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Complex derivatives

  1. Feb 23, 2007 #1
    I have to solve an ODE with variation of coefficient technique. It's pretty easy but I have no clue what is the first and second derivative of e^ix and e^-ix.
     
    Last edited: Feb 23, 2007
  2. jcsd
  3. Feb 23, 2007 #2

    arildno

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    If i had been a real number, what would the first and second derivatives have been then?
     
  4. Feb 23, 2007 #3
    e^ix
    first
    i*e^ix
    second
    i^2*e^ix

    e^-ix
    first
    -i*e^-ix
    second
    i^2*e^-ix

    p.s. I've read about the Cauchy-Riemann equation but just not sure how to apply it... should I split the exponential in a sin and a cos?
    p.s.s. There are probably rules, like exponential function are always derivable or something but I'm not fallowing any complex variables class right now so any insight is appreciated...
     
    Last edited: Feb 23, 2007
  5. Feb 23, 2007 #4

    arildno

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    EXACTLY!
    And that is precisely what holds when "i" is a complex/imaginary number as well! :smile:
     
  6. Feb 23, 2007 #5

    Gib Z

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    When dealing with these things, forget i is anything, just remember its a constant. Then after the actual differentiation, you can remember what it is.
     
  7. Feb 23, 2007 #6
    Yeah. If

    [tex]\exp(ix),\,\,\,x\in \mathbb{R},[/tex]

    (which is what it looks like you have) then it's what the above two said. But if you have

    [tex]\exp(iz),\,\,\,z\in \mathbb{Z},[/tex]

    you need to be more careful. Let us know if that is indeed what you have.
     
  8. Feb 23, 2007 #7

    mathwonk

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    what are you doing in a de course ifm you do not know the derivative of e^z?
     
  9. Sep 1, 2010 #8
    I'm doing the same derivative problem & i was wondering if you could give any tips on how to solve the derivative of e^ix? I would really appreciate it. A good reference website, anything assistance at all.
     
  10. Sep 2, 2010 #9

    HallsofIvy

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    That is exactly what has been answered in each of these responses. For any constant, a, the derivative of [itex]e^{ax}[/itex] is [itex]ae^{ax}[/itex].

    That is a result of the very basic fact that the derivative of [itex]e^x[/itex] is [itex]e^x[/itex] (world's easiest derivative!) and the chain rule.
     
  11. Sep 3, 2010 #10
    [tex]\frac{d}{dx}(e^{jx})=je^{jx}[/tex]
    [tex]\frac{d^2}{dx^2}(e^{jx})=-e^{jx}[/tex]
     
  12. Sep 3, 2010 #11

    mathwonk

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    Actually I myself was once in an ode course when I had forgot the derivative of e^x. My solution was to go get a Schaum's outline series of ode and do a lot of problems and review my $$$ off.
     
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