Re: Differentiation wrt functions in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg20385] Re: [mg20327] Differentiation wrt functions in Mathematica*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Mon, 18 Oct 1999 02:40:31 -0400*References*: <199910160020.UAA26378@smc.vnet.net.>*Sender*: owner-wri-mathgroup at wolfram.com

jonparker at my-deja.com wrote: > > I am trying to get Mathematica 3.0 to differentiate equations > containing Sin and Cos terms, with respect to Sin[x] and Cos[x]. On > investigation of the results I find I am not getting the answer I > expect. As a test I asked M to evaluate the following: > D[Sin[x],{Cos[x],1}] > and go the result 0. I would have expected to get -Cos[x]/Sin[x]. If > I reformulate the expression as: > D[(1-Cos[x]^2)^.5,{Cos[x],1}] > I then get the answer I expect. This work-around is not convenient for > the full expressions I would like to deal with. > > Is there a way of reminding M the Sin=(1-Cos^2)^.5? > Thanks, Jon > > Sent via Deja.com http://www.deja.com/ > Before you buy. Here are some possibilities. (i) Transform coordinates explicitly. In[25]:= InputForm[D[Sin[ArcCos[u]], u] /. u->Cos[x]] Out[25]//InputForm= -(Cos[x]/Sqrt[1 - Cos[x]^2]) This may be simplified a bit using Simplify, and then one might invoke PowerExpand to kill any branch cut distinctions. In[26]:= PowerExpand[Simplify[%25]] Out[26]= -Cot[x] (ii) Use the chain rule to define your own differentiation with respect to a function. Since df/du = (df/dx) * (dx/du) = (df/dx) / (du/dx) this may be done rather simply as below. In[27]:= functionalD[f_, g_, x_] := D[f,x] / D[g,x] In[28]:= functionalD[Sin[x], Cos[x], x] Out[28]= -Cot[x] Daniel Lichtblau Wolfram Research

**References**:**Differentiation wrt functions in Mathematica***From:*jonparker@my-deja.com