## University Calculus: Early Transcendentals (3rd Edition)

The graph of $f'(x)$ and the graph of $y=|x|/x$ are down below.
$$f(x)=|x|$$ For $x\ge0$, $f(x)=x$ and for $x\lt0$, $f(x)=-x$. Therefore, we divide $f(x)$ into 3 intervals to find its derivative: 1) On $(-\infty,0)$: $f(x)=x$ $$f'(x)=\lim_{z\to x}\frac{f(z)-f(x)}{z-x}=\lim_{z\to x}\frac{z-x}{z-x}=1$$ 2) On $(0,\infty)$: $f(x)=-x$ $$f'(x)=\lim_{z\to x}\frac{f(z)-f(x)}{z-x}=\lim_{z\to x}\frac{-z-(-x)}{z-x}=\frac{x-z}{-(x-z)}=-1$$ 3) At $x=0$: $f(0)=0$ $$f'(0)=\lim_{z\to0}\frac{f(z)-f(0)}{z-0}=\lim_{z\to0}\frac{f(z)-0}{z}=\lim_{z\to0}\frac{f(z)}{z}$$ - As $z\to0^+$, $|z|=z$, so $\lim_{z\to0^+}\frac{f(z)}{z}=\frac{z}{z}=1$ - As $z\to0^-$, $|z|=-z$, so $\lim_{z\to0^-}\frac{f(z)}{z}=\frac{-z}{z}=-1$ Therefore, $\lim_{z\to0}\frac{f(z)}{z}$ does not exist, so $f$ is not differentiable at $x=0$. The graph of $f'(x)$ and the graph of $y=|x|/x$ are down below. In the graph, we see only the blue lines, which is the graph of $y=|x|/x$. Well in fact, these two graphs overlap each other throughout the whole of their domains. In other words, $$f'(x)=\frac{|x|}{x}$$