How do you find the derivative of #arcsin(sqrt(1-x^2)) #?

1 Answer
Nov 29, 2016

# d/dx arcsin(sqrt(1-x^2))= -x/(|x|sqrt(1-x^2)) #

Explanation:

Let #y = arcsin(sqrt(1-x^2)) #
Then # siny = sqrt(1-x^2) = (1-x^2)^(1/2) #

Differentiating Implicitly and applying the chain rule;
# cosydy/dx = (1/2)(1-x^2)^(-1/2)(-2x) #
# :. cosydy/dx = -x/sqrt(1-x^2) ... [1]#

Then using the fundamental trig identity # sin^2A+cos^2A-=1# we have:

# cos^2y=1-sin^2y #
# cos^2y=1-((1-x^2)^(1/2))^2 #
# :. cos^2y=1 - (1-x^2)) #
# :. cos^2y=x^2 #
# :. cosy=+-x #
# :. cosy=|x| #

Substituting into [1] we get:
# :. |x|dy/dx = -x/sqrt(1-x^2) #

Leading to the solution:
# dy/dx = -x/(|x|sqrt(1-x^2)) #