Re: Differentiation wrt functions in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg20361] Re: [mg20327] Differentiation wrt functions in Mathematica*From*: "Andrzej Kozlowski" <andrzej at tuins.ac.jp>*Date*: Sun, 17 Oct 1999 03:01:49 -0400*Sender*: owner-wri-mathgroup at wolfram.com

Mathematica's D[f,x] ignores any implicit relationships between f and x. You can get what you want by using the total derivative Dt: In[2]:= Dt[Sin[x], Cos[x]] /. Dt[x, f_] -> 1/Dt[f, x] Out[2]= -Cot[x] This is better than what you get by using your "reformulation": In[3]:= D[(1 - Cos[x]^2)^(1/2), {Cos[x], 1}] // Simplify Out[3]= Cos[x] -(-------------) 2 Sqrt[Sin[x] ] The reason of course is that it is not true that Sin[x] is always (1 - Cos[x]^2)^(1/2), sometimes it is -(1 - Cos[x]^2)^(1/2). -- Andrzej Kozlowski Toyama International University JAPAN http://sigma.tuins.ac.jp http://eri2.tuins.ac.jp ---------- >From: jonparker at my-deja.com >To: mathgroup at smc.vnet.net >Subject: [mg20361] [mg20327] Differentiation wrt functions in Mathematica >Date: Sat, Oct 16, 1999, 9:20 > > 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. >