So i am here with my question , Question asks $(x-y)^2 = x + y - 1$.
$2 (x-y) (1- y') = 1 + y'$
I am a little confused when it comes to right here, can anyone clarify a little,
I know the step that should come next which is $2 (x-y) - 2 (x-y) y' = 1 + y'$ but how did they get $-2(x-y)y'$?
What the authors probably do is suppose that $y$ is a function of $x$. That way, you have that $$(x+y-1)' = (x)' + (y(x))' - (1)' = 1 + y' - 0$$ on the right side. The left side is best seen as a function of $x$ that maps $x$ to $(x-y(x))^2$. Differenting this gives $$2(x-y(x)) \cdot (x-y(x))'= 2(x-y(x))(1-y')$$