Another service from Omega

Computing Partials of inverse trig function


*****

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
sin = sqrt(1-cos^2)
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!


Link to the commands in this file
Carlos Rodriguez <carlos@math.albany.edu>
Last modified: Thu Mar 29 16:02:13 EST 2001