tt = Flatten[Table[{x, y, z, btot[x, y, z]}, {x, -1, 1, 0.1}, {y, -1, 1,0.1}, {z, -1, 1, 0.1}], 2]; ff = Interpolation[tt] Till here it is working fine as it is returning the values of the interpolated function at various {x,y,z} points.
Then I want to find the gradient of this interpolated function. But when I am using
ffd[x_,y_,z_]:= D[ff[x,y,z],{{x,y,z}}] I am not getting the gradient.
With
ffd[x_,y_,z_]:= D[ff[x,y,z],{{x,y,z}}] the values of x, y, and z are substituted as arguments causing differentation wrt. numbers, i.e. nonsense. Moreover, you are using SetDelayed, which differentiate once for every call, which rather should be once for all time.
The solution to both problem is replacing SetDelayed with Set:
ffd[x_,y_,z_]= D[ff[x,y,z],{{x,y,z}}] When you define your function like this, then e.g. ffd[.5, .5, .5] is really D[ff[.5, .5, .5], {{.5, .5, .5}}].
to avoid scoping issues with
x=5; ffd[x_, y_, z_] = D[ff[x, y, z], {{x, y, z}}] you can use
ffd = Evaluate[D[ff[#, #2, #3], {{#, #2, #3}}]] & & means there is a held Function so we have to use Evaluate to force computation of gradien so it is not repeated each time.
:=in your code with=and it should work $\endgroup$:=with=. $\endgroup$Remove[ffd]before trying with=. $\endgroup$