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Making this derivative look prettier

  1. Mar 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the 1st and 2nd derivative:

    [tex]f(x) = x^{1/3}* e^{-x^2}[/tex]

    3. The attempt at a solution

    [tex]f'(x) = x^{1/3} * -2xe^{-x^2} + e^{-x^2} * \frac {1}{3}x^{-2/3}[/tex]

    I simplified this to:
    [tex][e^{-x^2}]*[-2x^{4/3} + \frac {1}{3}x^{-2/3}][/tex]

    Also to find the x values is

    [tex]-2x^{4/3} + \frac {1}{3}x^{-2/3}[/tex]
    [tex]x^{-2/3}*[-2x^{2} + \frac {1}{3}][/tex]

    [tex]x^{2} = 1/6[/tex]
    [tex]x = +- \sqrt{1/6}[/tex]

    Okay now I'm trying to make the 2nd derivative be simplified so I can solve for the inflection points.

    [tex]f''(x) = [e^{-x^{2}}]*[\frac {-8}{3}x^{1/3} - \frac {2}{9}x^{-5/3}] + [-2x^{4/3} + \frac {1}{3}x^{-2/3}]*[-2xe^{-x^{2}}] [/tex]

    Im kind of stuck after this. I think it would involve taking out an [tex]e^{-x{2}}[/tex] but it's very confusing.
     
    Last edited: Mar 27, 2008
  2. jcsd
  3. Mar 27, 2008 #2

    tiny-tim

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    … just keep going …

    Hi Flappy! :smile:

    Just keep going … you should get [tex]}x^{-5/3} e^{-x^2} [/tex] times a quadratic in x^2. :smile:
     
  4. Mar 27, 2008 #3
    Hmm I tried to do this:

    [tex] e^{-x^{2}}[ -\frac {8}{3}x^{1/3} - \frac {2}{9}x^{-5/3} + 4x^{7/3} - \frac {2}{3}x^{1/3}][/tex]

    I multiplied the -2x and then took out the e^(-x^2). Would this be right? I'm not sure where you're getting x^(-5/3)
     
  5. Mar 27, 2008 #4
    Ah, i think i see it. There's a common factor of x^(-5/3)
     
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