Start by splitting into two separate bits and differentiating them one at a time (sorry for sticking in random letters all over the place - I have tried to avoid #y#, where a #y# is misleading!).
Suppose #k=s*sqrt(1-s^2)# and #m=arccos(s)# and #u=sqrt(1-s^2)#
First #d/(ds)(s*sqrt(1-s^2))#, is determined using the chain rule and product rule.
#(dk)/(ds)= s*d/(ds)(u(s)) + u*d/(ds)(s)#
Set #t=1-s^2# and #u=t^(-1/2)# then apply the chain rule #(du)/(ds) = (du)/(dt) * (dt)/(ds)#.
Hence, #dy/(ds) = -2s * 1/2*(1-s^2)^(-1/2)#
This simplifies down to #(du)/(ds) = -s/(sqrt(1-s^2)#.
Next, to apply the product rule to find the overall derivative.
We have #s# times that derivative, and the derivative of #s# times the undifferentiated expression:
#d/(ds)(s*sqrt(1-s^2)) = (-s^2)/sqrt(1-s^2) + sqrt(1-s^2)#.
When you find a common denominator by algebraic manipulation, hence:
#(dk)/(ds) = -(2s^2-1)/sqrt(1-s^2)#.
Then, there is the differentiation of the #arccos# function.
Apply #cos# to both sides to undo the the effect of the inverse #cos#. Hence #m=arccos(s)# becomes #s=cos(m)#.
Using the knowledge that #d/dx(cos(x)) = -sin(x)#, we can say that #(ds)/(dm)= -sin(m)#
Next, to remove the necessity of the #sin# function in there, we can use the Pythagorean identity to find the derivative in terms of #s# only.
Since #cos^2(theta) + sin^2(theta) = 1#, it follows that #sin(theta) = sqrt(1-cos^2(theta))#.
Hence, #(ds)/(dm) = -sqrt(1-s^2#, because #s=cos(m)#.
We know from the chain rule that #dy/dx = 1/(dx/dy)#, so finally #d/(ds)(arccos(s)) = -1/sqrt(1-s^2)#.
The simplification of the two together is quite delightful, as the already have the same denominator - so you can just add!
Finally, we have #dy/(ds) = -((2s^2) / sqrt(1-s^2))#.
Sorry for all the confusion here, I hope it helps!
