What is the derivative of this function #y=sin^-1(1-x)^(1/2)#?

1 Answer
Steve M
Nov 30, 2016

# dy/dx = -1/(2sqrt(x)sqrt(1-x) #

Explanation:

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

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

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

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

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

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

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