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

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Answer 1 · 2018-06-03T11:29:31

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

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!

How do you find d2y/dx2 by implicit differentiation where x^2y + xy^2 = 3xx2y+xy2=3xx^2y + xy^2 = 3x?