Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 42

The given function: f(x) =arcsec(2x) is in a form of an inverse trigonometric function.
For the derivative formula of an inverse secant function, we follow:
d/(dx)(arcsec(u))=((du)/(dx))/(|u|sqrt(u^2-1))
To be able to apply the formula, we let u =2x then u^2 = (2x)^2=4x^2 and
(du)/(dx) = 2 .
It follows that f(x) =arcsec(2x) will have a derivative of:
f'(x) = 2/(|2x|sqrt((2x)^2-1))
f'(x) = 2/(|2x|sqrt(4x^2-1))
Cancel out common factor 2 from top and bottom:
f'(x) = 1/(|x|sqrt(4x^2-1))

This can also be written as : f'(x)= 1/(sqrt(x^2)sqrt(4x^2-1)) since
|x| = sqrt(x^2)
Then applying the radical property: sqrt(a)*sqrt(b)= sqrt(a*b) at the bottom, we get:
f'(x) = 1/sqrt(x^2*(4x^2-1))
f'(x) = 1/sqrt(4x^4 -x^2)

The derivative of the function f(x) =arcsec(x) can be :
f'(x)= 1/(|x|sqrt(4x^2-1))
or f'(x)= 1/(sqrt(x^2)sqrt(4x^2-1))
or f'(x)= 1/(sqrt(4x^4-x^2))