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Calculus question- differentiation

  1. Sep 14, 2005 #1
    Could someone please help me with the method and steps to this question:

    Find dy/dx for the following:

    y = (5e^(2+ix))/(3e^(1-2ix))
     
  2. jcsd
  3. Sep 14, 2005 #2
    Simplify it so it is of the form constant*e^(something) and then carry out differentiation of exponentials as you would.
     
  4. Sep 14, 2005 #3
    This is the way I tried to do it:

    (Quotient rule)
    dy/dx = ((5ie^(2 + ix))(3e^(1-2x)) - (-6ie^(1 - 2ix))(5e^(2 + ix))) / ((3e^(1 - 2ix))^2)

    = (15ie^(3 - ix) + 30ie^(3 - ix)) / (9(e^(1 - 2ix))^2)

    = (45ie^(3 - ix)) / (9(e^91 - 2ix)^2)

    = (5ie^(3 - ix)) / ((e^(1 - 2ix))^2)

    And I'm not sure where to go from there its either wrong or it needs simplifying some more...
     
  5. Sep 14, 2005 #4
    Ah thanks, yeah your right, I've got it :) it all makes perfect sense now :)

    y = 5/3.e^(2 + ix - (1 - 2ix))

    =5/3.e^(1 + 3ix)

    dy/dx = 5ie^(1 + 3ix)

    :D thanks!
     
  6. Sep 14, 2005 #5
    You're welcome :-)
     
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