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Higher derivatives

  1. Feb 26, 2006 #1
    I'm a little stuck on this one

    Find [tex] f^{(n)}(x) = x^n [/tex]

    I know [tex] f'(x) = n(x^{n-1}) [/tex]
    and [tex] f''(x) = n(n-1)(x^{n-2}) [/tex] and so on

    But I can't seem to see the pattern that leads to the answer [tex] n! [/tex]
     
  2. jcsd
  3. Feb 26, 2006 #2

    VietDao29

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    Then just carry on differentiating some more times:
    f(x) := xn
    f'(x) = nxn - 1.
    f''(x) = (f'(x))' = (nxn - 1)' = n(n - 1)xn - 2.
    f'''(x) = (f''(x))' = (n(n - 1)xn - 2)' = n(n - 1)(n - 2)xn - 3.
    f''''(x) = n(n - 1)(n - 2)(n - 3)xn - 4.
    f(v)(x) = n(n - 1)(n - 2)(n - 3)(n - 4)xn - 5.
    ...
    Now if k < n, what can you say about f(k)(x)?
    If k = n then f(k)(x) = ? (i.e, what's f(n)(x))
    --------------
    Another hint:
    Consider g(x) := x2, so g''(x) = 2.
    h(x) := x3, so h'''(x) = 6.
    Can you go from here? :)
     
  4. Feb 26, 2006 #3

    arildno

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    Well, you could prove by induction that we have:
    [tex]f^{(l)}(x)=\frac{n!}{(n-l)!}x^{n-l}, l\leq{n}[/tex]
     
  5. Feb 26, 2006 #4
    I think I've got it. If I expanded it out far enough

    [tex] f^{(n)} (x) = n(n-1)(n-2)....(n-(n-1)) x^{n-n} [/tex]

    Since [itex] n-(n-1) = 1 [/itex] the coefficient becomes [itex] 1(2)(3)...(n-1)(n) = n! [/itex] and [itex] x^{n-n} = x^0 = 1 [/itex]

    So the whole things boils down to [tex] f^{(n)} (x) = n!(1) = n! [/tex]
     
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