Derivative of arcsec

[Euler]

I know that $\displaystyle \frac{d}{dx}(sec^{-1}x)=\frac{1}{|x|\sqrt{x^2-1}}$

But I do not know why the extra "x" exists outside of the square root.
I realize that the unit circle covers both a positive and negative quadrant, resulting in the absolute value symbol, but why the extra "x"?

 

[Unco]

Try deriving it.

arcsec(x) = y

x = sec(y)

Differentiate implicitly:

dy/dx = 1/(sec(y)*tan(y))

          /|
    x   /  |
      /    | sqrt(x^2-1)
    /      |
  /__y_____|
       1
    sec(y) = x
so tan(y)=sqrt(y^2-1)