Marc Hodapp wrote on 09/14/2005 06:22:16 AM:
> (%i1) expr1:diff(u(x),x)
> (%i2) subst(l*X,x,expr1)
> (%o2) d u(lX)
> ------------
> d(lX)
>
> How can I get
>
> 1 d u(X)
> -- ------------
> l d(X)
>
>
> from (%i2)? Do you have any idea?
>
> Many thanks.
>
> Marc
(%i6) load("pdiff")$
(%i7) diff(u(x/l),x);
(%o7) u[(1)](x/l)/l
(%i8) subst(x = l * X,%);
(%o8) u[(1)](X)/l
Barton