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

1 Answer
Steve M
Oct 29, 2016

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

Explanation:

Let # y = (arcsinx)^2 => y^(1/2) = arcsinx#
# :. sin(y^(1/2)) = x #

Differentiating wrt #x#;
# :. cos(y^(1/2))(1/2y^(-1/2))dy/dx = 1 #

# :. dy/dx = 2/(cos(y^(1/2))(y^(-1/2))) #

# :. dy/dx = (2y^(1/2))/(cos(y^(1/2))) #

Now # sin^2A + cos^2A -= 1 => cosA = sqrt(1 - sin^2A #
So, # cos(y^(1/2)) = sqrt(1 - sin^2(y^(1/2)) #
# :. cos(y^(1/2)) = sqrt(1 - x^2) #

Hence, # dy/dx = (2arcsinx)/sqrt(1 - x^2) #

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