Try this:

restart;
VectorCalculus[DirectionalDiff](x*exp(y)/(3*z^2+1),
<1,-2,3>,[x,y,z]);

However, if you have several directional derivatives to compute you could do this:

restart;
f := (a,b,c)->VectorCalculus[DirectionalDiff](x*exp(y)/(3*z^2+1),<a,b,c> ,[x, y, z] ):
f(1,-2,3);
f(4,-5,6);

Hope this helps,

J. Tarr

The directional derivative will evaluate to a simple numerical answer if the derivatives of the function are constants. For example:

restart;
f := (a,b,c)->VectorCalculus[DirectionalDiff](2*x+3*y+4*z+1,<a,b,c> ,[x, y, z] ):
f(1,-2,3);
f(4,-5,6);

Hope this helps,

J. Tarr

Sorry, I misread your question. The directional derivative can be evaluated at any given point by substituting the value of that point's coordinates in the derivative. For example:

restart; with(VectorCalculus):
sph := (x-a)^2+(y-b)^2+(z-c)^2-r^2;
dd := DirectionalDiff( sph , <v1,v2,v3>, [x,y,z] );
v1,v2,v3 := 1,2,3;
dd;
x,y,z := 4,5,6;
dd;
a,b,c := 0,0,0;
dd;

You can, of course, use eval, or subs, if that is more convenient.

J. Tarr

You could do something like this:

restart; with(VectorCalculus):
sph := (x-7)^2+(y-8)^2+(z-9)^2-r^2;
eval(DirectionalDiff( sph , <1,2,3>, [x,y,z] ), {x=4,y=5,z=6});

or a bit more flexibly like this:

restart;
f := (V1,V2,V3,X,Y,Z)->eval(VectorCalculus:-DirectionalDiff(g,<i,j,k> ,[x, y, z] ),{i=V1,j=V2,k=V3,x=X,y=Y,z=Z}):
g := (x-7)^2+(y-8)^2+(z-9)^2-r^2;
ans := f(1,2,3,4,5,6);

Hope this helps,

J. Tarr

Is the DirectionalDiff command the tool I should use to find the gradient of a vector.  MAPLE HELP has a gradient as well as NABLA command but it seems they can only be used on scalar functions to convert them into a gradient vector.  Instead of this I am seeking the gradient of a vector which then yields a 2nd order tensor.  Is there something else I should use?

It would seem that the MAPLE help menu would have a link to the Jacobian from the gradient help page.  That would have helped me save some time.