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August 23, 2014

August 23, 2014

Posted by **renee** on Wednesday, June 8, 2011 at 8:59pm.

x^2y+y^2x=0

at x=-2, x=2

- calculus -
**bobpursley**, Wednesday, June 8, 2011 at 10:01pm2xydx+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 10:04pmx^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

- calculus -
**drwls**, Wednesday, June 8, 2011 at 10:08pmDo 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 10:13pmI messed up in the calculation of the point

when x = -2

4y - 2y^2 = 0

2y(2-y)=0

y = 0 or y = 2

So the second point is (-2,2) not (-2,-2)

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