Can someone help me with finding the derivative of this function?

Question

y = -( csc x)2 (cos-1(1-x2))

Answer 1

product rule and chain rule -- if

y = uv
y' = u'v + uv'
Here u=csc^2 x and v = arccos(1-x^2)

y' = -2 cscx cscx cotx arccos(1-x^2) + csc^2 x 1/√(1-(1-x^2)^2) * -2x

kind of messy, no? well, recall that if cos u = 1-x^2, sin u = x, so we really have

y = -csc^2(x) arcsin(x)
y' = csc^2(x) (2arcsin(x)cot(x) - 1/√(1-x^2))