Posted by renee on Wednesday, June 8, 2011 at 8:59pm.
find dy/dx implicit differentiation
x^2y+y^2x=0
at x=2, x=2

calculus  bobpursley, Wednesday, June 8, 2011 at 10:01pm
2xydx+x^2 dy + 2yxdy+y^2 dx=0
dy/dx(x^2+2yx)=(y^2/2xy)
ok, at the given point, you can figure dy/dx. I am not certain what you mean for the point to be.

calculus  Reiny, Wednesday, June 8, 2011 at 10:04pm
x^2(dy/dx) + 2xy + y^2 + 2xy(dy/dx) = 0
(dy/dx)(x^2 + 2xy) =  y^2  2xy
dy/dx = (y^22xy)/(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

calculus  drwls, Wednesday, June 8, 2011 at 10:08pm
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

correction  calculus  Reiny, Wednesday, June 8, 2011 at 10:13pm
I messed up in the calculation of the point
when x = 2
4y  2y^2 = 0
2y(2y)=0
y = 0 or y = 2
So the second point is (2,2) not (2,2)
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