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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 15,407

*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?

Character is who you are when no one is looking.

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**darthradius****Member**- Registered: 2005-11-28
- Posts: 97

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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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

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

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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