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

##### 1 Answer

Jun 8, 2017

#### Explanation:

Note that:

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

So for any

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

Note that for any real

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

Further note that if

#cos theta >= 0#

Hence:

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

for any

Outside this interval

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)#