Answers: 1
![\bf \stackrel{f(x)}{y}=|x^2-1|\implies y=\sqrt{(x^2-1)^2}\implies y=[(x^2-1)^2]^{\frac{1}{2}} \\\\\\ \cfrac{dy}{dx}=\stackrel{chain~rule}{\cfrac{1}{2}[(x^2-1)^2]^{-\frac{1}{2}}\cdot 2x}\implies \cfrac{dy}{dx}=x[(x^2-1)^2]^{-\frac{1}{2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{x}{[(x^2-1)^2]^{\frac{1}{2}}}\implies \cfrac{dy}{dx}=\cfrac{x}{\sqrt{(x^2-1)^2}} \\\\\\ \left. \cfrac{dy}{dx}=\cfrac{x}{|x^2-1|} \right|_{x=1}\implies \cfrac{1}{|1-1|}\implies \cfrac{1}{0}\implies und efined](/tpl/images/0146/2989/f1c83.png)
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