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

*To*: mathgroup at smc.vnet.net*Subject*: [mg25559] RE:[mg25495] ? D[f,{x,n}]*From*: "Ersek, Ted R" <ErsekTR at navair.navy.mil>*Date*: Sat, 7 Oct 2000 03:36:07 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

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 Get Mathematica tips, tricks from http://www.verbeia.com/mathematica/tips/Tricks.html