You see the minus sign in front of \left( x^2 -1 \right) \cdot 1, right? That is just short notation for
- \left( x^2 -1 \right) \cdot 1=\left( -1 \right) \cdot \left( x^2 -1 \right)= (-1) \cdot x^2 + (-1) \cdot (-1) = -x^2+1
Agreed? Or are you pulling my leg here?
EDIT: Ah, good. I was...
Alright, for a function
f(x) = \frac{g(x)}{h(x)}
the quotient rule says
f'(x) = \frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)}
In your case, g(x)=x^2-1 and h(x)=x. Thus g'(x)=2x and h'(x)=1. Then you get
\frac{\mathrm{d}}{\mathrm{dx}}\frac{x^2-1}{x}=\frac{2x\cdot x - \left( x^2 -1 \right) \cdot...