TITLE: How to disprove this fallacy that derivatives of $x^2$ and $x+x+x+\dots\quad(x\text{ times})$ are not same.
QUESTION [5 upvotes]: Possible Duplicate:
Where is the flaw in this argument of a proof that 1=2? (Derivative of repeated addition) 

\begin{align*}
x^2 &= \underbrace{x + x + x + \dots + x}_{x \text{ times}}, \\
\therefore \frac{\mathrm{d}}{\mathrm{d}x} (x^2)
&= \frac{\mathrm{d}}{\mathrm{d}x} (\underbrace{x + x + x + \dots + x}_{x \text{ times}}) \\
&= \underbrace{1 + 1 + 1 + \dots + 1}_{x \text{ times}} \\
&= x.
\end{align*}
But we know that
$$ \frac{\mathrm{d}}{\mathrm{d}x} (x^2) = 2x. $$
So what is the problem?
My take is that
we cannot differentiate both sides because $\underbrace{{x+x+x+\cdots+x}}_{x \text{  times}}$ is not fixed and thus $1$ is not equal to $2$.

REPLY [17 votes]: Simply because "$x \text{ times}$" is also a "function" of $x$. One mistake is not considering that in the derivation.

REPLY [8 votes]: You say "$x\text{ times}$".  The number of "times" you add it up---the number of terms in the sum---keeps changing as $x$ changes.  An what if $x=1.6701$?  How do you add up $x$ $1.6701$ times?