-----"Viktor T. Toth" <vttoth at vttoth.com> wrote: -----
>Presently, it is defined as
>
> W' = exp(-W)/(W+1)
>
>which is equivalent I believe to
>
> W' = W/[x(W+1)],
>
>since W(x)/x = 1/[x/W(x)] = 1/exp(W(x)) = exp(-W(x)).
>
>However, I believe the latter form is preferable as it contains no
>exponential term and thus lends itself to algebraic simplifications more
>easily. Is there a reason that I may not be aware of why the first form of
>the derivative was used?
Yes, there is a good reason: exp(-W)/(W+1) doesn't have a removable
discontinuity at zero, but W/[x(W+1)] does. With W' = W/[x(W+1)] we'll
get a mess for a Taylor expansion of lambert_w centered at zero. See
specfun.lisp CVS revision 1.23. Thanks for asking :)
Barton
P.S. lambert_w isn't in the user documentation :(