Why can't we take derivative when the number of terms depends on $x$?

Question

See this answer.

You cannot take the derivative of $\underbrace{x + x + x + \dots + x}_{\text{repeated $x$ times}}$ with respect to $x$ one term at a time because the number of terms depends on $x$.

In this case, why can't we take derivative one term at a time when the number of terms depends on $x$?

This supposes $x$ can only be a natural number. What would a derivative mean in this context? — Bernard, Sep 06 '20 at 11:23
Sure, but ‘repeated $\lfloor x\rfloor$ times’ is not the same as ‘repeated $x$ times’. — Bernard, Sep 06 '20 at 11:41
You wrote the second phrasing, which does not mean the same as the first. — Bernard, Sep 06 '20 at 11:57
Not clear what you mean when you say $x$ times for non natural $x$. if you mean $(x[x])'=[x]$ where that's defined, that would be right. — Macavity, Sep 06 '20 at 12:03

zwim · Accepted Answer · 2020-09-06T12:01:46.303

Well you can it is by the chain rule : $$\underbrace{(1+1+\cdots+1)}_{x \text{ times}}+\underbrace{(x+x+\cdots+x)}_{1\text{ times}}=x+x=2x$$

More seriously, derivating requires a 'continuous' variable, it makes no sense for integers. I talk about integers because the underbrace $\underbrace{\hphantom{---}}_{n\text{ times}}$ implicitely supposes $n$ is an integer.

However the continuous version of the sum $\sum\limits_{k=0}^n f(k)$ is the integral $\displaystyle \int_0^x f(t)\mathop{dt}$

And by the fundamental theorem of calculs this integral has derivative $f(x)$, so we can somehow give a (very twisted I admit) meaning to the derivative of the underbrace with an $x$ variable.

See Mickael E2's answer in the link you provided yourself https://math.stackexchange.com/a/3643749/39926

I understood your answer but not the point of your answer. Please explain. — Niraj Raut, Sep 06 '20 at 11:51

Why can't we take derivative when the number of terms depends on $x$?

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