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### Derivative of Inverse Cosine

```Date: 05/24/2005 at 12:22:56
From: shivraj
Subject: Derivatives of inverse trignometric functions

Given f(x) = arccos[(a*sin(x) + b*cos(x))/(a*a + b*b)^(1/2)], find
f'(x).  I did this work:

b/((a*a + b*b)^(1/2)) = cos(Y) and a/((a*a + b*b)^(1/2)) = Sin(Y), so

f(x) = arccos[cos(x)*cos(Y) + sin(x)*Sin(Y)]
f(x) = arccos[Cos(x-Y)]   OR   f(x) = arccos[cos(Y-x)]
f(x) = x - Y              OR   f(x) = Y - x
f'(x) = 1                  OR  f'(x) = -1

I'm not sure whether f'(x) is 1 or -1, and don't understand how it can
be both.

```

```
Date: 05/24/2005 at 14:01:37
From: Doctor Rick
Subject: Re: Derivatives of inverse trignometric functions

Hi, Shivraj.

Let's think about the argument of arccos for a moment.  As you've
shown, this function is a phase-shifted cosine function.  The amount
of the phase shift, which you've called Y, depends on the values of
parameters a and b.  Furthermore, we know that cos(x) = cos(-x) for
any x, therefore cos(x-Y) and cos(Y-x) are identical.

The complication therefore must arise when we apply the arccos
function.  Let's examine the function arccos(cos(x)).

It is not true that arccos(cos(x)) is the identity function f(x) = x.
There are many values of x that have a given value of cos(x); the
arccos function must be single-valued, so it selects one of those
values, namely the one in the range [0,pi).  Thus the composition
arccos(cos(x)) can be represented like this:

arccos(cos(x)) = x,       0 <= x <  pi
pi-x,   pi <= x < 2pi
x-2pi, 2pi <= x < 3pi
x-3pi, 3pi <= x < 4pi
etc.

If you plot this function, it is a sawtooth:

pi +             *                           *
|           /   \                       /   \
|         /       \                   /       \
|       /           \               /           \
|     /               \           /               \
|   /                   \       /                   \
| /                       \   /                       \
+-------------+-------------+-------------+-------------+---
0             pi           2pi           3pi           4pi

You want to take the derivative of the function arccos(cos(x-Y)).
This function is the sawtooth function with a phase shift: it is zero
at x = Y + 2k*pi, for any integer k.  By inspection we see that the
derivative is 1 or -1, depending on the value of x:

d/dx(arccos(cos(x-Y)) =  1,       Y <= x - 2k*pi < Y + pi
-1,  Y + pi <= x - 2k*pi < Y + 2pi

for some integer k (such that x - 2k*pi is in one of these ranges).

derivative is 1 or -1, but not both at the same time.  The value of
the derivative depends on the value of x (as derivatives usually do!).

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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