I have noticed a weird behavior of the "Derivative" operator, and I would like to know to why is this happening and if it is intended, as I'm using this function quite a bit in my code and I would like for it to work. Here is a code snippet that reproduces de problem :

f1[λ_, v_] := f2[λ, vectrans[v]]
f2[λ_, v_] :=λ (v[[1]] + v[[2]])
vectrans[v_] := ( {
    {0, 2},
    {2, 0}
   } ) . v
f1[λ, {a, b}]
Derivative[1, 0][f1][λ, {a, b}]
D[f1[λ, {a, b}], λ]

Now if you look at the output of this code, I coded the function f1 to get a scalar and a vector as input, and spit out $λ(2a+2b)$, if the inputs are $λ$ and $v=\{a,b\}$. Then, in the first attempt I use the operator Derivative[1,0] to do the partial derivative w.r.t. $\lambda$, but I get the following output with mathematica 13 :

{{{a, b}, {2 + a, 2 + b}}, {{2 + a, 2 + b}, {a, b}}}

Which honestly I have no idea how it comes about.

If we use instead the operator D, we get the expect result $2a+2b$.

Why is that ?

  • 2
    $\begingroup$ Derivative, in effect, differentiates f1[\[Lambda], x] and then replaces x by {a, b}. D differentiates the result of f1[\[Lambda], {a, b}]. $\endgroup$
    – Michael E2
    Jun 14, 2022 at 15:36
  • $\begingroup$ I see. I will try to include conditions on the parameters of f1 to see if it changes anything. $\endgroup$
    – Frotaur
    Jun 14, 2022 at 15:37
  • $\begingroup$ Indeed, adding a ListQ to the v parameter just prevents Derivative from computing the derivative altogether, as f1 just doesn't evaluate... Apparently using "Derivative[1,{0,0}]" works, as it then knows that the second parameter is a vector. $\endgroup$
    – Frotaur
    Jun 14, 2022 at 15:40

1 Answer 1


So the problem as pointed in the comments was that Derivative will assume all parameters of the function to be differentiated are scalars, in other words it differentiates f[λ,x] and then replaces x by {a,b}, which gives the unexpected result.

One way around this, while still using Derivative is as follows :


This was indeed written in the Documentation, but I had missed it :

Derivative[{n1,n2,…}][f] represents the derivative of f[{x1,x2,…}] taken $n_i$ times with respect to $x_i$. In general, arguments given in lists in f can be handled by using a corresponding list structure in Derivative.


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