MathGroup Archive 1992

[Date Index] [Thread Index] [Author Index]

Search the Archive

D[myFunc[g[x]


In Re: the following...

> I want to give the function 'myFunc' 
> the property of being 'transparent' to differentiation.
> i.e., I want to have 
> 
>   d myFunc[g[x]]                       d g[x]
>   --------------    give me    myFunc[ ------ ]                (1)
> 	dx                               dx
> 
> where g is an arbitrary function.  "Easy" I hear you say...



Actually, I didn't say that.  But here is one simple solution.


--------------------------------------------------------------------


In[1]:= Unprotect[D]

Out[1]= {D}

In[2]:= D[MyFunc[a_],b_]:=MyFunc[D[a,b]]

In[3]:= Protect[D]

Out[3]= {D}

	(*  That's it!
	    Other functions are unchanged...  *)

In[4]:= D[Sin[x^n],x]

           -1 + n      n
Out[4]= n x       Cos[x ]


	(* MyFunc now behaves as advertised.  *)


In[5]:= D[MyFunc[x^n],x]

                  -1 + n
Out[5]= MyFunc[n x      ]

In[6]:= D[MyFunc[x^n],n]

                n
Out[6]= MyFunc[x  Log[x]]

In[7]:= Quit



There may be many other solutions.


Keith  (clay at galileo.phys.washington.edu)





  • Prev by Date: Mathematica 2.0 and solving ODE's
  • Next by Date: Re: D[myFunc[g[x]]] -> myFunc[D[g[x],x]: How to do in ch
  • Previous by thread: Mathematica 2.0 and solving ODE's
  • Next by thread: Re: D[myFunc[g[x]]] -> myFunc[D[g[x],x]: How to do in ch