differentiating, knowing derivative of arcsinx is 1/(root(1-x^2))?

Question

how do i use the chain rule to find:

d/dx arcsin(x/2)

d/dx arcsin(tanx)

d/dx 1/(arcsinx^2)

thanks for your help!

Answer 1

d/dx arcsin(x/2)
Let u = x/2, du/dx = 1/2

d/dx arcsin(tanx)
Let u = tanx, du/dx = sec^2x

d/dx 1/(arcsinx^2)
Rewrite as [arcsinx^2]^-1
Let u = x^2, du/dx = 2x
v = arcsinu, dv/du = 1/(root(1-u^2))