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.

## Comments

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TopNewestFirst off, note that equation is valid only if \(x\) is an

integer. As the equationdoesn'thold in any neighbourhood of a positive integer, youcannot differentiate the given equation on both sides!Hence, the result of differentiating that equation is wrong! – Deeparaj Bhat · 9 months, 3 weeks ago

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– Aditya Kumar · 9 months, 3 weeks ago

That's right. It is just like differentiating xy wrt x and then substituting y=x.Log in to reply

– Deeparaj Bhat · 9 months, 3 weeks ago

Nice example.Log in to reply

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 · 8 months, 1 week ago

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Thank you I found my mistake. – Priyamvad Tripathi · 9 months, 3 weeks ago

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It's there in there – Vicky Vignesh · 9 months, 3 weeks ago

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