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I have a question in Mathematica, I would appreciate it if you possibly answered me. What can I do for defining a special variable for derivative of arbitrary variable. For example, imagine I have a function as following

F[a_,b_]:= a  + b

Then I want to define

D[a,c] == R1

Define D[b,c] == R2

and at the end I get:

D[F[a,b],c] = R1 + R2 

can I do something like this in Mathematica? Thanks

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One possibility: assign UpValues to symbols a and b and useDt instead:

Clear[a, b, f, r1, r2]
a /: Dt[a, c] := r1
b /: Dt[b, c] := r2
f[a_, b_] := a + b

Then,

Dt[f[a, b], c]
(* r1 + r2 *)

To see more clearly what's going on with Dt, consider:

Clear[g]
Dt[g[a, b]]
(* Dt[b] D[g[a, b], b] + Dt[a] D[g[a, b], a] *)
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  • $\begingroup$ Thanks alot, and one more thing, if the function f[c1_, c2_] and g[c1_,c2_], is there any possiblity to do the same thing for g and f for example f[c1, c2] /: Dt[f[c1, c2], g[c1,c2]] := r1 $\endgroup$ – amin bk Nov 24 '15 at 1:21
  • $\begingroup$ @aminbk: Once you start mixing the definitions like this, it's probably best to go Pillsy's route and use replacement rules. $\endgroup$ – march Nov 24 '15 at 3:15
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I like @march's suggestion to use Dt (as well as the implicit suggestion to use lower-case symbol names), but one alternative approach is to use replacement rules. If they work for your use case, I find them much clearer than setting UpValues:

Clear[a, b, f, r1, r2];

f[a_, b_] := a + b;

Dt[f[a, b]] 

(* Dt[a, c] + Dt[b, c] *)

% /. {Dt[a, c] -> r1, Dt[b, c] -> r2}

(* r1 + r2 *)
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  • $\begingroup$ Thanks, but just one more thing, Now if I have something like this f[a_,b_] and g[a_,b_] how can I define Dt[F[a,b],g[a,b]] ->r1 ?? t $\endgroup$ – amin bk Nov 24 '15 at 16:19

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