How to get diff to yield result in terms of original function

Hello everyone,

Happy new year. I wonder if there's a way to get something like

  f(x) := exp(g(x))$
  diff(f(x), x);

to yield f(x)*'diff(g(x),x).

The above expressions, as they stand, yield %E^g(x)*'diff(g(x),x).
Which is correct, but I'd like to see if I can get the other form.

I can do subst(''(f(x)),f(x),%)$ and that does yield
f(x)*'diff(g(x),x).
However, in the problem that I'm working on, which is a little bit
more complicated, the expression gets rearranged somewhat and the
substitution fails. I don't think I can count on an after-the-fact
substitution always acting as expected.

Is there a flag to tell diff to express derivatives of exponentials
in terms of the original function or expression? I read through
the documentation and skimmed the mailing list archives, but I
couldn't find anything.

Thanks for any light you can shed on this topic.

Robert Dodier

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