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/
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