Hi Cal,
this one would work:
Plot[{x, D[y, y] /. y -> x}, {x, 0, 5}]
It calculates the derivative first and then substitutes the value for x. This is trivial in this case, of course, as D[y,y] is identical 1. But it gives the same plot as
f[_x]:=x; Plot[{f[x],f'[x]},{x,0,5}]
Likewise,
Plot[{x, Evaluate[D[x, x]]}, {x, 0, 5}]
forces the evaluation of D[x,x] first and gives the desired result.
Cheers,
Marco
