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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 $[\pi,3\pi/2)$"?

user182601
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GambitSquared
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  • Usually ones defines $\sec^{-1}$ be restricting $\sec$ to the domain $[0,\pi] \setminus {\pi/2}$ and hence it has positive and negative values, that is why the absolute value on the derivative – Alonso Delfín Feb 24 '16 at 22:54
  • Do you know what $[0/\pi/2]$ means and what $[\pi,2\pi/2]$ means? Notice that the top of the lower interval and the bottom of the upper interval among these two has a gap between them --- and interval of positive length. That is what it means to call them separated. Do you recall the definitions of $\arcsin$ and $\arccos$? The sine and cosine functions are not one-to-one and so do not have inverse functions, but if you restrict the sine function to the interval $[-\pi/2,\pi/2]$ then that has an inverse. I wonder if what you need to do is review some basic trigonometry. $\qquad$ – Michael Hardy Feb 24 '16 at 23:35

1 Answers1

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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.

egreg
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