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January 28, 2015

January 28, 2015

Posted by **Fredrick E.** on Tuesday, November 1, 2011 at 2:35pm.

I know that you have to find the derivative when it equals -1

i got

y'=(2x*y^2-y)/(2x^2*y+x)

(i don't know if it is correct or not)

but i don't know how to go from here. any help would be good

Thanks

- calculus -
**bobpursley**, Tuesday, November 1, 2011 at 3:21pmtry changing variables. let z=xy in the origianal equation, solve for z

z^2+z-2=0

(z+2)(z-1)=0

z=1 or -2 or z is a constant.

Then z=xy or

z'=y+xy'=0

but y'=1

y=-x and xy=-1 or 1

xy=-x^2

-2=-x^2

x=sqrt2, y=-x =-sqrt2

check: original line 4-2=2 checks.

1=-x^2

x=i, y=-x=-i

check: originalline 1=1=2 checks.

solutions:

pointA: x=i,y=-1

PointB: x=sqrt2, y=-sqrt2

- calculus -
**Anonymous**, Tuesday, November 1, 2011 at 6:44pmCan you explain using implicit differention?

- calculus -
**bobpursley**, Tuesday, November 1, 2011 at 10:07pmNo, I tried it that way, it bogged down in a third degree equation after I substituted, so I gave up that way.

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