Subject: eq:''diff(x + y =a, x);solve([eq],[dy/dx]);
From: Jorge Barros de Abreu
Date: Sat, 23 Apr 2005 15:29:33 -0300
Exactly Albert.
But this is not work in:
diff(ev(y, solve(x^.5+y^.5=a^.5, y)), x);
Maxima ask me:
Is SQRT(a) - SQRT(x) positive, negative, or zero?
How I say it to maxima?
How I say to maxima that a>=0 and 0<=x<=a and 0<=y<=a?
[]´s
Em Sat 23 Apr 2005 14:17, Albert Reiner escreveu:
> [Jorge Barros de Abreu , Sat, 23 Apr 2005 13:57:02
-0300]:
> > Hi for all.
> > How I make the following situation?
> >
> > depends(y,x);
> > eq:''diff(x + y =a, x);
>
> This differentiates both sides of the equation with respect to x. The
> result is the equation 1 = 0.
>
> > solve([eq01],[dy/dx]);
>
> I suspect you want something like
>
> diff(ev(y, solve(x+y=a, y)), x); ==> -1 //
>
> HTH,
>
> Albert.
>
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