$y=\operatorname{arcsec}x$ can be defined in two ways. The first restricts the domain of $\sec y$ to $[0,\pi], y\neq\frac{\pi}{2}$. So the range of $y$ goes between $[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$ and the slope of the function is always positive. The derivative is
$$\frac{dy}{dx}=\frac{1}{|x|\sqrt{x^2-1}}$$
And when the domain of $\sec y$ function is defined as $[0,\frac{\pi}{2})\cup[\pi,\frac{3\pi}{2})$ the $\operatorname{arcsec}x$ has a negative slope for $x\leq-1$. That's why the derivative is
$$\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}$$
(It's just different in the absolute value). But what's the proof?
In my attempt to prove it, I'm using the theorem for derivate an inverse function, which is
$$\frac{d}{dx}[f^{-1}(x)]=\frac{1}{f'(f^{-1}(x))}$$
So, we have
$$\frac{d}{dx}[\operatorname{arcsec}(x)]=\frac{1}{\sec (\operatorname{arcsec}(x))\cdot\tan (\operatorname{arcsec}(x))}$$
At this point, I think is safe to say that $\sec (\operatorname{arcsec}(x))=x$ because this identity works for $|x|\geq1$, and that's the domain of $y$.
For the $\tan$, I'm using the right triangle diagram
Where the hypotenuse is $x$, the adjacent side is 1 (so $\sec y=x$) and the opposite side is $\sqrt{x^2-1}$. The $\tan y$ results in $\sqrt{x^2-1}/1=\sqrt{x^2-1}$. Substituting we get
$$\frac{d}{dx}=\frac{1}{x\sqrt{x^2-1}}$$
But I don't understand where does the absolute value come from.
Reading other questions and watching some videos, I realize that the absolute value comes from the identity $\tan^2x+1=\sec^2x$, but as you can see, this prove doesn't use that identity.
Best Answer
The right triangle with hypotenuse $x$ and legs $1$ and $\sqrt{x^2-1}$ is a useful mnemonic device, but strictly speaking it is only valid for $0 < y < \frac\pi2,$ because by definition the non-right angles in a right triangle are positive and acute.
When $\frac\pi2 < y \leq \pi,$ it turns out that actually $\tan y < 0.$ The correct formula in that case is $\tan y = -\sqrt{x^2-1},$ which you can find from the facts that $\tan y < 0$ and $$ \tan^2 y = \sec^2 y - 1. $$
So if you define the arc secant function on $(-\infty,-1] \cup [1,\infty)$ so that its range is $\left[0,\frac\pi2\right) \cup \left(\frac\pi2,\pi\right]$ then $$ \tan(\operatorname{arcsec}(x)) = \begin{cases} \sqrt{x^2-1} & x \geq 1, \\ -\sqrt{x^2-1} & x \leq 1. \end{cases} $$ On the other hand, you are correct when you say that $\sec (\operatorname{arcsec}(x) = x$ for all $x$ in the domain of the arc secant. When you use these equations to make substitutions in the formula $$ \frac{d}{dx}[\operatorname{arcsec}(x)] = \frac{1}{\sec(\operatorname{arcsec}(x)) \tan(\operatorname{arcsec}(x))}, $$ you get $$ \frac{d}{dx}[\operatorname{arcsec}(x)] = \begin{cases} \dfrac{1}{x \cdot \sqrt{x^2-1}} & x \geq 1, \\ \dfrac{1}{x \cdot (-\sqrt{x^2-1})} & x \leq 1. \end{cases} $$ Now observe that if $x \geq 1$ then $\lvert x\rvert = x$ and therefore $$\lvert x\rvert \cdot \sqrt{x^2-1} = x \cdot \sqrt{x^2-1},$$ whereas if $x \leq -1$ then $\lvert x\rvert = -x$ and therefore $$\lvert x\rvert \cdot \sqrt{x^2-1} = -x \cdot \sqrt{x^2-1} = x \cdot (-\sqrt{x^2-1}).$$ So both cases (positive and negative $x$) are correctly represented by $\lvert x\rvert \sqrt{x^2-1}.$
If we define the arc secant instead so that its range is $\left[0,\frac\pi2\right) \cup \left(\pi,\frac32\pi\right]$, then $\tan y > 0$ on all parts of the range and your derivation is completely correct.