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Homework Help: Differentiate e^(a^2 z) with respect to z

  1. Feb 8, 2006 #1
    I am to differentiate with respect to z, where a is independent of z... (am assuming that that a is a constant?)

    a. [tex]e^{a^2 z}[/tex]=diff=>[tex]a^2 e^{a^2 z}[/tex]

    b.[tex]e^{ia z}=cos(az)+isin(az)[/tex]=diff=>[tex]-sin(az)+icos(az)[/tex]

    c. [tex](e^{-i z})^2=cos(2z)-isin(2z)[/tex]=diff=>[tex]-sin(2z)-icos(2z)[/tex]

    d. [tex]e^{-iz^2}[/tex]=diff=>[tex]-sin(z^2)-icos(z^2)[/tex]

    e. [tex]e^{az}+e^{-az}[/tex]=diff=>[tex]ae^{az}+-ae^{-az}[/tex]

    do these look alright?
     
  2. jcsd
  3. Feb 8, 2006 #2

    quasar987

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    Whatever it is, it does not depend on z, so when z changes, a does not change. That's the definition of "constant".

    Double check b,c,d; you're forgetting the derivative of the interior (chain rule).

    note: you don't need to write e^ix in temrs of cis(x) b4 taking the derivativ; complex exponentials differentiate like regular ones, i.e.

    (e^iax)' = ia(e^iax)
     
    Last edited: Feb 8, 2006
  4. Feb 8, 2006 #3
    for b.
    [tex]-azsin(az)+azicos(az)[/tex]

    for c.
    [tex]2isin(2z)-2cos(2z)[/tex]

    for d.
    [tex]-4z(icos(z^2)+sin(z^2))+2(cos(z^2)-isin(z^2))[/tex]

    how does it look?
     
  5. Feb 8, 2006 #4
    let me correct myself:
    for b:
    [tex]-asin(az)+aicos(az)[/tex]

    for c:
    [tex]-2sin(2z)-i2cos(2z)[/tex]

    for d:
    [tex]-2z(sin(z^2)+icos(z^2))[/tex]
     
  6. Feb 9, 2006 #5

    VietDao29

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    Looks good. But as quasar987 said before, just try to differentiate it without changing to cis function.
    [tex](e ^ {iaz})' = ia e ^ {iaz}[/tex], which is a lot easier, right? This answer is exactly the same as yours:
    [tex]ia e ^ {iaz} = ia (\cos (az) + i \sin (az)) = ia \cos (az) - a \sin (az)[/tex]
     
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