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). |
|
|
Solution:
| 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! |