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Differentiate this

  1. May 3, 2004 #1
    f(x) = x.e^(-pi.x^2)

    this is how I tried to do it...

    f'(x) = x.(-pi.2.x.e^(-pi.x^2)) + 1.e^(-pi.x^2)
    f'(x) = (1 - 2.pi.x^2).e^(-pi.x^2)

    can anyone see anything wrong with this?

  2. jcsd
  3. May 3, 2004 #2


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    Dearly Missed

    Seems correct to me, at least
  4. May 3, 2004 #3


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    Well, you could try this way:
    [tex]x \times e^{ - \pi x^2}=e^{- \pi x^2 \ + ln x}[/tex]
    so you get:
    [tex](\frac{1}{x} - 2\pi x) e^{- \pi x^2 \ + ln x}[/tex]
    which is the same.

    Looks good to me.
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