I am trying to define a function f[x1,x2,...,xn] with n integer but not specified. And then I would like this funktion to behave properly under the derivative operation, so that D[f[x1,x2,...,xn],xi] actually results in an expression which unambiguously displays that it is the first derivative in respect to the i-th variable. Up until now I had little success in employing the Mathematica function D[x_,y_] and have been simply substituting it for a place holder d[x_,y_]. However, this is not very convenient, since those placeholders do not respect commutativity. Maybe there is such functionality already, which I do not know of? Or maybe one can implement it? Thank you for any suggestion!


Please note, I am not looking to define custom functions of a finite number of variables tied to the derivative operation. What I am actually interested in, is defining a function which is dependent on an unspecified number of variables n. And introducing a formalism such that taking derivatives in respect to one of these variables of an unspecified index i still works consistently.


To explain what I mean in more detail, I would like to be able to define a function f[v] in which I would like to treat v as a 1-dimensional list v={v1,v2,v3,...,vn}. However, I would like to keep the amount of elements in v arbitrary (without specifying to a certain length). And then I would like to define a derivative operation, which would allow expressions like D[f[v],v1] or D[f[v],v2] and even D[f[v],vi] which should not evaluate to zero, but to some meaningful expression, from which I still could extract the information which derivative it is, and with respect to which of the n variables.

  • 1
    $\begingroup$ Remember that you can use Derivative[n][f] $\endgroup$ Commented Oct 26, 2014 at 4:01
  • $\begingroup$ When you try D[f[x1, x2, xn], x2] you do get an unambiguous result, right? So I think you have to state the problem more clearly with an example. $\endgroup$
    – Jens
    Commented Oct 26, 2014 at 17:32
  • $\begingroup$ The answer in the tagged question above does not answer my actual question. I will edit my question to make it more clear. $\endgroup$
    – Kagaratsch
    Commented Oct 27, 2014 at 17:30
  • $\begingroup$ This may be related to or perhaps is a duplicate of symbolic-derivative-of-n-term-product. $\endgroup$ Commented Oct 27, 2014 at 18:49
  • 1
    $\begingroup$ Derivative[n][f] denotes an unspecified number of derivatives with respect to a single variable. What I need is a notation that denotes an unspecified number of derivatives with respect to an unspecified number of variables unambiguously, which clearly cannot be captured by merely Derivative[n][f]. $\endgroup$
    – Kagaratsch
    Commented Oct 27, 2014 at 22:35

1 Answer 1


I hope this toy example helps.

First I define some function of an arbitrary number of arguments, such that

f[v1, v2, v3,...] == v1 + 2 v2 + 3 v3 + ...

I do this with

f[v__] := Plus @@ MapIndexed[#1 First@#2 &, {v}, {1}]

I can then, without knowing the exact number of arguments, take a derivative with respect to a given argument by

i = 3
(* 3 & *)

Unfortunately, it doesn't allow a purely symbolic approach. I can't clear the i and do this:

(* i & <--- ain't happening *)

Also problems arise if the derivative is dependent on the number of arguments. e.g.

f[v__] := Times @@ MapIndexed[Exp[#1] First@#2 &, {v}, {1}]
Derivative[0, 1][f]
(* 2 E^(#1 + #2) & *)
Derivative[0, 1, 0][f]
(* 6 E^(#1 + #2 + #3) & *)
  • $\begingroup$ And now I realize, I'm necroposting. Hopefully my activity will attract someone with a better solution. $\endgroup$
    – LLlAMnYP
    Commented Apr 27, 2015 at 14:21
  • $\begingroup$ Necroposting is certainly not frowned upon here, long as you make a meaningful contribution. $\endgroup$ Commented May 27, 2015 at 14:39
  • $\begingroup$ @LLlAMnYP I think you expressed very clearly what I think the OP is asking for when you say you "can't clear the i"? If we could clear the i that would be the solution I think. Maybe it would be worth adding that into the question? Just trying to get some attention to this as I would love to have a solution. $\endgroup$
    – tortortor
    Commented Jan 27, 2018 at 10:58

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