Question

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

Answer 1

`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)`