# can you please explain how to to find dy/dx for the function x^2 y+ Y^2 x = -2

I need to find dy/dx of this function and evaluate the derivative at the point (2,-1)

x^2 y+y^2 x = -2

I need to find dy/dx of this function and evaluate the derivative at the point (2,-1)

x^2 y+y^2 x = -2
solve the equation to be  = 0

x^2y + y^2x +2 = 0 find partial derivatives for dy and dx,

the first term has two parts

x^2y has two partial derivatives 2xy dx + x^2 dy

the second term has 2 parts

y^2x has 2 partial derivatives y^2 dx + 2yx dy

and 2 has no derivative

2xy dx + y^2 dx + x^2 dy + 2yx dy = 0

2xy + y^2 dx = (-x^2 - 2xy) dy

dy/dx = (2xy + y^2)/(-x^2 - 2xy)

use the point (2,-1)

dy/dx = [2*2*(-1) + (-1)^2]/[-1(2)^2 -2*2*(-1)]

dy/dx = (-4 + 1) / (-4 + 4)  there is a 0 in the de=nominator therefore it is undefined

answered Dec 20, 2011 by Level 9 User (43,640 points)

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