How to disprove this fallacy that derivatives of $x^2$ and $x+x+x+\dots\quad(x\text{ times})$ are not same.

Question

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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 $2$.

Answer 1

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

Answer 2

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?