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N'th derivative of f(x)^n

  1. Jul 27, 2010 #1


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    does anyone know a general formula for

    I couldn't find in any tables I had and I have trouble figuring out the pattern.
    Code (Text):
    Array[D[f[x]^#, {x, #}] &, {4}]
    in Mathematica gives first four terms:



    [tex]18f(x)f^{(1)}(x)f^{(2)}(x)+3f(x)^2f^{(3)}(x)+6 f^{(1)}(x)^3[/tex]

    [tex]144f(x)f^{(1)}(x)^2 f^{(2)}(x)+36f(x)^2f^{(2)}(x)^2+48f(x)^2f^{(1)}(x)f^{(3)}(x)+4f(x)^3f^{(4)}(x)+24f^{(1)}(x)^4[/tex]

    Thanks for any help.
    Last edited: Jul 27, 2010
  2. jcsd
  3. Jul 27, 2010 #2
    Far Out!
    Lovely mathematical induction,
    (when i say test i mean substitute)
    they way they tought us back in school was
    1. test for 1, (yes the numerical value 1)
    2. assume with K (that's just fill n with K and normally don't simplify it)
    3. test for K+1 (substitute and solve)

    if that works were on our first step (should give you the pattern)

    now i'm from a country where physics and mathematics aren't appreciated, so i must ask when it states the letter d is d a proneumeral or does it mean derivative, and is that a derivative on the approaching power of n, it's just we use both d and ' .
    if we can get that out of the way we can get a good foot in solving the problem.
  4. Jul 27, 2010 #3


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    The very first thing I did was try induction (I even did some other proofs to make sure I remember how to do it correctly :) ) but I still can't figure it out. I'm probably missing something obvious here, hence my question on the forum.
    As for the latter part of your post, far out to you too, man :)
  5. Jul 30, 2010 #4


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    Homework Helper

    There's a theorem:

    [tex]\frac{d^n}{dx^n}(h(x)g(x)) = \sum_{k=0}^n {_n}C_k h^{(n-k)}(x)g^{(k)}(x)[/tex]
    where [itex]{_n}C_{k}[/itex] is the binomial coefficient

    [tex]{_n}C_k =\frac{n!}{k!(n-k)!}.[/tex]

    Choosing h(x) and g(x) appropriately may help with your problem.
  6. Jul 31, 2010 #5


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    I found a formula I missed earlier in the tables that gives
    [tex]\frac{d^n}{dx^n}\left[f(x)^p\right]=p\left(\begin{matrix}n-p&\\n\\\end{matrix}\right)\sum_{j}^{} (-1)^j \left(\begin{matrix}n&\\j\\\end{matrix}\right) \frac{f(x)^{p-j}}{p-j}\frac{d^n}{dx^n}\left[f(x)^j\right][/tex]

    but it didn't really help much because of it's recursive definition. Turns out using http://mathworld.wolfram.com/FaadiBrunosFormula.html" [Broken]. So that

    [tex]\frac{d^n}{dx^n}\left[f(x)^n\right]=\sum_{k=1}^{n}(-1)^k (-n)_{k}f(x)^{n-k} B_{n,k}\left(\frac{d}{dx}\left[f(x)\right],\,...\, ,\frac{d^n}{dx^n}\left[f(x)\right]\right)[/tex]

    where [tex](a)_{k}[/tex] is Pochhammer symbol. Strangely Mathematica does not implement multivariate version of Bell polynomials, so if anyone ever needs one:

    Code (Text):

    B[n_, k_, X_] := Module[{
        p = Table[Permutations[PadRight[p, n-k+1]], {p, IntegerPartitions[k]}],
        testf = (Sum[j #[[j]], {j, 1, n-k+1}]==n)&&(Sum[#[[j]], {j, 1, n - k + 1}]==k) &
       Expand[n!Sum[Product[Power[X[[j]]/(j!), pk[[j]]]/(pk[[j]]!), {j, 1, n-k+1}],
          {pk, Select[Flatten[p, 1], testf]}

    (* eg: *)
    B[6, 3, Array[Subscript[x,#]&, {6}]]
    Last edited by a moderator: May 4, 2017
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