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Proof - the derivative of a scalar multiple

  1. Sep 22, 2008 #1
    1. The problem statement, all variables and given/known data

    I am confused how the scalar multiple is divided out of the proof of this rule without taking an h with it in the denominator, which would get very tiny meaning the entire thing would go to infinity or negative infinity or zero, you can't tell.

    Start with: f(x) = k g(x) End: f'(x) = k g'(x)



    2. Relevant equations

    This is the proof I was given:

    f'(x) = lim(h->0) [k g(x+h) - k g(x)] / h

    f'(x) = lim(h->0) [k {g(x+h) - g(x)}] /h

    Next step I do not agree with: (Never mind -this is legal, right?)
    f'(x) = lim(h->0) k [{g(x+h) - g(x)}/h]

    f'(x) = k lim(h->0) [{g(x+h) - g(x)}/h]

    f'(x) = k g'(x)

    3. The attempt at a solution

    This is what I think would happen at the step I disagree with:

    f'(x) = lim(h->0) k/h * [{g(x+h) - g(x)}/h]

    f'(x) = lim(h->0) k/h * lim(h->0) [{g(x+h) - g(x)}/h]

    f'(x) = ??? * g'(x)
     
    Last edited: Sep 22, 2008
  2. jcsd
  3. Sep 22, 2008 #2

    Dick

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    (k/h)*(g(x+h)-g(x))/h=(k*g(x+h)-k*g(x))/h^2. That's not equal to (k*g(x+h)-k*g(x))/h. Why would you do that?
     
  4. Sep 22, 2008 #3
    Check out the presentation of the formal definition of "limit" (the "delta-epsilon" formality):
    http://mathforum.org/library/drmath/view/53403.html

    That doesn't answer your question, I know. But it's an unusually interesting discussion of what is really meant by "limit".

    jf
     
  5. Sep 22, 2008 #4

    HallsofIvy

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    That's just bad algebra. You now have two "h" s in the denominators.

     
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