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Differential coefficient of \(x^2\) if \(x\) is a positive integer

We all know that the differential coefficient of \(x^2\) is \(2x\). But if we write
\(x^2 = \underbrace{x+ x+x+x+\cdots + x}_{x \text{ times}} \)
On differentiating both sides with respect to \(x\) we get: \(d(x^2)/dx = \underbrace{1+ 1+1+1+\cdots + 1}_{x \text{ times}} \) = \(x\)
Why has the differential coefficient changed?

Similarly one can find differences in differential coefficients of \(x^3\),\(x^4\),\(x\),etc. if \(x\) is a positive integer.

Please help in finding my mistake.

Note by Priyamvad Tripathi
5 months ago

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First off, note that equation is valid only if \(x\) is an integer. As the equation doesn't hold in any neighbourhood of a positive integer, you cannot differentiate the given equation on both sides!

Hence, the result of differentiating that equation is wrong! Deeparaj Bhat · 5 months ago

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@Deeparaj Bhat That's right. It is just like differentiating xy wrt x and then substituting y=x. Aditya Kumar · 5 months ago

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@Aditya Kumar Nice example. Deeparaj Bhat · 5 months ago

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If you look at it correctly, one way of interpreting is that the sum is only valid for for integer values. But another way of seeing it is that in writing the sum the numbers of times is x occurs is also a function in x, 'x times'. So it is possibly an implicit function of two variables. Like sum x y number of times. Now you can differentiate it wrt to x. Abhi Kumbale · 3 months, 3 weeks ago

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Thank you I found my mistake. Priyamvad Tripathi · 5 months ago

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