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

1 Answer
Jun 3, 2018

#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!