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Differentiating an exponential with a complex exponent

  1. Feb 18, 2016 #1
    Hello, folks. I'm trying to figure out how to take the partial derivative of something with a complex exponential, like
    [tex] \frac{\partial}{\partial x} e^{i(\alpha x + \beta t)} [/tex]
    But I'm not really sure how to do so. I get that since I'm taking the partial w.r.t. x, I can treat t as a constant term and thus pretend it's something like
    [tex] \frac{\partial}{\partial x} e^{i(\alpha x +\beta)} [/tex]
    But then my confusion comes from me not being able to separate the exponent into some suitable form like
    [tex] e^{\alpha + \beta i} [/tex]
    I guess I could separate it into two separate ones, like
    [tex] e^{i\alpha x}e^{i\beta} [/tex]

    How should I deal with this, any pushes in the right direction?
     
  2. jcsd
  3. Feb 18, 2016 #2

    fresh_42

    Staff: Mentor

    You should apply the chain rule of differentiation to ##e^{f(x)}##.
     
  4. Feb 18, 2016 #3
    Thanks, lemme take a shot.

    [tex]\frac{\partial}{\partial x}e^{i(\alpha x +\beta t)} = e^{i (\alpha x + \beta t)} \cdot \frac{\partial}{\partial x}[i(\alpha x + \beta t)]= i \alpha e^{i (\alpha x + \beta t)}[/tex]
     
  5. Feb 18, 2016 #4
    Is this correct, I guess? I got the same thing by the product rule.

    Edit: it must be, because when I applied this bit to the larger problem I was working on I got it right! Thanks so much!
     
    Last edited: Feb 18, 2016
  6. Feb 18, 2016 #5

    fresh_42

    Staff: Mentor

    Yes, except you lost the ##t## in your product rule. It's going to zero (in one term) as it is viewed as a constant by partial differentiation but you must not just drop it before and then again pull it out of the hat again.
     
  7. Feb 18, 2016 #6
    Uh oh, I don't get what you just said, let me work it out again both ways and see where I'm making the mistake.
     
  8. Feb 18, 2016 #7

    fresh_42

    Staff: Mentor

    ##t## has gone.
    And you said you got the same result by product and by chain rule, so there is a ##t## again of the exponent in the result.
     
  9. Feb 18, 2016 #8
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