Question

How do you find d2y/dx2 by implicit differentiation where `x^2y + xy^2 = 3x`?

Answer 1

`y''=(2*x^4y+3x^3y^2+3x^2y^3+xy^4+x^3y+3x^2y^2+3xy^4+2y^4-3x^3+3x^2y-3x^2-9xy-6y^2-9x)/(x^3(x+2y)(x+2xy)^2)`

Explanation:

Differentiating
`x^2y+xy^2=3x`
with respect to `x` we get

`2xy+x^2y'+y^2+2xyy'=3`
so we get

`y'=(3-y^2-2xy)/(x^2+2xy)`

for the second derivative we obtain

`y''=((-2yy'-2y-2xy')(x+2xy)-(3-y^2-2xy)(2x+2y+2xy'))/(x^2+2xy)^2`

Now plug the result for `y'` in this equation!