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

##### 1 Answer

Oct 21, 2016

#### Explanation:

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

Assuming that we want to consider this as a Real valued function of Real numbers, the domain of

When

#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

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)#