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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 $x$, $1$?

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  • $\begingroup$ Can you find the left-hand derivative at zero? Is it equal to $1$? $\endgroup$ – user539887 Jul 15 '18 at 21:24
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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 $0$.

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  • $\begingroup$ why is it that we can only make deductions about the left and right derivatives? if the case includes zero then surely you could write it as an interval, say ( ∞ , 0 ] and this would include the derivative at 0. $\endgroup$ – user571032 Jul 15 '18 at 21:54
  • $\begingroup$ Because the assertion $x\geqslant0\implies|x|=x$ only contains information about what happens on $[0,+\infty)$. SInce $0$ is the extreme left of this interval, the onlt information we can extract is therefore about the right derivative at $0$. That's all. $\endgroup$ – José Carlos Santos Jul 15 '18 at 21:57
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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.

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