looking at the piecewise definition why isn't the absolute value of x differentiable at 0?

Question

I understand how simple it is to see that the absolute value function is not differentiable at zero however by definition we have that the absolute value of $x$ is equal to $x$ when $x$ is greater than or equal to $0$, but that why is the derivative at zero not the derivative of $x$, $1$?

Answer 1

Because from the fact that $x\geqslant0\implies|x|=x$, the only thing that you can deduce about the differentiablity of the aboslute function at $0$ is that the right derivative at $0$ is $1$. And $x\leqslant0\implies|x|=-x$, from which you can deduce that the left derivative at $0$ is $-1$. Since the left and the right derivatives at $0$ are different, the absolute value function is not differentiable at $0$.

Answer 2

You could just as easily have said:

however by definition we have that the absolute value of $x$ is equal to $-x$ when $x$ is less than or equal to 0, but then why is the derivative at zero not the derivative of $-x$, $-1$?

In more detail: the derivative is (by definition) a limit of a difference quotient, i.e. with $f(x)=|x|$ we have $f'(0)$ defined by $$f'(0) =\lim_{h\to 0}\frac{|h|-0}{h}$$

I think many of us, especially when we are new to Calculus, tend to have an unconscious "right-handed bias": when we think about an expression like $f(a+h)$ we naturally think of it as calculating the value of $f(x)$ at a point slightly to the right of $x=a$. But the limit is a two-sided operation, and there's no reason to think of $h$ as a positive number. In this case, it's true that as you approach $h=0$ from the right, the difference quotient is $1$. However, as you approach $h=0$ from the left, the difference quotient is $-1$. Because these values don't agree, the limit at $h=0$ does not exist.