calculus
posted by Fredrick E. on .
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^2y)/(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

try changing variables. let z=xy in the origianal equation, solve for z
z^2+z2=0
(z+2)(z1)=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 42=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 
Can you explain using implicit differention?

No, I tried it that way, it bogged down in a third degree equation after I substituted, so I gave up that way.