Could someone explains what's gone wrong here?

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SUMMARY

The discussion centers on a flawed argument regarding the derivative of the function f(x) = x². The incorrect reasoning suggests that rewriting x² as the sum of x added to itself x times leads to the conclusion that the derivative is x, rather than the established fact that the derivative is 2x. The mistake lies in misunderstanding the definition of the derivative and failing to account for the variable nature of x in the summation. This oversight is further illustrated by the example of f(x) = 5x, which incorrectly evaluates to zero under the same flawed reasoning.

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Firepanda
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I just read this on another forum

'We know that the derivative of x^2 with respect to x is 2x. However, what if we rewrite x^2 as the sum of x x's, and then take the derivative:

d/dx[ x^2 ] = 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 shows that the derivative of x^2 with respect to x is actually x. So what's going on here?'

Wheres the mistake? :P
 
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What happens if x is not a positive integer? What if x is negative or a fraction like 1/2? How would it evaluate?

However, that's not all that's wrong with this argument. Are you familiar with the definition of the derivative? If you are, then take a closer look at the definition to see if this method is still valid under the original fundamental definition.
 
The function
[tex]f(x)= 5x= (5+ 5\c+ 5 \cdot\cdot\cdot+ 5)[itex]{x times} <br /> is a function of x: your method would say <br /> [itex]f'(x)= (0+ 0+ 0\cdot\cdot\cdot+ 0)= 0[/itex] <br /> which is, of course, not true. <br /> [itex]f(x)= x^2= (x+ x+ \cdot\cdot\cdot+ x)[/itex] {x times} <br /> <br /> is a function of x in <b>two</b> ways. You have not taken into account the "x times" part of the function.[/itex][/tex]
 

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