# Uniqueness of derivative

I don't know how to phrase the title better.

We say that a function is "growing" (don't know the English term) if

$f'(x)\ge0, \forall x$

However, if we want to say that the function is "strictly growing", we also need an extra statement saying that

$S=\{x | f'(x)=0\}$

contains no intervals. This means that if the derivative of a function is 0 in one isolated point, the function can still described as "strictly growing".

That got me thinking. If we had a function like the following:

$f(x)=x$

It's derivative is obviously

$f'(x)=1$

But why wouldn't the function

$g(x)=\begin{cases} 1 &\mbox{if } x \ne 2 \\ 0 & \mbox{if } x = 2 \end{cases}$

Also be it's derivative? I've chosen the number 2 at random but the point is that only one of the points differs which I don't think is enough for the function to "change it's angle". So, what gives?

Furthermore, what does the function look like if it's derivative is something like

$g(x)=\begin{cases} 1 &\mbox{if } x \in \mathbb{Q} \\ 0 & \mbox{if } x \not\in \mathbb{Q} \end{cases}$

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– oks Mar 5 '13 at 12:50
– oks Mar 5 '13 at 12:58

The derivative is the limit of $$\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$ When a function is differentiable the limit exists, and limits must be unique, so $g(x)$ is not the derivative.
There is no function whichs derivative is $g(x)$ as $g(x)$ is discontinuous in every point.