posted by renee .
find dy/dx implicit differentiation
at x=-2, x=2
2xydx+x^2 dy + 2yxdy+y^2 dx=0
ok, at the given point, you can figure dy/dx. I am not certain what you mean for the point to be.
x^2(dy/dx) + 2xy + y^2 + 2xy(dy/dx) = 0
(dy/dx)(x^2 + 2xy) = - y^2 - 2xy
dy/dx = (-y^2-2xy)/(x^2 + 2xy)
when x = -2 in original
4y + 2y^2 = 0
2y(2 + y)=0
y = 0 or y = -2
now plug in (-2,0) and (-2,-2) into dy/dx
do the same using x = 2
Do you mean x = -2, y = 2?
Do you mean x^2*y + y^2*x = 0 ?
What are the exponents?
2 or 2x and 2y?
x cannot have two values unless you want the derivative at two different points.
Anyway, implicit differentiation with respect to x results in:
x^2*dy/dx + y*2x + y^2 + 2y = 0
I messed up in the calculation of the point
when x = -2
4y - 2y^2 = 0
y = 0 or y = 2
So the second point is (-2,2) not (-2,-2)