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MathGroup Archive 2003

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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


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