TITLE: Computing Partials of inverse trig function
# !b2 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).
# !b3 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
<center> sin = sqrt(1-cos^2) </center>
Of course we could have just done,
> 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!

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