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I have to calculate the derivatives of the function K defined in the code:

<< VectorAnalysis`
K = δ[t]*Exp[-(r/L)]/r;
SetCoordinates[Cartesian[x, y, z]];

r = Sqrt[(x - x0)^2 + (y - y0)^2 + (z - z0)^2];
Hes = D[K, {{x, y, z}, 2}];
der = Grad[K];
FullSimplify[der[[1]]]
FullSimplify[Hes[[1, 2]]]
FullSimplify[Hes[[1, 1]]]

This is part the output just for FullSimplify[der[[1]]]:

-((E^(-(Sqrt[(x - x0)^2 + (y - y0)^2 + (z - z0)^2]/ L)) (x - x0) (L + 
    Sqrt[(x - x0)^2 + (y - y0)^2 + (z - z0)^2]) δ[t])/( L ((x - x0)^2 + (y - y0)^2 + 
  (z - z0)^2)^(3/2)))

I would like the function K to appear in the result instead of that "mess": is it possible to do it?

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1 Answer 1

up vote 1 down vote accepted

I'm not sure what's your goal since you've expressed K and switched coordinates.

Maybe this could be what you're after?

Simplify[(der /. r^2 -> rad^2), rad > 0] /. E^(-(rad/L)) -> newK rad/δ[t]

enter image description here

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Yes it is. I thought it was more complicated... thank you! –  mattiav27 Apr 23 at 10:31
    
@mattiav27 I'm glad you like it. It is good to hold on with an accept a day or two, there are better ways of workng with vector analisys and someone may provide an answer :) Let's do not discourage them :) You can still upvote this answer if you want. –  Kuba Apr 23 at 10:35

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