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MathGroup Archive 2006

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Re: Defining a derivative that distributes for a function

  • To: mathgroup at smc.vnet.net
  • Subject: [mg64246] Re: Defining a derivative that distributes for a function
  • From: dh <dh at metrohm.ch>
  • Date: Wed, 8 Feb 2006 03:53:32 -0500 (EST)
  • References: <ds9mqp$48$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi Andreas,
both expressions are equivalent. The first one is obtained if the chain 
rule is used, the seconde one if your rule is used.
You can also directly verify the equivalence by noting that the solution 
to your definition is: f[x_]= f[1] x.

Daniel

Andres Corrada-Emmanuel wrote:
> Hello,
> 
> I'm trying to define a derivative for a function that distributes:
> 
> D[f[expr_],x_] ^:= f[D[expr,x]]
> 
> This gives me:
> 
> D[f[x^2],x] = f[2x]
> 
> and
> 
> D[f[x^4],x] = f[4x^3]
> 
> But D[f[x^2] + f[x^4],x] = 2xf'[x^2] + 4x^3f'[x^4] instead of the
> desired:
> 
> D[f[x^2] + f[x^4],x] = f[2x] + f[4x^3]. Why? And how do I get the desired behaviour.
> 
> --------------------------------------
> Andres Corrada-Emmanuel
> Lecturer in Physics
> Physics Department
> University of Massachusetts at Amherst
> --------------------------------------
> 


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