Question

How do you find the derivative of `Cos[arcsin(x)]`?

Answer 1

`d/(dx) cos(arcsin(x)) = -x/sqrt(1-x^2)`

Explanation:

Note that:

`cos^2 theta + sin^2 theta = 1`

So for any `theta`, we have:

`cos(theta) = +-sqrt(1-sin(theta))`

Note that for any real `x in [-1, 1]`, by definition:

`arcsin(x) in [-pi/2, pi/2]`

Further note that if `theta in [-pi/2, pi/2]` then:

`cos theta >= 0`

Hence:

`cos(arcsin(x)) = sqrt(1-x^2)`

for any `x in [-1, 1]`

Outside this interval `cos(arcsin(x))` takes non-real complex values.

So if we are dealing with this purely as a real valued function, then this is the only definition we need.

Then:

`d/(dx) cos(arcsin(x)) = d/(dx) (1-x^2)^(1/2)`

`color(white)(d/(dx) cos(arcsin(x))) = 1/2(1-x^2)^(-1/2)*(-2x)`

`color(white)(d/(dx) cos(arcsin(x))) = -x/sqrt(1-x^2)`