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RE:? D[f,{x,n}]


Jack Goldberg had trouble making it so 
D[f, x]   and   D[f, {x, n}] 
perform some "simplification" on (f) 
before the built in D is used.

I give some code below that does this. 
Notice if you do these examples without  
changing D you get factored polynomials.

Since I don't give a rule for D[f, {x, n}] this 
seems to prove that D[f, {x, n}]  evaluates 
  Nest[ D[#, x]&, f, n ]
or something equivalent.
-----------------------------

In[1]:=
Unprotect[D];
flag=True;

D[f_,x_]/;flag:=
  Block[{flag}, D[Expand[f],x] ]


In[4]:=
D[(x+2)^3,x]

Out[4]=
12 + 12*x + 3*x^2


In[5]:=
D[(x+2)^4,{x,2}]

Out[5]=
48 + 48*x + 12*x^2

--------------------
Notice, inside Block I didn't have to set (flag=False) 
We only care that flag is not True.

--------------------
On a related note, the Help Browser says 
"Derivative is generated when you apply D to a function  
whose derivative Mathematica does not know."  We see Derivative 
in the FullForm below.

In[6]:=
f'[x] //FullForm

Out[6]=
Derivative[1][f][x]


In[7]:=
f''[x] //FullForm

Out[7]=
Derivative[2][f][x]


--------------------
Regards,
Ted Ersek

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