Subject: eq:''diff(x + y =a, x);solve([eq],[dy/dx]);
From: Albert Reiner
Date: 23 Apr 2005 19:17:49 +0200
[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.