Consider the implicit equation 2xy-1=(x+y+1)^2

Maybe my derivative isn't correct in the first place? I used the quadratic formula to solve for x and got x being equal to -1 but when I plug that into the equation I get 0/0, which certainly means I'm doing something wrong. Any ideas?

Any ideas anyone? I need to have this turned in by tomorrow, any help is greatly appreciated

2xy-1=(x+y+1)^2
2y + 2xy' = 2(x+y+1)(1+y')
2y + 2xy' = 2(x+y+1) + 2(x+y+1)y'
y' (2x - 2(x+y+1)) = 2x+2
y' = -(x+1)/(y+1)

So, your derivative is ok.

when y = -1, x = -1, so y' = 0/0

so, what happens near (-1,-1)?
If y = -1+h, x = -1+h+√(2h)
so, y' = -(h+√(2h))/h
as h->0, y' -> -1 -√(2/h)
So, it looks like a vertical tangent.

But it still bothers me. If we look back at the original equation,

(x+y+1)^2 = 2xy-1
x^2+y^2+1+2xy+2x+2y - 2xy+1 = 0
x^2+2x+1 + y^2+2y+1 = 0
(x+1)^2 + (y+1)^2 = 0

The only point which satisfies this equation is (-1,-1)!

So, there is no tangent line at all, since f(x,y) is a single point!

That is the second strange tangent line problem today. The first one was horizontal :)

I had that on the back of my head when I was doing this but thought that wouldn't be possible, thanks so much!