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!
