Let's go through the derivation for the general form of the derivative of #arccos(x)#

Consider #y = cos^(-1)(x)#

#implies x = cos(y)#

We can construct a triangle from this. Recall that cosine is adjacent/hypotenuse. This means that adjacent to angle #y#, we have side of length #x# and that the hypotenuse is of length #1#. By Pythagoras' the length of the opposite side is #sqrt(1-x^2)#.

Now, differentiate our expression:

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

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

But from our triangle we can work out #sin(y)!#. Sine is opposite/hypotenuse so

#sin(y) = sqrt(1-x^2)#

#therefore (d)/(dx)(cos^(-1)(x)) = -1/(sqrt(1-x^2))#

That's the general form, but here we have a function of x inside the function, ie #y(u(x))#. This calls for the chain rule.

#(dy)/(dx) = (dy)/(du)(du)/(dx)#

#u=x/2 implies (du)/(dx)=1/2#

#y = cos^(-1)(u) implies (dy)/(du) = -1/(sqrt(1-u^2))#

#therefore (dy)/(dx) = -1/(sqrt(1-(x/2)^2))*1/2 = -1/(sqrt(4-x^2))#