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### Derivatives

```
Date: 12/07/98 at 02:43:15
From: Rakesh
Subject: Derivatives

We know the derivative of x^2 with respect to x is 2*x.

But write x^2 as x+x+x+x+...(x times) and we get the derivative as x.
This should be true at least for positive integers but there are an
infinite number of examples which counter this.

How I tried:

d/dx(x^2) = 2x

Since:

d/dx(x^n) = n*x^(n-1)

Also:

d/dx(x^2) = d/dx(x+x+x+...x times)
= d/dx(x)+d/dx(x)+...x times
= 1+1+1+...(x times)
= x
```

```
Date: 12/07/98 at 10:35:43
From: Doctor Rob
Subject: Re: Derivatives

The problem is that you cannot write x^2 = x+x+...+x (x times) unless
x is a positive integer. The function on the right is only defined on
positive integers, so it is not continuous anywhere, not to mention
differentiable. Thus when you try to differentiate it, you get

By the way, dividing by x here gives a fallacious proof that 1 = 2,
using calculus!

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Functions

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