Definition of $\operatorname{arcsec}(x)$ My calculus book gives for the derivative of $\operatorname{arcsec}(x)$:
$$\frac{d}{dx} \sec^{-1}x=\frac{1}{|x|\sqrt{x^2-1}}$$
Then it continues to state that: 

"Some authors prefer to define $\sec^{-1}$ as the inverse of the restriction of $\sec(x)$ to the separated intervals $[0,\pi/2]$ and $[\pi,3\pi/2]$ because this prevents the absolute value from appearing in the formula for the derivative"

Question: what does this mean? What is: "the inverse of the restriction of $\sec(x)$ to the separated intervals $[0,\pi/2]$ and $[\pi,3\pi/2]$"?
 A: Since the secant function is not invertible, you have to restrict its domain in such a way that it takes on each value only once.
The most natural choice is to restrict it to the domain
$$
[0,\pi/2)\cup(\pi/2,\pi]
$$
so that it agrees with the restriction of the cosine normally used for defining the arccosine.
In this case the derivative of the arcsecant will be determined by the usual procedure: if $x\in(-\infty,-1)\cup(1,\infty)$ we have, by definition
$$\def\arcsec{\operatorname{arcsec}}
\sec\arcsec x=x
$$
so
$$
\sec'\arcsec x\arcsec' x=1
$$
Since $\sec y=1/\cos y$, we have
$$
\sec'y=\frac{\sin y}{\cos^2y}
$$
and $\sin\arcsec x\ge0$ because of how we chose the range of $\arcsec$; therefore $\sin\arcsec x=\sqrt{1-\cos^2\arcsec x}$ and
$$
\arcsec'x=\frac{\cos^2\arcsec x}{\sqrt{1-\cos^2\arcsec x}}=
\dfrac{1}{\dfrac{1}{|\cos\arcsec x|}\sqrt{\dfrac{1}{\cos^2\arcsec x}-1}}
=\frac{1}{|x|\sqrt{x^2-1}}
$$
If we instead restrict the secant to $[0,\pi/2)\cup(\pi,3\pi/2]$, we can notice that for $x\in(1,\infty)$ the computation is the same, but when $x\in(-\infty,-1)$, the sine will be negative and this compensates for the sign when $\cos\arcsec x$ is pulled into the square root, so the derivative will not have the absolute value also in this case.
Nothing really to worry about.
