How do you find the derivative of the function: sin[arccos(x)]?

1 Answer
George C.
Oct 21, 2016

d/(dx) sin(arccos(x)) = -x/(sqrt(1-x^2)

Explanation:

f(x) = sin(arccos(x))

Assuming that we want to consider this as a Real valued function of Real numbers, the domain of f(x) is [-1, 1]

When x in [-1, 1] then arccos(x) in [0, pi] and hence:

sin(arccos(x)) >= 0

Then from Pythagoras, we have:

cos^2 theta + sin^2 theta = 1

Hence:

sin(arccos(x)) = sqrt(1 - x^2)

using the non-negative square root since we have already established that sin(arccos(x)) >= 0

So:

d/(dx) sin(arccos(x)) = d/(dx) (1-x^2)^(1/2)

color(white)(d/(dx) sin(arccos(x))) = 1/2(1-x^2)^(-1/2)*(-2x)

color(white)(d/(dx) sin(arccos(x))) = -x/(sqrt(1-x^2)

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