How do you derive -2arccos(2x)?

please say the result and how you find it

do you mean differentiate?

From first principles

y = -2arccos(2x)

arccos(2x) = -y / 2

take cos of both sides

2x = cos(-y / 2)

now use implicit differentiation

d / dx(2x) = d / dx( cos(-y / 2))

2 = -sin(-y/2) [(-1/2) dy/dx)]

dy / dx = 4 / [sin (-y/2)]

now recall that
sin^2 + cos^2 = 1
so sin^2(-y/2) = 1 - cos^2(-y/2)
sin(-y/2) = sqrt(1 - cos^2(-y/2))

so

dy / dx = 4 / [sqrt(1 - cos^2(-y/2))]

earlier we had

2x = cos(-y / 2)

so

dy / dx = 4 / [sqrt(1 - (2x)^2)]

= 4 / [sqrt(1 - 4x^2)]

here y = -2arccos(2x) we must use chain rule... it is so

y´= 4/sqrt(1- 4x^2) OK!