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

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Answer 1 · 2016-11-14T21:46:47

The derivative is #f'(x) = (2arcsinx)/sqrt(1 - x^2)#.

Let #y = u^2# and #u = arcsinx#. By the chain rule, you have:

#dy/dx = dy/(du) xx (du)/dx#

Let's start by finding the derivative of #u = arcsinx#.

#u = arcsinx -> sinu = x#

Differentiate implicitly:

#cosu((du)/dx) = 1#

#(du)/dx = 1/cosu#

Rearrange the identity #sin^2x + cos^2x = 1# for #cosx# to get #cosx = sqrt(1 - sin^2x)#.

#(du)/dx= 1/sqrt(1 - sin^2u)#

Since #sinu = x#, we have:

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

Now to #dy/(du)#. By the product rule, we have, #dy/(du) = 2u#.

Call the function #y = (arcsinx)^2# #f(x)#.

#f'(x) = dy/(du) xx (du)/dx#

#f'(x) = 2u xx 1/sqrt(1 - x^2)#

#f'(x) = (2arcsinx)/sqrt(1- x^2)#

Hopefully this helps!