# 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>