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))#
