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))
