I am trying to define a derivative that acts on a custom function, $f$, such that when it acts, it increments one of the arguments of by 1. f is not defined; I just want to use it to keep track of derivatives. I know I could define a separate operator to do this, but this function, $f$, will be multiplied by other functions that need differentiating, so I thought it easier to define how derivatives act on f. My attempt thus far:

Derivative[1, 0, 0, 0][f[a,b,c,d]][x1, x2, x3, x4] = f[a,b,c,d]

However, when I go to evaluate this derivative, I get

Input: D[f[a,b,c,d][x1, x2, x3, x4], x1]
Output: f[a,b,c,d]^{1,0,0,0}[x1, x2, x3, x4]

whereas, I would hope to get something like:

Input: D[f[a,b,c,d][x1, x2, x3, x4], x1]
Output: f[a+1,b,c,d][x1, x2, x3, x4]

The derivative definition only works specifically for the exact syntax in the definition, and not with any other arguments apart from a,b,c,d:

Input: D[f[a,b,c,d][x1, x2, x3, x4], x1]
Output: f[a+1,b,c,d]

Any suggestions on how to obtain the second listed output?

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    – bbgodfrey
    Jul 27, 2015 at 21:03
  • 1
    $\begingroup$ Clear all your previous definitions, try Derivative[1, 0, 0, 0][f] = CKB[#1 + 1, #2, #3, #4] &, and report back. $\endgroup$ Jul 27, 2015 at 21:12
  • $\begingroup$ You're using Slot the wrong way. Also, I don't see why you expect the output to be in terms of f if you define the right-hand side to be CKB in the first line. $\endgroup$
    – Jens
    Jul 27, 2015 at 21:15
  • $\begingroup$ @Jens My mistake - CKB should be f. Edited. $\endgroup$ Jul 27, 2015 at 21:42
  • $\begingroup$ @Guesswhoitis. Using Slot isn't necessary - since I don't fully understand how it works, I'm editing the question to functions without Slot. $\endgroup$ Jul 27, 2015 at 21:44

1 Answer 1


It seems to be that all you need is SetDelayed (:=) instead of Set (=):

Derivative[1, 0, 0, 0][f][a_, b_, c_, d_] := f[a + 1, b, c, d]

Now let us evaluate some derivatives:

D[f[x, y, z, p], {x, 1}] (* with respect to that first argument *)
D[f[x, y, z, p], {y, 1}] (* with respect to any other argument *)

f[1 + x, y, z, p]
Derivative[0, 1, 0, 0][f][x, y, z, p]

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