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Re: How to calculate the partial derivative?
Posted:
Nov 21, 2012 5:17 AM
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Dear Dr. M. Abbasi,
Thank you very much for your kindly reply. Your second method still can't simplify the express as I wanted if I put '/6/V' after Li.
Thanks,
Tang Laoya
On Wednesday, November 21, 2012 5:58:45 PM UTC+8, Nasser M. Abbasi wrote:
> On 11/21/2012 2:17 AM, Nasser M. Abbasi wrote:
>
> > On 11/21/2012 12:15 AM, 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
>
> >>
>
> >
>
>
>
>
>
> > The problem is that you did not use equations, but used expresions.
>
> > Hence in your case L1 is now 'a1+b1*x+c1*y' and similarly with L2.
>
> >
>
> > To do this right, I'd write things as equations, and use rules, like
>
> > this:
>
> >
>
> > --------------------------
>
> > Clear[L1, L2, a1, b2, x, c1, y, a2, b2, c2, lhs, rhs]
>
> > eq1 = L1 == a1 + b1*x + c1*y;
>
> > eq2 = L2 == a2 + b2*x + c2*y;
>
> > lhs[eq_] := eq /. (lhs_ == rhs_) -> lhs;
>
> > rhs[eq_] := eq /. (lhs_ == rhs_) -> rhs;
>
> >
>
> > NN = rhs[eq1]*rhs[eq2];
>
> > DNx = D[NN, x]
>
> > ----------------------------
>
> >
>
> > which gives what you showed
>
> >
>
> > b2 (a1+b1 x+c1 y)+b1 (a2+b2 x+c2 y)
>
> >
>
> > Now it is easy to use the rules to get what you want
>
> >
>
> > ------------------
>
> > DNx/.{rhs[eq1]->lhs[eq1],rhs[eq2]->lhs[eq2]}
>
> > ---------------------
>
> >
>
> > which gives
>
> >
>
> > b2 L1+b1 L2
>
> >
>
>
>
> After thinking more about this, another way to do this,
>
> and to keep from having to write rhs[eq] and rhs[eq] all
>
> the time, which is not as nice looking as L1 and L2 etc..
>
> and can be harder to read if there are many equations, is
>
> to use rules.
>
>
>
> Like this:
>
>
>
> -----------------------------
>
> Clear["Global`*"];
>
>
>
> rule1 = {L1 -> a1 + b1*x + c1*y, L2 -> a2 + b2*x + c2*y}
>
> rule2 = (Last[#] -> First[#]) & /@ rule1
>
> NN = (L1*L2) /. rule1;
>
> DNx = D[NN, x]
>
> -----------------------
>
>
>
> Which gives
>
>
>
> b2 (a1 + b1 x + c1 y) + b1 (a2 + b2 x + c2 y)
>
>
>
> Now to get back, use rule2 above, which is the reverse
>
> of rule1
>
>
>
> -------------------------------
>
> In[185]:= DNx/.rule2
>
> Out[185]= b2 L1+b1 L2
>
> -----------------------------
>
>
>
> The advantage of this method again is that one used L1 and L2,
>
> instead of rhs[eq1], rhs[eq2] etc.. which is more clear.
>
>
>
> --Nasser
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