How do you differentiate #(arcsinx)^2#?

If you are studying maths, then you should learn the Chain Rule for Differentiation, and practice how to use it:

If # y=f(x) # then # f'(x)=dy/dx=dy/(du)(du)/dx #

I was taught to remember that the differential can be treated like a fraction and that the "#dx#'s" of a common variable will "cancel" (It is important to realise that #dy/dx# isn't a fraction but an operator that acts on a function, there is no such thing as "#dx#" or "#dy#" on its own!). The chain rule can also be expanded to further variables that "cancel", E.g.

# dy/dx = dy/(dv)(dv)/(du)(du)/dx # etc, or # (dy/dx = dy/color(red)cancel(dv)color(red)cancel(dv)/color(blue)cancel(du)color(blue)cancel(du)/dx) #

So with # y = (arcsinx)^2#, Then:

# { ("Let "u=arcsinx, => , (du)/dx=1/sqrt(1-x^2) "(see below)"), ("Then "y=u^2, =>, dy/(du)=2u ) :}#

Using # dy/dx=(dy/(du))((du)/dx) # we get:

# dy/dx = (2u)(1/sqrt(1-x^2)) #
# :. dy/dx = 2arcsinx/sqrt(1-x^2) #

NOTE - Derivative of #arcsin(x)#
We can easily find #d/dx(arcsinx) # using implicit differentiation:
Let # y = arcsinx <=> siny=x #

# :. cosydy/dx=1 #
Using #sin^2y+cos^2y-=1 => cos^2y=1-x^2 #
#:. cosy=sqrt(1-x^2) #
#:. sqrt(1-x^2)dy/dx=1 #
#:. dy/dx=1/sqrt(1-x^2) #