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

1 Answer
George C.
Jun 8, 2017

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)

Related questions
See all questions in Differentiating Inverse Trigonometric Functions
Impact of this question
7969 views around the world
You can reuse this answer
Creative Commons License