First, let's see what the derivative of the #arccos# function is by using implicit differentiation:

Let #y = arccos(x)#

#=>cos(y) = x#

#=>d/dxcos(y) = d/dxx#

#=>-sin(y)dy/dx = 1#

#=>dy/dx = -1/sin(y)#

#=-1/sin(arccos(x))#

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

(For the last step, try drawing a right triangle where #cos(theta) = x# and then see what #sin(theta)# is equal to)#

With that, the remainder of the problem can be done using the chain rule:

#d/dxsqrt(arccos(x^2-1))#

#= 1/2(arccos(x^2-1))^(-1/2)(d/dxarccos(x^2-1))#

#=1/(2sqrt(arccos(x^2-1)))*(-1)/sqrt(1-(x^2-1)^2)(d/dx(x^2-1))#

#=(2x)/(2sqrt(arccos(x^2-1))sqrt((1+x^2-1)(1-x^2+1))#

#=x/(|x|sqrt(arccos(x^2-1))sqrt(2-x^2))#