I type the following command line

Table[D[#1,varx[[i]]] &,{i,3}]

where varx is a vector of dummy variables

varx = {x1, x2, x3}

According to this link NonCommutativeMultiply, I want to obtain a vector like this

$\left\{\frac{\partial }{\partial \text{x1}},\frac{\partial}{\partial \text{x2}},\frac{\partial}{\partial \text{x3}}\right\}$

But I got this

enter image description here

It seems like the index i in the command Table does not work. I don't know why this happens. I try not to use the placeholder '#1' like this

Table[D[2 x1^2 + 4 x2^3 + x3^4, varb[[i]]], {i, n}]

And this works all fine

{4 x1, 12 x2^2, 4 x3^3}

If I want to keep the placeholder, what should I do?


2 Answers 2


Use With:

Table[With[{v = varx[[i]]}, D[#1, v] &], {i, 3}]`
(* {D[#1, x1] & , D[#1, x2] & , D[#1, x3] & } *)

See the section "Scope" of the documentation page for With. Note that Function (&) has the attribute HoldAll, so that the value of varx[[i]] needs to be inserted into the function.

The above gives a list of operators. An alternate interpretation of what is sought is a single operator that evaluates to a list. Here's a way:

grad = Block[{D}, Evaluate[Table[D[#1, varx[[i]]], {i, 3}]] &]
(* {D[#1, x1], D[#1, x2], D[#1, x3]} & *)

Here's another method for this particular example, based on behavior of D[f, {{x1, x2, ...}}]:

grad = With[{v = varx}, D[#, {v}] &]

Both yield

grad[x1 + x2^2 x3]
(* {1, 2 x2 x3, x2^2} *)

You can use Slot (#) but the pure function (&) should be at a different position, i.e.

varx = {x1, x2, x3};

Table[List[#1, varx[[i]]], {i, 3}] & @@@ {{f, g}, {h, i}}

{{{f, x1}, {f, x2}, {f, x3}}, {{h, x1}, {h, x2}, {h, x3}}}


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