Let's go through the derivation for the general form of the derivative of arccos(x)arccos(x)
Consider y = cos^(-1)(x)y=cos−1(x)
implies x = cos(y)⇒x=cos(y)
We can construct a triangle from this. Recall that cosine is adjacent/hypotenuse. This means that adjacent to angle yy, we have side of length xx and that the hypotenuse is of length 11. By Pythagoras' the length of the opposite side is sqrt(1-x^2)√1−x2.
Now, differentiate our expression:
(dx)/(dy) = -sin(y)dxdy=−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))
