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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 nonsensical answers. 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/ |
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