Noting that
#d/dx arccos(x) = -1/sqrt(1-x^2)#
and
#d/dx arcsin(x) = 1/sqrt(1-x^2)#
(derivations are included at the bottom)
Applying the chain rule gives us
#d/dxarccos(arcsin(x)) = -1/sqrt(1-arcsin^2(x))(d/dxarcsin(x))#
#=-(1/sqrt(1-x^2))/sqrt(1-arcsin^2(x))#
#=-1/(sqrt(1-x^2)sqrt(1-arcsin^2(x))#
.
.
.
To derive the derivatives used above, we can use implicit differentiation.
Let #y = arccos(x)#
#=> cos(y) = x#
#=> -sin(y)dy/dx = 1#
#=> dy/dx = -1/sin(y)#
#=-1/sin(arccos(x))#
#=-1/sqrt(1-x^2)#
(Try drawing a right triangle where #cos(theta) = x# and calculate #sin(theta)# for the final step)
Let #y = arcsin(x)#
#=> sin(y) = x#
#=> cos(y)dy/dx = 1#
#=> dy/dx = 1/cos(y)#
#=1/cos(arcsin(x))#
#=1/sqrt(1-x^2)#
