Problem:Find the partial derivatives w.r.t. x and y for 
> f := (x,y) > arccos(x*y);
f := (x, y) > arccos(x y)
evaluate the partials at (0.5,0.5). 

Don't remember the derivative of "arccos"? I never remember it myself either. But here is how you can compute it. Use the fact that if v() is the inverse function of u() then, 
> v(u(x)) = x:
So by the chain rule we get 
> diff(v(u(x)),x) = 1;
/d \ D(v)(u(x))  u(x) = 1 \dx /
and if we call y=u(x) and x=v(y) we can write: 
> ;
d 1  v(y) =  dy d  u(v(y)) dx
so that's the general formula. Now use u=cos and v=arccos to obtain, 
> Diff(arccos(y),y) = 1/(sin(arccos(y)));
d 1  arccos(y) =   dy 2 1/2 (1  y )
the last simplification follows from the fact that the derivative of cos is
sin and the fact that

> diff(arccos(y),y);
1   2 1/2 (1  y )
but that's no fun. 
Now the partials. The partial w.r.t. the first ([1]) argument of f evaluated at (0.5,0.5) is given by: 
> D[1](f)(0.5,0.5);
.5163977795
ghosh that was easy! Now the other partial w.r.t. y the second ([2]) argument of f. 
> fy := D[2](f);
x fy := (x, y) >   2 2 1/2 (1  x y )
now evaluate: 
> fy(0.5,0.5);
.5163977795
Done! 