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Regards. I would like to pour out something that I found on Youtube, and test my view on this.

$$ f(x) = x^{2} $$

$$ f'(x) = 2x $$

$$ x^{2} = \: x \: + \: x \: + \: ... \: + \: ... \: + \: x \: \: (\text{as many as x})$$

$$ \frac{d(x^{2})}{dx} = \: 1 \: + \: 1 \: + \: .... \: + \: 1 \: = 1 \cdot x $$

So we get from the results above : $$ 2 = 1 \: \: (??) $$

But in another way

$$ x^{2} = x \cdot x $$

$$ \frac{d(x^{2})}{dx} = (1 \cdot x) + (x \cdot 1) = 2x $$

This refers to the video :

https://www.youtube.com/watch?v=DV9I4fm2GaQ&t=191s

The mistake in the 2nd paradox in the video is clear, but for the 1st one .. some insightful thoughts?

My argument is that the 'summation' $$ x \: + \: x \: + \: ... \: + \: ... \: + \: x \: \: (\text{as many as x}) $$ may not be seen as a function of $x$. But also because this only holds when $x$ is a positive integer, and we count the limit as below

\begin{align*} \frac{d(x^{2})}{dx} &= \lim \frac{(x + \triangle x)^{2} - (x)^{2}}{\triangle x} \\ &= \lim \frac{(x^{2} + 2 x \triangle x + \triangle x^{2}) - (x)^{2} }{\triangle x} \\ &= \lim \frac{(x + x) \triangle x + \triangle x^{2} }{\triangle x} = (x+x) = 2x \end{align*}

Thanks, all the best.

Redsbefall
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