In the video lecture that I am watching, the instructor compared an example of finding the derivative of the $f(x) = |x|$ vs $f(x) = x|x|$, where $f(x) = x|x|$, derivative is $0$, whereas $f(x) = |x|$, does not have a derivative.
The instructor highlighted that the absolute value function does not have a derivative compared to $f(x) = x|x|$.
If I would to apply the power rule: $f(x) = nx^{n-1}$, where $f(0) = 1*0^{1-1}$. Because 0 to the power of 0 is undefined.
Is it because $0$ to the power of $0$ is undefined, therefore the absolute value function does not has a derivative?
Best Answer
Recall the definition of the derivative as the limit of the slopes of secant lines near a point.
$$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$
The derivative of a function at $x=0$ is then
$$f'(0) = \lim_{h\to 0}\frac{f(0+h)-f(0)}{h}= \lim_{h\to 0}\frac{f(h)-f(0)}{h}$$
If we are dealing with the absolute value function $f(x)=|x|$, then the above limit is
$$\lim_{h\to 0}\frac{|h|-|0|}{h} = \lim_{h\to 0}\frac{|h|}{h}$$
If $h$ approaches $0$ from the left, it is negative, so that $|h| = -h$ and the above limit is $-1$. But if $h$ approaches $0$ from the right, it is positive, so that $|h|=h$ and the limit is $1$. The left and right hand limits disagree, so the derivative does not exist at $0$.