Differentiating $\sec^{-1}(\sqrt{1+x^2})$ for $x \in (-1, 1)$ Problem
Differentiate the following inverse trig. function for $x\in (-1, 1)$:
$$f(x) = \sec^{-1}\left(\sqrt{1+x^2}\right)$$
Attempting to solve
Putting $x = \tan(\theta)$ so that $\theta \in \left(\dfrac{-\pi}{4}, \dfrac{\pi}{4}\right)$
$$\begin{align}\implies f(x) & = \sec^{-1}\left(\sqrt{1+\tan^2\theta}\right)\\&=\sec^{-1}\left(\sqrt{\sec^2\theta}\right)\\&=\sec^{-1}\left(|\sec\theta|\right)\tag{$*$}\end{align}$$
This is equal to $\theta$ only when $\theta \in \left(0, \pi\right)$. I'm not getting it... what to do when $\theta$ doesn't lie in that interval?
I'm having great difficulties in defining $\sec^{-1}\left(\sec x\right)$ for different intervals.

Edit
By some sort of graphing, I obtained the solution as,
$$f'(x) = \begin{cases}\dfrac{1}{1 + x^2},\quad {\rm if\: } x\in(0, 1)\\\\\dfrac{-1}{1+x^2},\quad{\rm if\ }x \in(-1,0)\end{cases}$$
which implies that the above expression $(*)$ can be defined as,
$$\sec^{-1}\left(|\sec\theta|\right) = \begin{cases} \theta + C_1, \quad {\rm if \:} \theta \in \left(0, \dfrac{\pi}{4}\right)\\\\-\theta 
 + C_2, \quad {\rm if \:} \theta \in\left(-\dfrac{\pi}{4},0\right) \end{cases}$$
---though I've obtained the derivative, but still, I'm not understanding it :(
 A: Let $\phi = \sec^{-1}(\sec \theta)$. Then $\sec(\phi)=\sec\big(\sec^{-1}(\sec \theta)\big)= \sec(\theta)$, because $\sec(\sec^{-1} x)=x$ is always true (although, as you point out, $\sec^{-1}(\sec x)$is not always equal to $x$). This means that $\phi = \theta + 2n\pi$, for some $n$, or $\phi = 2n\pi - \theta$, for some $n$. You choose the multiple of $2\pi$ and whether to add or subtract $\theta$ so that the result lies in $[0,\pi]$. For example, when $x>0$, $\theta=\tan^{-1} x$ is in $(0, \tfrac{1}{2}\pi)$, so $\phi = \sec^{-1}(\sec \theta) = +\tan^{-1} x$, but when $x<0$, $\theta=\tan^{-1} x$ is in $(-\tfrac{1}{2}\pi, 0)$, so $\phi = \sec^{-1}(\sec \theta) = -\tan^{-1} x$.
A: The easiest way is through implicit differentiation.
Suppose that $y=\sec^{-1}{\sqrt{1+x^{2}}}$
This implies that $\sec{y}=\sqrt{1+x^{2}}$
Taking the derivative with respect to $x$, you get:
$\sec{y}\tan{y}\frac{dy}{dx}=\frac{x}{\sqrt{1+x^{2}}}$, by the product rule.
Isolating $\frac{dy}{dx}$, you get
$\frac{dy}{dx}=\frac{1}{\sec{y}\tan{y}}\frac{x}{\sqrt{1+x^2}}$
Note that $\tan{y}=\sqrt{\sec^{2}{y}-1}=\sqrt{x^2}=x$
So, $\frac{dy}{dx}=\frac{x}{x(x^2+1)}=\frac{1}{x^2+1}$, for $x\in[0,1]$.
You can also note that, $\frac{dy}{dx}=\frac{-1}{x^2+1}$, for $x\in[-1,0)$.
If you want to combine this into one formula, with some thought, you can see that, if you want to find the sign of the input, you can take $\frac{x}{|x|}$, giving a final solution of $\frac{dy}{dx}=\frac{x}{|x|}\frac{1}{x^2+1}$, for $x\in[-1,1]$
A: I get $${dy\over dx}={1\over1+x^2}$$ for the derivative of $\sec^{-1}(\sqrt{1+x^2})$. I'll put the steps here.
$$y=\sec^{-1}(\sqrt{1+x^2})\\
\sec y=\sqrt{1+x^2}\\
{\sin y\over\cos^2 y}{dy\over dx}={x\over\sqrt{1+x^2}}
\\
\sqrt{x^2\over1+x^2}(1+x^2){dy\over dx}={x\over\sqrt{1+x^2}}\\
{dy\over dx}={x\over|x|}{1\over1+x^2}
$$
A: 

Differentiate the following inverse trig. function for $x\in (-1, 1)$:
$$f(x) = \sec^{-1}\left(\sqrt{1+x^2}\right)$$

Putting $x = \tan(\theta)$ so that $\theta \in \left(\dfrac{-\pi}{4}, \dfrac{\pi}{4}\right)$

Quibble: you mean "such that" rather than "so that".
Okay, continuing with your substitution method, but choosing a more suitable restriction:
Let $\theta \in [0,\pi){\setminus}\left\{\frac\pi2\right\}$ and $x = \tan\theta.$ Then

$$\begin{align} f(x) & = \sec^{-1}\left(|\sec\theta|\right)\tag{$*$}\end{align}$$

$$=\begin{cases}\theta,&{\rm if\: } \theta\in[0, \frac\pi2)\\\\\pi-\theta,&{\rm if\ }\theta \in(\frac\pi2,\pi)\end{cases}.$$

So, $$f'(x)=\begin{cases}\dfrac{\mathrm d\theta}{\mathrm dx},&{\rm if\: } \theta\in[0, \frac\pi2)\\\\-\dfrac{\mathrm d\theta}{\mathrm dx},&{\rm if\ }\theta \in(\frac\pi2,\pi)\end{cases},$$ giving

$$f'(x) = \begin{cases}\dfrac{1}{1 + x^2},\quad {\rm if\: } x\in(0, 1)\\\\\dfrac{-1}{1+x^2},\quad{\rm if\ }x \in(-1,0)\end{cases},$$

noting that $f$ has no derivative at $x=0$ since

*

*$\operatorname{arcsec}$ has no derivative at $\pm1,$

*and $\sqrt{1+x^2}=\pm1\iff x=0.$
A: We can recognize six $ \theta$ related inverse function  in six ways by drawing the originating triangle in trigonometry:
$$\theta=\sec^{-1}\sqrt{1+x^2} = \tan^{-1}x =\cos^{-1}\frac{1}{\sqrt{1+x^2} }=\sin^{-1}\frac{x}{\sqrt{1+x^2} } \text{ &..}$$
So we then can recognize to have a directly more common derivative of arctan
$$ \frac{d\theta}{dx}=\pm\frac{1}{1+x^2}; $$
where the radical sign in $\sec^{-1}\sqrt{1+x^2} $ admits negative sign also .

