`f(x) = arcsec(2x)` Find the derivative of the function

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))`

...

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))`