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Homework Help: Could someone explains what's gone wrong here?

  1. May 25, 2008 #1
    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. jcsd
  3. May 25, 2008 #2

    Defennder

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    Homework Helper

    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.
     
  4. May 25, 2008 #3

    HallsofIvy

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

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