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Consider $f(x)=x^2$
Then $f'(x) = 2x$
But $f(x) = x+x+x + ...+x$ ($x$ times)
then $f'(x)=1+1+1+...+1$
  i.e. $f'(x)$ $=x \ne 2x$
So this implies that if we see derivative as an operator, its proper usage needs that the term that we differentiating must NOT be variable in terms of size. The representation should be fixed wrt to the variable that we are diffrentiating i.e the ones that consist of $f(x)=\sum_{i=lo}^{hi}g(i)$ with $lo,hi$ independent of the variable $x$.
Is there any better explanation?

maverick
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    Rewriting $x^2=\sum_{i=1}^x x$ only works for natural number values of $x$, and you can't differentiate a function only defined on $\mathbb{N}$. – florence May 22 '17 at 04:08
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    I don't see the paradox. You are not taking the derivative correctly when considering $f(x)=x+x+...$, $x$ times. – Vladimir Vargas May 22 '17 at 04:08
  • @VladimirVargas I understand your point. I have changed the title. I was curious to see what went wrong. – maverick May 22 '17 at 04:11
  • @florence That makes sense. So does this mean that we can NEVER take derivative when the summation limits are a function of $x$? – maverick May 22 '17 at 04:13
  • The number of terms is a variable, which you are differentiating. You can extend such a sum to $\mathbb{R}$ I'm sure, but you would still have to differentiate with respect to the number of terms and I don't see how you could accomplish that. – Kaynex May 22 '17 at 04:13
  • See the comment of florence. Anyway, to say what's wrong "intuitively", when taking derivative, not only does the $x$ in the sum changes, but your "number of times" also changes. – velut luna May 22 '17 at 04:14
  • @velutluna I understand. As I have asked above, does this mean that we can NEVER take derivative when the summation limits are a function of $x$? – maverick May 22 '17 at 04:17
  • We probably can with SOME method, but it definitely wouldn't be what you did above. – Kaynex May 22 '17 at 04:23
  • @Kaynex What can be that "SOME" method? I want to know. – maverick May 22 '17 at 04:24

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