Hi Martin
Thanks for the explanations. I really don't want to be nit-picking, 
however, you hit the nail:

> Maybe the confusion comes from the difference between D and f_1:
> f_1(a,b) is the *result* of D(f(a,b), a). 
> D is a FriCAS function, whereas f_1(a,b) is an expression.
  

F(x,y)==D(y*sin(x),x)
-> InputForm (* y (cos x))

That is, the information that y*cos(x) comes from a derivative is lost 
(although not really because F(y,y)::INFORM -> (+ (sin y) (* y (cos y))) 
(before your patch). 

So why not expecting F(u,v)==D(f(u,v),u) as f_1(u,v)? Then F(u,u)=f_1(u,u) 
naturally. On the other hand when using macro (==>):

F(x,y) ==> D(y*sin(x),x)

it's clear to me that we have a substitution F(x,x)->D(x*sin(x),x). But you 
are right of course, one may/should see it as you mentioned:

> D is a FriCAS function, whereas f_1(a,b) is an expression.

Again, thanks for the trouble!
Kurt

On Tuesday, 19 February 2019 07:28:07 UTC+1, Martin R wrote:
>
> Dear Kurt!
>
> As far as I can see, eval, D, == and := are behaving exactly as 
> specified.  With
>
> f(x,y) == D(f(x,y),x)
>
> the symbol "f" is a FriCAS function, and NOT a function in the usual 
> mathematical sense!
> In particular, f(a,b) returns the expression (in the pre-calculus sense)
>
> f_1(a,b),
>
> first derivative of with respect to the first argument, evaluated at a and 
> b.
> Note that it is somewhat sloppy to write f_x(a,b) in calculus, because it 
> is
> unclear what x refers to.
>
> Therefore, eval(f(a,b), b=a) must return f_1(a,a).
>
> On the other hand, f(a,a) should return the expression D(f(a,a), a), which
> is the derivative of f(a,a) with respect to a.
>
> Maybe the confusion comes from the difference between D and f_1:
>
> f_1(a,b) is the *result* of D(f(a,b), a). 
>
> D is a FriCAS function, whereas f_1(a,b) is an expression.
>
> Martin
>

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