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Re: Problem with Function (there isn't any...)


If you further define

In[1]:= h=Function[x,x + -x^2]

Out[1]:= Function[x, x- x^2]

In[2]:= h'[2]

Out[2]:= -3

The trick is that Subtract is not associative and hence

In[3]:= D[Funcion[x, x-x^2],x]

                             (1,0)     2            (0,1)    2
Out[3]:= Function[x, Subtract     [x, x ] + Subtract    [x, x  ] 2 x]


All in all, this is the reason why languages like APL have a different
symbol which is used for negation and for subtraction. They are different
concepts.

Peace
Bob Korsan


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