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Nth derivative 2

  1. Nov 29, 2005 #1

    daniel_i_l

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    The question is to find the nth derivative of:
    [tex]
    f(x) = (ax - b)^-1
    [/tex]
    the answer in the book is:
    [tex]
    f^[n](x) = n!(-a)^n(ax-b)^{-1-n}
    [/tex]

    now I know that the nth derivative of:
    [tex]
    f(x) = x^-1
    [/tex]
    is:
    [tex]
    (-1)^nn!x^{-1-n}
    [/tex]
    but why would the nth of ax be a^n? Shouldn't it be 0 for every n over 1 which would make the whole term 0?
     
  2. jcsd
  3. Nov 29, 2005 #2

    Tide

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    You're not being asked to find the [itex]n^{th}[/itex] derivative of [itex]ax[/itex] nor is it even suggested by the "answer in the book."
     
  4. Nov 29, 2005 #3

    daniel_i_l

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    but by the chain rule don't I have to multiply by the nth derivative of
    (ax + b) ? I thought that the a^n was implied because that is the term that is missing from my answer: a^n * (-1)^n = (-a)^n .
     
  5. Nov 29, 2005 #4

    Tide

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    Each time you take a derivative you reduce the power of (ax-b) and obtain an additional factor of a.
     
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