eq:''diff(x + y =a, x);solve([eq],[dy/dx]);

That is OK.

I only erase (%i5):

(%i1)algebraic : true;
(%i2)depends(y,x);
(%i3)diff(x^(1/2) + y^(1/2) = a^(1/2), x);
(%i4)solve(%,diff(y,x));
(%i5)rat(%);

The (%i5) output now is
(%o5)dy/dx=-(SQRT(x) SQRT(y))/x

How I transform (SQRT(x) SQRT(y)) in SQRT(xy)???

Thanks.


Em Sun 24 Apr 2005 20:31, Barton Willis escreveu:
> Maybe you want something like this
>
> (%i1) algebraic : true;
> (%o1) true
> (%i2) depends(y,x);
> (%o2) [y(x)]
> (%i3) diff(x^(1/2) + y^(1/2) = a^(1/2), x);
> (%o3) 'diff(y,x,1)/(2*sqrt(y))+1/(2*sqrt(x))=0
> (%i4) solve(%,diff(y,x));
> (%o4) ['diff(y,x,1)=-sqrt(y)/sqrt(x)]
> (%i5) %*2*sqrt(y);
> (%o5) [2*sqrt(y)*('diff(y,x,1))=-(2*y)/sqrt(x)]
> (%i6) rat(%);
> (%o6) [2*sqrt(y)*('diff(y,x,1))=-(2*sqrt(x)*y)/x]
>
>
> Barton
>
> -----maxima-admin@math.utexas.edu wrote: -----
>
> >To: maxima 
> >From: Jorge Barros de Abreu
> >
> >Sent by: maxima-admin@math.utexas.edu
> >Date: 04/24/2005 12:35PM
> >Subject: eq:''diff(x + y =a,
> >x);solve([eq],[dy/dx]);
> >
> >Changing the words:
> >
> >How I make the following situation?
> >
> >(%i2) depends(y,x);
> >(%i3) diff(x^(1/2) + y^(1/2) = a^(1/2), x);
> >(%i4) %-(1/(2*sqrt(x)));
> >(%i5) %*2*sqrt(y);
> >
> >I need for (%o5) output with no square root in
> >denominator an without (%i4)
> >and (%i5) manual input.