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Derivative of e^(x-2)

  1. Nov 13, 2004 #1

    quasar987

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    I am asked to find the derivative function of [itex]f(x)=e^{x-2}[/itex] using the definition. That is to say, I have to evaluate this limit, if it exists:

    [tex]\lim_{x\rightarrow x_0}\frac{e^{x-2} - e^{x_0-2}}{x-x_0} = \lim_{x\rightarrow x_0}\frac{e^{x} - e^{x_0}}{e^2 (x-x_0)}[/tex]

    How can this undeterminate form be simplified? Thanks.

    (The answer is [itex]f'(x_0)=e^{x_0-2}[/itex].)
     
    Last edited: Nov 13, 2004
  2. jcsd
  3. Nov 13, 2004 #2

    mathman

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    Factor out exp(x0) from the numerator. Then expand exp(x-x0) in a power series. The rest is obvious.
     
  4. Nov 13, 2004 #3

    quasar987

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    Power serie not allowed, sorry.
     
  5. Nov 13, 2004 #4

    StatusX

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    im used to defining derivatives in a slightly different way, but if you need to it should be easy to convert to your way:

    [tex]\frac{d}{dx}(e^{(x-2)}) = \lim_{h \rightarrow 0} \frac{e^{(x-2)+h} - e^{(x-2)}}{h}[/tex]

    [tex]= \lim_{h \rightarrow 0} e^{(x-2)} \frac{e^{h} -1}{h}[/tex]

    now noting that:

    [tex]e = \lim_{h \rightarrow 0} (1+h)^{1/h}[/tex]

    raising both sides to h (this is the only step I'm not sure about, but it gives the right answer):

    [tex]\lim_{h \rightarrow 0} e^h = \lim_{h \rightarrow 0} ((1+h)^{1/h})^h = \lim_{h \rightarrow 0} 1+h [/tex]

    so:
    [tex]\lim_{h \rightarrow 0} \frac{e^{h} -1}{h}= \lim_{h \rightarrow 0} \frac{1 + h -1}{h}= 1[/tex]

    which gives the answer.
     
  6. Nov 13, 2004 #5

    cepheid

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    lol, nobody is defining derivatives in different ways...only the notation differs

    [tex] h \equiv \Delta x \equiv x - x_0 [/tex]
     
  7. Nov 13, 2004 #6

    StatusX

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    yea, thats all i meant.
     
  8. Nov 13, 2004 #7

    quasar987

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    This looks nice Status, but isn't there a [itex]e^{(x-2)}[/itex] remaining?
     
  9. Nov 13, 2004 #8
    Yes, which is multiplied by 1, giving the answer, as is easily verified using the chain rule.
     
  10. Nov 14, 2004 #9

    cepheid

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    oh ok, sorry
     
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