Question

second derivative of y3+y-2 x2 y=12

Answer 1

\[y^3+y-2x^2y=12\]
Do you need to find y''?

Answer 2

with respective to x & y both

Answer 3

If so,
\[3y^2y'+y'-2(x^2y'+2xy')=0\]\[3y^2y'+y'-2x^2y'-4xy=0\]
Hence.
\[y'=\frac{ 4xy }{ 3y^2-2x^2+1 }\]
Second derivative.
\[y''\frac{ (3y^2-2x^2+1)[4(xy'+y)]-(4xy)[6yy'-4x] }{ (3y^2-2x^2+1)^2 }\]

Answer 4

thank q