# Could someone explains what's gone wrong here?

1. May 25, 2008

### Firepanda

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

2. May 25, 2008

### Defennder

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.

3. May 25, 2008

### HallsofIvy

Staff Emeritus
The function
[tex]f(x)= 5x= (5+ 5\c+ 5 \cdot\cdot\cdot+ 5)${x times} is a function of x: your method would say [itex]f'(x)= (0+ 0+ 0\cdot\cdot\cdot+ 0)= 0$
which is, of course, not true.
$f(x)= x^2= (x+ x+ \cdot\cdot\cdot+ x)$ {x times}

is a function of x in two ways. You have not taken into account the "x times" part of the function.