# calculus

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find all points on the curve x^2y^2+xy=2 where the slope of the tangent line is -1

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 -

try 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 -

Can you explain using implicit differention?

• calculus -

No, 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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