I am not pasting my input code here since the website modifies my code. Instead I am attaching a screenshot of my notebook.

Issue: I enter two separate commands (asking Mathematica to differentiate in both case). I expected the answers to be '2' and '0' respectively, but it spits out some weird thing in the second case. Please tell me what went wrong?

In short, I expect Mathematica to identify 'x' and 'x_{p}' (where '_' stands for subscript) as two independent variables while performing the differentiation but it looks like it has some trouble in identifying variable name with a subscript component.

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  • 2
    $\begingroup$ This is perhaps the commonest problem associated with subscripts. See this site for multiple explanations $\endgroup$
    – mikado
    Sep 4, 2016 at 18:55
  • $\begingroup$ @mikado, you did not give the website. Please do so. And thanks! $\endgroup$ Sep 4, 2016 at 19:45
  • $\begingroup$ This site means this site, Mma.SE :) -- Search this site for subscript, Symbolize and the Notation package, for instance. $\endgroup$
    – Michael E2
    Sep 4, 2016 at 20:04
  • 2
    $\begingroup$ Please note that $x_p$ in your second example is not a symbol. Rather, it is an expression that is a function of two separate symbols Subscript[x,p], which merely formats as $x_p$. Therefore, Mathematica uses the chain rule to differentiate, giving the correct result. Never ever EVER use subscript to denote symbols. Use xp instead. $\endgroup$
    – QuantumDot
    Sep 4, 2016 at 23:23
  • 1
    $\begingroup$ If you insist on Mathematica interpreting $x_p$ as a totally independent symbol, then use the following two lines: MakeBoxes[xp, StandardForm] := SubscriptBox["x", "p"]; MakeExpression[SubscriptBox["x", "p"], StandardForm] := MakeExpression["xp", StandardForm]. Then in all subsequent evaluations in the session, xp will display as $x_p$, but will be interpreted as xp behind-the-scenes. Only then, would D[$x_p$,x] yield zero as expected. (Try also running FullForm[$x_p$] to understand what's going on). $\endgroup$
    – QuantumDot
    Sep 4, 2016 at 23:31

2 Answers 2


A subscripted symbol is an expression involving the symbol. It is not a new symbol. Here is what you are asking Mma to do:

D[Times[2, Subscript[x, p]], x]

This is an expression in x, so Mma does exactly what you ask it to: it differentiates an expression in x with respect to x.


You can use the Notation` package to get $x_p$ treated as a symbol.

Symbolize[ParsedBoxWrapper[SubscriptBox["x", "p"]]]
D[2*x⎵Subscript⎵p^2, x⎵Subscript⎵p]

This looks rather strange in input form, but in standard form it looks as you would expect it to.


Or even more generally,


Which is the same as



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