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