Stavros,
Many thanks for your reply. I'm reviewing my college calculus (A. W.
Goodman, 1st edition - yeah I'm an old guy with nothing much to do!). I like
to arrive at the same answers provided in the text. Maxima always provides a
correct answer, but I find that trying to achieve the same form as the book
is often instructive both in the Calculus but also in Maxima.
I am truly impressed by the support the developers of Maxima provide. Thanks
again.
Ron
-----Original Message-----
From: macrakis at gmail.com [mailto:macrakis at gmail.com] On Behalf Of Stavros
Macrakis
Sent: Wednesday, May 28, 2008 2:09 PM
To: Ronald Modesitt
Cc: maxima at math.utexas.edu
Subject: Re: [Maxima] More Chain Rule questions
I'm not sure I understand why you want the form you want. Is it for
teaching?
If so, you will want more control over Maxima's operations than you
get by using the default representation and algorithms. In
particular, internally, Maxima does not represent divisions as
"/"(a,b) but as "*"(a,"^"(b,-1)), and (a/b)^7 is represented as
a^7/b^7 and so really as "*"("^"(a,7),"^"(b,-7)).
For more control, you could do something like the following:
gradef(div(a,b),1/b,-a/b^2)$ <<< define derivatives for
your own division operator
Now,
diff( div(a*x+b,c*x+d)^7 , x) =>
7 * (a/(c*x+d)-c*(a*x+b)/(c*x+d)^2) * div(a*x+b,c*x+d)^6
(notice the treatment of the two-argument function div in the chain rule)
and subst("/",div,%) =>
7 * (a*x+b)^6 * (a/(c*x+d)-c*(a*x+b)/(c*x+d)^2) / (c*x+d)^6
There are several more simplification steps between this and the form
you want, but this is what you get when you really do follow the chain
rule step by step.
-s
*That* is the correct result using the chain rule.