Computation of dparx(x)

> dparx1:=collect(subs(d(y)=0,df),p);

The next three steps are a bit of 'sleight of hand' to get Maple to remove the appropriate terms... namely those with a p in the numerator, and no p's in the denominator.

> dparxromeo:=simplify(p*dparx1);

> dparxlist:=convert(dparxromeo,list);

> dparxjuliet:=remove(has,dparxlist,{p^2,p^3,p^4});

> dparx2:=convert(dparxjuliet,`+`)/p;

Now we simplify dparx2, set it equal to zero, and solve for d(x). Something which is now easily done since dparx2 is linear in d(x).

> dparxx:=solve(dparx2=0,d(x));

And we are done.... the next section we repeat this for dpary.

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