${f'_ + }(1) = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\left| {{x^2} - 1} \right| - 0}}{{x - 1}} $
$= \mathop {\lim }\limits_{x \to {1^ + }} \frac{{(x - 1)(x + 1)}}{{x - 1}} = 2$
${f'_ - }(1) = \mathop {\lim }\limits_{x \to {1^ - }} \frac{{\left| {{x^2} - 1} \right| - 0}}{{x - 1}} $
$= \mathop {\lim }\limits_{x \to {1^ - }} \frac{{ - (x - 1)(x + 1)}}{{x - 1}} = - 2$
${f'_ - }(1) \ne {f'_ + }(1)$ پس تابع مشتقپذیر نمیباشد