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This question already has an answer here:

I'm trying to define the following function

Df[k_, t_, x_, M_] = 1/Factorial[k] D[Exp[-(4 M x t)/(1 - t)], {t, k}]

I want to differentiate that exponential with respect to t k times but I also want to attribute a value to t after the derivation. But when I call this function with the arguments

Df[1,0,1,1]

for example I get a message saying that zero is not a valid variable. I think it is making t=0 before the derivation. Is there anyway around this?

Thank you very much.

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marked as duplicate by J. M. will be back soon Oct 12 '16 at 15:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Define it like this: Df[k_, t_, x_, M_] := 1/Factorial[k] D[Exp[-(4 M x t0)/(1 - t0)], {t0, k}] /. t0 -> t. (This is a dupe, but I have not time to search.) $\endgroup$ – march Oct 12 '16 at 15:41
  • $\begingroup$ Try SeriesCoefficient[Exp[-(4 M x \[FormalT])/(1 - \[FormalT])], {\[FormalT], t, k}]. $\endgroup$ – J. M. will be back soon Oct 12 '16 at 15:45
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Perhaps this is what you're after:

Df[k_, t_, x_,  M_] := (1/Factorial[k] D[Exp[-(4 M x tt)/(1 - tt)], {tt, k}]) /.  tt -> t

Test it:

In=  Df[1, 0, 1, 1]
Out= (* -4 *)
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  • $\begingroup$ Like a charm. Thank you very much. $\endgroup$ – Gabu Oct 12 '16 at 15:48
  • $\begingroup$ @Gabu If it suits you, please accept by clicking the "tick" below the vote count. You can also add an upvote if you think it deserves it. $\endgroup$ – anderstood Oct 16 '16 at 1:47

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