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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).

Does this answer your question?  The short answer is yes, the 
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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