Answer
$g^{'}(1)=1$
Work Step by Step
$g(x)=x^{2}-x$
$g^{'}(c)=\lim\limits_{x \to c} \frac{g(x)-g(c)}{x-c}$
$g^{'}(1)=\lim\limits_{x \to 1}\frac{x^{2}-x-((1)^{2}-1)}{x-1}$
$g^{'}(1)=\lim\limits_{x \to 1}\frac{x^{2}-x}{x-1}$
$g^{'}(1)=\lim\limits_{x \to 1}\frac{x(x-1)}{x-1}$
$g^{'}(1)=\lim\limits_{x \to 1} x$
$g^{'}(1)=1$
