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It seems reasonable to write $x^2=x+x+...+x$ ($x$ times) but we run into a problem with derivatives if we do this.

The derivative of $x^2$ is $2x$ but the derivative of the sum on the right hand side turns out to be $x$, which can't be possible..

My question is where do we commit a mistake here? Is it while writing $x^2$ as the sum or while differentiating that sum?

Lalit Tolani
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Sisyphus
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    1) The usual rules for derivation apply only to sums with a *fixed* number of terms. 2) the identity $x^2=x+\ldots+x$ only makes sense at *integers* (ie discrete points), so it makes no sense to differentiate it. – Aphelli Oct 28 '21 at 10:00
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    Try instead $x^{2}=\int_{0}^{x}x\,dy$. – Ali Oct 28 '21 at 10:17
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    Try a more extreme example: $x = 1+\cdots+1$, but the derivative of the RHS is zero! The problem is that we need to apply the chain rule properly, and Ali's comment points to how we can make this rigorous. – Slade Oct 28 '21 at 11:28
  • For $x=3$ I can write the sum $x^2=x+x+x.$ What do I write for $x=3.24$? – David K Oct 28 '21 at 12:45

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The problem lies in the $(x \text{ times})$. It's difficult to express the right hand side of your equation purely algebraically. Because in the definition of the derivative we have

$$\frac{df}{dx} = \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$

and we need $f(x+h)$ as part of evaluating the limit, typically by replacing $x$ with $x + h$. In many attempts to formulate the function, it turns into $(x+h \text{ times})$ and then $f(x+h)=(x+h)(x+h)$ so it all works out. (Try it!)

Moreover, applying product rule to $xx$ also leads to $2x$.

okzoomer
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