substitution in differentiation



Barton Willis wrote:

>>I've got a function u depending on a variable x.  So I declared:
>>
>>(%i1)      depends(u,x);
>>
>>I now want to differentiate u with respect to x.  So:
>>
>>(%i2)      diff(u,x);
>>
>>Now I want to substitute the independent variable x with l*X,  where  l 
>>is  a constant parameter.  What I want to obtain is:
>>
>>d u        1    d u
>>------   = ---   -------
>>d x         l     d X
>>
>>    
>>
>
>My favorite way to do things like this is to use the positional
>derivative package:
>
>(%i1) load("pdiff")$
>(%i2) diff(u(x/l),x);
>(%o2) u[(1)](x/l)/l
>
>On the screen, u[(1)] is u subscript (1). For user documentation, 
>see the pdf in /share/contrib/pdiff.
>
>Barton
>  
>
Hi Barton,

thanks for the hint. I wasn't aware of that package. It's true, it 
improves the clarity. But unfortunately, it doesn't solve my problem.
What I have done so far, is

(%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