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Re: Dt@x@1

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  • Subject: [mg119428] Re: Dt@x@1
  • From: Roland Franzius <roland.franzius at>
  • Date: Thu, 2 Jun 2011 19:12:21 -0400 (EDT)
  • References: <is4tgc$b3t$>

Am 01.06.2011 10:33, schrieb Chris Chiasson:
> Why does Dt@x@1 return zero? I would expect it to return unevaluated.

The chain rule for Dt acting on a chain of functions of a single 
argument says

Dt@x@1 = x'[1] Dt[1]  ~  Dt[1]=0


Trace[Dt[x[y[w[u]]], Constants -> {u, v}]] // TreeForm

Trace[Dt[x[y[w[z]]], Constants -> {u, v}]] // TreeForm

to see that Dt[mostinnnerargument]->0 is used as a rule without 
calulating superflous inner derivatives x', y', w' first.

Its of course the simplifying use of those general cancelling rules, 
easy to recognize and to apply, that makes the CAS working at all (in a 
limited collection of cases in finite time ;-( ).


Roland Franzius

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