derivative of $y = |x|$ and contradiction with the intermediate value property The Unit step function(derivative of $y = |x|$) does not take the values from $(0, 1)$. It is said that this function cannot be a derivative of any real valued function. But isn't $y = |x|$ a real valued function, whose domain and range are both real numbers? So, how is this unit step function not a derivative of any real valued function?
 A: $\DeclareMathOperator{\abs}{abs}$By definition, a real-valued function $f$ defined on some non-empty open set $I$ is differentiable on $I$ if $f$ is differentiable at $x$ for every $x$ in $I$.
Let $\abs(x) = |x|$ for $x$ real. The difference quotient at $0$ is
$$
\frac{\abs(h) - \abs(0)}{h} = \frac{|h|}{h},\quad h \neq 0.
$$
Since the difference quotient has no limit as $h \to 0$ (the one-sided limits exist but are unequal), $\abs'(0)$ does not exist. That is, the absolute value function is not differentiable at $0$.
Consequently, the absolute value function is "not differentiable" on any open interval containing $0$, e.g., is not differentiable on the set of real numbers. The fact that $\abs'(x)$ exists for all $x \neq 0$ is immaterial.
Careful examination of the intermediate value property for derivatives reveals that the domain of $f$ must be an interval. The absolute value function does not contradict the intermediate value property: the function $\abs'(x) = |x|/x$ for $x \neq 0$ fails to satisfy the intermediate value property on a non-empty open real interval $I$ if and only if $0 \in I$, if and only if $\abs$ is not differentiable on $I$.
