Answer
$\frac{dy}{dx} = 2x-1 $
Work Step by Step
$y = x^2 - x $
$\frac{dy}{dx} = \lim\limits_{∆x \to 0} \frac {f(x+∆x) -f(x)}{∆x} $
$\frac{dy}{dx} = \lim\limits_{∆x \to 0} \frac {[(x+∆x)^2 - (x+∆x)]-[x^2-x]}{∆x} $
$\frac{dy}{dx} = \lim\limits_{∆x \to 0} \frac {x^2+2x∆x+(∆x)^2-x-∆x-x^2+x}{∆x} $
$\frac{dy}{dx} = \lim\limits_{∆x \to 0} \frac {2x∆x+(∆x)^2 -∆x}{∆x} $
$\frac{dy}{dx} = \lim\limits_{∆x \to 0} \frac {∆x(2x+∆x-1)}{∆x} $
$\frac{dy}{dx} = \lim\limits_{∆x \to 0} 2x+∆x-1 $
$\frac{dy}{dx} = 2x+0-1 $
$\frac{dy}{dx} = 2x-1 $