y - x^2)^2}$

Since $y'=  {2xy \over 3y - x^2}$,

$y''= {(2x\big({2xy \over 3y - x^2}\big) + 2y)(3y - x^2) - 2xy(3y\big({2xy 
\over 3y - x^2}\big) - 2x)\over (3y - 
x^2)^2}$





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