Question

Find the derivative of f(x) = |x2 – 1| at the point (1, 0).

Answer 1

`\bf \stackrel{f(x)}{y}=|x^2-1|\implies y=√((x^2-1)^2)\implies y=[(x^2-1)^2]^{(1)/(2)} \\\\\\ \cfrac{dy}{dx}=\stackrel{chain~rule}{\cfrac{1}{2}[(x^2-1)^2]^{-(1)/(2)}\cdot 2x}\implies \cfrac{dy}{dx}=x[(x^2-1)^2]^{-(1)/(2)} \\\\\\ \cfrac{dy}{dx}=\cfrac{x}{[(x^2-1)^2]^{(1)/(2)}}\implies \cfrac{dy}{dx}=\cfrac{x}{√((x^2-1)^2)} \\\\\\ \left. \cfrac{dy}{dx}=\cfrac{x} \right|_(x=1)\implies \cfrac{1}1-1\implies \cfrac{1}{0}\implies und efined`

meaning, the function is not differentiable at x = 1, graph wise, it simply means is not a "smooth curve", instead is an abrupt edge, or a "cusp" or a "spike", so the graph may be continuous at x = 1, however, is the extremum is not a smooth curve, notice the picture below.