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Topic: How to calculate the partial derivative?
Replies: 6   Last Post: Nov 21, 2012 5:17 AM

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Tang Laoya

Posts: 23
Registered: 11/21/12
Re: How to calculate the partial derivative?
Posted: Nov 21, 2012 4:08 AM
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Dear Dr. M. Abbasi,

Thank you very much for your kindly reply. I tested your code, it works!
Howerver, I still didn't understand your code, could you give me some more explains? In addition, I write similar code according to your format, it doesn't work, could you please help me to take a look at it?

Thanks,
Tang Laoya


Clear["Global`*"];
eq1 = L1 == (a1 + b1 x + c1 y + d1 z)/6/V;
eq2 = L2 == (a2 + b2 x + c2 y + d2 z)/6/V;
eq3 = L3 == (a3 + b3 x + c3 y + d3 z)/6/V;
eq4 = L4 == (a4 + b4 x + c4 y + d4 z)/6/V;
lhs[eq_] := eq /. (lhs_ == rhs_) -> lhs;
rhs[eq_] := eq /. (lhs_ == rhs_) -> rhs;

nod = 10;
NN = Table[0, {nod}];
NN[[1]] = rhs[eq1] (2 rhs[eq1] - 1);
NN[[2]] = rhs[eq2] (2 rhs[eq2] - 1);
NN[[3]] = rhs[eq3] (2 rhs[eq3] - 1);
NN[[4]] = rhs[eq4] (2 rhs[eq4] - 1);
NN[[5]] = 4 rhs[eq1] rhs[eq2];
NN[[6]] = 4 rhs[eq2] rhs[eq3];
NN[[7]] = 4 rhs[eq3] rhs[eq1];
NN[[8]] = 4 rhs[eq3] rhs[eq4];
NN[[9]] = 4 rhs[eq4] rhs[eq1];
NN[[10]] = 4 rhs[eq2] rhs[eq4];

DNx = D[NN, x];
DNx = DNx /. {rhs[eq1] -> lhs[eq1], rhs[eq2] -> lhs[eq2],
rhs[eq3] -> lhs[eq3], rhs[eq4] -> lhs[eq4]}

DNy = D[NN, y];
DNy = DNy /. {rhs[eq1] -> lhs[eq1], rhs[eq2] -> lhs[eq2],
rhs[eq3] -> lhs[eq3], rhs[eq4] -> lhs[eq4]}

DNz = D[NN, z];
DNz = DNz /. {rhs[eq1] -> lhs[eq1], rhs[eq2] -> lhs[eq2],
rhs[eq3] -> lhs[eq3], rhs[eq4] -> lhs[eq4]}



On Wednesday, November 21, 2012 2:15:07 PM UTC+8, Tang Laoya wrote:
> Dear all,
>
>
>
> I am trying to calculate the partial derivative by mathematica, I have the following commands:
>
> L1=a1+b1*x+c1*y;
>
> L2=a2+b2*x+c2*y;
>
> L3=a3+b3*x+c3*y;
>
>
>
> NN=L1*L2;
>
>
>
> DNx=D[NN,x];
>
>
>
> I got the following result:
>
> DNex=b2 (a1+b1 x+c1 y)+b1 (a2+b2 x+c2 y)
>
>
>
> How to do to have the following result?
>
>
>
> DNex=b2*L1 + b1 * L2
>
>
>
>
>
> Thanks,
>
> Tang Laoya





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