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## #1 2006-03-02 18:37:52

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,812

### Differentiation conundrum

I found this browsing the net. Any comments?

The derivative of x², with respect to x, is 2x.  However, suppose we write x² as the sum of x x's, and then take the derivative:

Let f(x) = x + x + ... + x  (x times)

Then f'(x) = d/dx[x + x + ... + x]  (x times)
= d/dx[x] + d/dx[x] + ... + d/dx[x]  (x times)
= 1 + 1 + ... + 1  (x times)
= x

This argument appears to show that the derivative of x², with respect to x, is actually x.  Where is the fallacy?

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## #2 2006-03-03 04:06:24

Member
Registered: 2005-11-28
Posts: 97

### Re: Differentiation conundrum

I don't know...the only thing I can think is that
x+x+x+.....+x  (x times) is only defined for x in Z+

So that is isn't continuous anywhere or differentiable, really...

The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
-Bertrand Russell

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## #3 2006-03-03 04:31:32

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: Differentiation conundrum

Also the equation cannot be written down without ellipsis (...) in it.  Or you could write a summation.
But the number of terms changes with x, so the equation keeps changing, plus for negative x or fractional x's,
how many terms are there?

igloo myrtilles fourmis

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## #4 2006-03-03 06:03:02

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: Differentiation conundrum

The fallacy is that you're treating the (x times) bit as a constant.

What you're actually doing there is differentiating kx to get k.

Why did the vector cross the road?
It wanted to be normal.

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