On 1/13/2012 12:36 PM, Hans W. Hofmann wrote:
> (%i2) f(x):=-k*x^3+3*k^2*x^2;
> 3 2 2
> (%o2) f(x) := (- k) x + 3 k x
>
> (%i3) df(x):=diff(f(x),x)$ df(x);
> 2 2
> (%o3) 6 k x - 3 k x
>
> (%i4) df(k)=3;
> 3
> (%o4) 8 k = 3
>
> whats this? should have
>
> (%i5) 6*k^2*x-3*k*x^2 = 3, x=k;
> (%o5) 3*k^3 = 3
>
> using define all works fine....
>
> Gru? HW
>
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no, f(k) is -k*k^3+3*k^2*k^2 or 2*k^4.
compute its derivative with respect to k, and one gets 8*k^3.
RJF