You can do this calculation using gradef.
First do this
gradef(f(u,v),df1(u,v), df2(u,v));
This means that we use the (unknown) function df1(u,v) to represent the
derivative of f with respect to its first argument.
And similarly for df2 and second argument.
Then write diff (f(t,s(t)), t).
If you wish to define s(t):=t/e, or just set s: t/e, you can also do
that.
Df1 here means the same as Barton's dt[1,0].
RJF
_____
From: maxima-bounces at math.utexas.edu [mailto:maxima-bounces at math.utexas.edu]
On Behalf Of ????
Sent: Friday, November 10, 2006 8:54 AM
To: Tomek
Cc: maxima at math.utexas.edu
Subject: Re: [Maxima] simple chain rule
hi toko
you may calc this with pdiff package .(written by Barton)
(%i16) load(pdiff);
(%o16) /usr/share/maxima/5.9.3/share/contrib/pdiff/pdiff.lisp
(%i17) pdiff:on;
from cvs diff forms get differential forms package
(%i18) batch("cartan_init.bat");
......
(%i33) f_star([t],(x:t,y:t/%e,depends(f,[x,y]),d(f(x,y))));
- 1 - 1 - 1
(%o33) Dt f (t, %e t) + %e Dt f (t, %e t)
(1, 0) (0, 1)
thanks
Gosei Furuya
2006/11/11, Tomek <toko at os.pl>:
Hallo,
I would like Maxima to do something like:
s(t) := t / e
Now I would like the following
diff (f (t, s), t)
to be resolved as
df/dt + 1/e * df/fs
How can I make it work ?
Thanks,
Tomek
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