Re: Derivative of a funtion evaluated at a point in 3D

*To*: mathgroup at smc.vnet.net*Subject*: [mg44502] Re: Derivative of a funtion evaluated at a point in 3D*From*: poujadej at yahoo.fr (Jean-Claude Poujade)*Date*: Wed, 12 Nov 2003 08:02:06 -0500 (EST)*References*: <boign7$oj1$1@smc.vnet.net> <bopd02$oq6$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

mroc_1000 at hotmail.com (mroc) wrote in message news:<bopd02$oq6$1 at smc.vnet.net>... > mroc_1000 at hotmail.com (mroc) wrote in message news:<boign7$oj1$1 at smc.vnet.net>... > > Hello, I am totally new to Mathematica and trying to program a simple > > FEM-type problem. I am trying to take the partial derivative > > (symbolically) of an expression that is a function of four points in > > 3D. All I can think of to do is F[p1,p2,p3,p4] = stuff then > > D[F,p1,p2,p3,p4] . But I keep getting a zero expression as a result. > > (where stuff is a nasty combination of these points) Any thoughts? > > Anyone? What am I missing? Is this question too easy or too hard? Maybe you should have given an explicit example? I hope this can help you : In[1]:=(* scalar function example : *) scalarFunc[p1_,p2_,p3_,p4_]:=(p1+a*p2).(p3+b*p4); In[2]:= scalarValue=scalarFunc[{p11,p12,p13},{p21,p22,p23},{p31,p32,p33},{p41,p42,p43}] Out[2]=(p11+a p21)(p31+b p41)+(p12+a p22)(p32+b p42)+(p13+a p23)(p33+b p43) In[3]:=D[scalarValue,a] Out[3]=p21(p31+b p41)+p22(p32+b p42)+p23(p33+b p43) In[4]:=(* vector function example : *) vectorFunc[p1_,p2_,p3_,p4_]:=(p1+a*p2+p3+b^2*p4); In[5]:= vectorValue=vectorFunc[{p11,p12,p13},{p21,p22,p23},{p31,p32,p33},{p41,p42,p43}] Out[5]={p11 + a p21 + p31 + b^2 p41, p12 + a p22 + p32 + b^2 p42, p13 + a p23 + p33 + b^2 p43} In[6]:=D[vectorValue,b] Out[6]={2 b p41, 2 b p42, 2 b p43} --- jcp