interpolation - Derivatives (divergence, gradient, curl) of interpolated 3D data

How can I calculate a vector derivative (i.e. divergence, gradient, curl) of interpolated data? For sample data, you can use:

f[x_, y_, z_] := Exp[I z] {1, 0, 0}
testdata=Flatten[Table[N@{x,y,z,f[x,y,z]},{x,0,4 Pi,Pi/10},{y,0,4 Pi,Pi/10},{z,0,4 Pi,Pi/10}],2];

intf = Interpolation[testdata]

I know that for 1D data, like

dim1 = Table[N@{z, f[0, 0, z]}, {z, 0, 2 Pi, Pi/10}];
int1 = Interpolation@dim1;

you can do D[int1[x],x]. However, I can't seem to get convince MMA that the interpolated function intf actually returns a "vector" quantity (i.e. Length@intf[x,y,z]->3), so that I can do something like:

curl = {D[#[[3]],y]-D[#[[2]],z],-(D[#[[3]],x]-D[#[[1]],z]),D[#[[2]],x]-D[#[[1]],y]}&;
curl@intf[x, y, z]
(* Out[] := {0,0,0} *)

Similarly, this should work for gradient:

grad = {D[#[[1]], x], D[#[[2]], y], D[#[[3]], z]} &;

and divergence:

div = (D[#[[1]], x] + D[#[[2]], y] + D[#[[3]], z])&;

I've posted my best attempt in an answer below, but I'm wondering what other solution approaches there are.

Answer

Here's another possibility. First compute the derivatives (this should use formal symbols for safety, but I've used x,y,z here to avoid cluttering the code)

intfd[x_, y_, z_] = D[intf[x, y, z], {{x, y, z}}];

then define the curl by

intfcurl[x_, y_, z_] := Module[{q = intfd[x, y, z]},
  {q[[2, 3]] - q[[3, 2]], q[[3, 1]] - q[[1, 3]], q[[1, 2]] - q[[2, 1]]}]

The results seem okay:

intfcurl[0.4, 0.2, 0.3]
(*  {0., -0.297934 + 0.954217 I, 0. + 0. I}  *)

and the timing is reasonable:

intfcurl @@@ testcoords; // Timing
(*  {7.863, Null}  *)

---

You can similarly define the gradient and the divergence using intfd:

intfgrad[x_, y_, z_] := Module[{q = intfd[x, y, z]}, Diagonal[q]]
intfdiv[x_, y_, z_] := Module[{q = intfd[x, y, z]}, Tr[q]]