d/dx

Solving it using the quotient rule of derivatives we get $\frac{2x(x-1)-x^2+1}{(x-1)^2}= \frac{(x-1)^2}{(x-1)^2}=1$

$x^2-1=(x+1)(x-1)$ , so divide out a factor $x-1$ whenever $x\neq 1$ . Now $\frac{d}{dx}(x+1)=1$ . The value is undefined at $x=1$ , the problem should exclude 1 from the domain.

$\frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1$ . The derivative of this is simply the slope, or 1.