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Is there a way to declare that subscripted letters are not functions of the letter? By default, this happens:

In[5]:= D[Subscript[t, 0], t] Out[5]= (Subscript^(1,0))[t,0]

which I can understand but is slightly eye-roll-inducing. Mathematical practice would like this t_0 to be constant.

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    $\begingroup$ You can use Symbolize. There are already a few questions about this. $\endgroup$ Commented Jan 23, 2024 at 16:26
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    $\begingroup$ Point 3: "Avoid subscripted symbols." That said, try setting Attributes[Subscript] = {Constant, NHoldAll}. $\endgroup$ Commented Jan 23, 2024 at 16:43
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    $\begingroup$ Instead of using Subscript[t, 0] use t0 and format it to look like Subscript[t, 0], i.e., Format[t0] = Subscript["t", 0]; Then D[t0, t] evaluates to 0 $\endgroup$ Commented Jan 23, 2024 at 17:55
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    $\begingroup$ Does this answer your question? Solve with v9 (issues with Subscript, Overscript, Superscript etc) $\endgroup$ Commented Jan 24, 2024 at 2:59
  • $\begingroup$ This is a simple mistake? $\endgroup$ Commented Jan 24, 2024 at 18:31

1 Answer 1

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You can use a variation of the following idea where I fix a similar issue with Solve:

$Notations = Alternatives[ Subscript, Superscript, Subsuperscript, SubPlus, SubMinus, SubStar, SuperPlus, SuperMinus, SuperStar, SuperDagger, Overscript, Underscript, Underoverscript, OverBar, OverVector, OverTilde, OverHat, OverDot, UnderBar ]; Unprotect[D]; D[a__] /; !FreeQ[{a}, $Notations[__]] := Block[{CompressedData}, With[{z=Unevaluated[D[a]] /. s:$Notations[__] :> CompressedData[Compress[s]]}, z /; !MatchQ[z,_D] ] ] Protect[D];

Then:

D[t Subscript[t,0] + t^2, t]

2 t + Subscript[t, 0]

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