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

> a diff-like function which quotes its 
> argument. It would be enough for present purposes if it could 
> recognize arguments like f(x) where f(x):=foo(g(x)) and foo 
> is exp, log, sin, etc.

I see what you want to do in the exp and * case, but what did you have
in mind for the log, sin, etc. cases?

   f(x):=exp(g(x))$
   qdiff(f(x),x) => f(x)*diff(g#(x),x)
     -- use g#(x) to represent (verbify(g))(x)

   f(x):= fff(x)$
   g(x):= ggg(x)$
   qdiff(f(x)*g(x),x) => f(x)*diff(g#(x),x) + g(x)*diff(f#(x),x)

   f(x):=sin(g(x))$
   qdiff(f(x),x) => sqrt(1-f(x)^2)*diff(g#(x),x)  --Is this what you
want?
   
   f(x):=log(g(x))$
   qdiff(f(x),x) => ???

By the way, you can do the product case by simply using noun forms.  The
question is: where exactly do you want to expand, and where do you not?
I suppose you could use substpart with apply_nouns to do this
selectively.

Then there's RJF's logarithmic derivative trick.

       -s