Implicit derivative of $(x - y)^2 = x + y - 1$

Question

I changed the function from:

$(x - y)^2 = x + y - 1$ to:

$(x^2) - (y^2) - x - y = -1$ All I did was move the variables to one side.

When moved, I get $\displaystyle \frac{dy}{dx} = \frac{2x - 1}{2y + 1}$

Otherwise, I get $\displaystyle \frac{dy}{dx} = \frac{2y - 2x + 1} {2y - 2x - 1}$

Anyone else having this problem? And which one is correct?

Answer 1

But $(x-y)^2 \ne x^2-y^2$ it is $x^2-2xy+y^2$

Answer 2

$$(x - y)^2 = x + y - 1$$ $$2(x-y)(1-y')=1+y'$$ $$2(x-y)-1=y'(2(x-y)+1)$$ $$y'=\frac{2(x-y)-1}{2(x-y)+1}=\frac{2y-2x+1}{2y-2x-1}$$