Thread: d/dx e^x

  1. #1
    l'Anziano DavidP's Avatar
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    Question d/dx e^x

    Why does

    d/dx e^x = e^x

    instead of

    xe^(x-1)
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  2. #2
    Pursuing knowledge confuted's Avatar
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    Because you can't use the power rule with functions raised to a variable.

    d/dx k^x = k^(bx)*ln k*b
    Away.

  3. #3
    l'Anziano DavidP's Avatar
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    you mean constants raised to a variable?
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  4. #4
    Mayor of Awesometown Govtcheez's Avatar
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    Originally posted by DavidP
    you mean constants raised to a variable?
    Either one.

    Instead of using a shortcut rule, actually put it through the long definition of a derivative and work it out.

  5. #5
    Registered User whistlenm1's Avatar
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    for me I did not understand it until I found a proof, so unless your going to look at a proof take it as is.
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  6. #6
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    just take it as it is. but if you want to know why, as cheez said, put it through the long definition of the derivative.

  7. #7
    Pursuing knowledge confuted's Avatar
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    http://archives.math.utk.edu/visual....definition.12/ The definition of a derivative can be found there if you want to have a go at it
    Away.

  8. #8
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    so is this setup what cheez and you guys meant:


    (e^(x+h) - e^(x)) / (h)
    Lim h->0

    I saw what confuted put for the answer but I'm not 100% sure the steps you have to take to get there.

    EDIT: does b stand for base above? (change of base)
    Last edited by Silvercord; 11-03-2003 at 04:38 PM.

  9. #9
    Pursuing knowledge confuted's Avatar
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    b and k were arbitrary constants.

    (e^(x+h) - e^(x)) / (h)
    Lim h->0

    (e^((x+h)/x)) / (h)
    Lim h->0

    You'll have to take an ln() in there to get that to something you can work with.
    Away.

  10. #10
    Toaster Zach L.'s Avatar
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    Use the Maclaurin series for e^x, and it becomes really obvious:

    e^x = sum(k=0, infinity) [ x^k / k! ]

    So for each term,
    d[x^k / k!]/dx = k*x^(k-1) / k! = x^(k-1) / (k-1)

    That is, when you take the derivative, each term becomes its predecessor, so the series remains unchanged.
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