Hmm, implementing the chain rule is trickier than i thought.  My
straightforward plan of attack was to write a function that
differentiates a symbolic expression as usual but when it comes to a
composition f o g, it uses the chain rule and returns the appropriate
entry of the matrix (Df o g)Dg.  Problems:

(a) How do you split apart a symbolic expression to scan for
compositions?
(b) How do you construct Df so that you can compose it with g?

Both thwart me and my white belt Sage-fu.

Any helpful suggestions for (a), (b), or the general project?

Alex


On Apr 23, 1:43 pm, Alex Raichev <tortoise.s...@gmail.com> wrote:
> Never mind.  I'll just right a short recursive function.  It's easy
> enough.
>
> Alex
>
> On Apr 23, 11:10 am, Alex Raichev <tortoise.s...@gmail.com> wrote:
>
> > Hi all:
>
> > Do any of you know how to get Sage to use the chain rule and expand
> > the derivative of a composition involving one or two callable symbolic
> > functions?  Here's an example with one callable symbolic function.
>
> > ----------------------------------------------------------------------
> > | Sage Version 3.4, Release Date: 2009-03-11                         |
> > | Type notebook() for the GUI, and license() for information.        |
> > ----------------------------------------------------------------------
> > sage: var('x,y,t')
> > (x, y, t)
> > sage: f= function('f',x,y)
> > sage: g= exp(I*t)
> > sage: diff(f(g,g^2),t).expand()
> > diff(f(e^(I*t), e^(2*I*t)), t, 1)
>
> > ------------------------------------------------------------------------
>
> > The reason i ask is that i have to take higher-order derivatives of a
> > composition f o g of two callable symbolic multivariate functions. I
> > want the expanded form so that i can evaluate at a certain point  c
> > and solve a linear system to get the derivatives of f at g(c).  (I
> > know the values of the derivatives f o g and g at c.)  I could write a
> > Sage function to expand the derivatives of f o g using Faà di Bruno's
> > formula, but before i do so, i was wondering if there's an easier
> > way.
>
> > Alex
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