Find Derivative of $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$? 
Question. If $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$  then $\frac{d}{dx}=?$

Answer: $\displaystyle\frac{dy}{dx}=-\frac{1}{2|x|\sqrt{x^2-1}}$
My 1st attempt- I followed the simple method and started by taking darivative of  tan inverse and the following with chain rule and i got my answer as ($-\frac{1}{2x\sqrt{x^2-1}}$), which is not same as the above correct answer. 2nd method is that you can substitute $x=\sec\left(\theta\right)$ and while solving in last step we will get $\sec^{-1}\left(\theta\right)$ whose derivative contains $\left|x\right|$, but still i searched and don't know why its derivative has $\left|x\right|$
Here's my attempt stepwise
$\displaystyle\frac{dy}{dx}=\frac{1}{1+\left(\sqrt{\frac{x+1}{x-1}}\right)^2}\cdot\frac{1}{2\sqrt{\frac{x+1}{x-1}}}\cdot\frac{\left(x-1\right)-\left(x+1\right)}{\left(x-1\right)^2}$
$\displaystyle=\frac{\left(x-1\right)}{\left(x-1\right)+\left(x+1\right)}\cdot\frac{1\sqrt{x-1}}{2\sqrt{x+1}}\cdot-\frac{2}{\left(x-1\right)^2}$
$\displaystyle=-\frac{1}{2x}\cdot\frac{\left(x-1\right)\sqrt{x-1}}{\left(x-1\right)^2}\cdot\frac{1}{\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x-1}\sqrt{x+1}}$
$\displaystyle=-\frac{1}{2x\sqrt{x^2-1}}$
Can you tell what i am doing wrong in my 1st attempt?
 A: $$\dfrac{d\arctan\sqrt{\dfrac{x+1}{x-1}}}{dx}=\dfrac1{1+\dfrac{x+1}{x-1}}\cdot\dfrac{d\sqrt{\dfrac{x+1}{x-1}}}{dx}$$
For $x-1\ne0,$
$\sqrt{\dfrac{x+1}{x-1}}=\sqrt{\dfrac{\sqrt{x^2-1}}{(x-1)^2}}=\dfrac{\sqrt{x^2-1}}{|x-1|}$
$F=\dfrac{d\sqrt{\dfrac{x+1}{x-1}}}{dx}=\dfrac{d\dfrac{\sqrt{x^2-1}}{|x-1|}}{dx}$
If $x-1>0,F=\dfrac x{(x-1)\sqrt{x^2-1}}-\dfrac{\sqrt{x^2-1}}{(x-1)^2}=\dfrac{x(x-1)-(x^2-1)}{(x-1)\sqrt{x^2-1}}=\dfrac1{\sqrt{x^2-1}}$
What if $x-1<0?$
A: You start right:
\begin{align}
\frac{dy}{dx}
&=\frac{1}{1+\left(\sqrt{\dfrac{x+1}{x-1}}\right)^2}
  \frac{1}{2\sqrt{\dfrac{x+1}{x-1}}}
  \frac{(x-1)-(x+1)}{(x-1)^2}\\
&=\frac{1}{2}\frac{x-1}{(x-1)+(x+1)}
  \sqrt{\dfrac{x-1}{x+1}}
\frac{-2}{(x-1)^2}
\end{align}
but then make a decisive error in splitting the square root in the middle and then use the wrong fact that $t=\sqrt{t^2}$, which only holds for $t\ge0$.
You can go on with
$$
=-\frac{1}{2x}\sqrt{\dfrac{x-1}{x+1}}\frac{1}{x-1}
$$
and now you have to split into the cases $x>1$ and $x<-1$ or observe that $x(x-1)=|x|\,|x-1|$ (when $|x|>1$) so you can write
$$
=-\frac{1}{2|x|}\sqrt{\dfrac{x-1}{x+1}}\frac{1}{|x-1|}
=-\frac{1}{2|x|}\sqrt{\dfrac{x-1}{x+1}\frac{1}{(x-1)^2}}
$$
and finish up.

You might try simplifying the expression, before plunging in the computations.
You have, by definition,
$$
\sqrt{\frac{x+1}{x-1}}=\tan y
$$
so that
$$
x+1=x\tan^2y-\tan^2y
$$
that yields
$$
x=\frac{1+\tan^2y}{\tan^2y-1}=\frac{1}{\cos^2y}\frac{\cos^2y}{-\cos2y}=-\frac{1}{\cos2y}
$$
Hence $\cos2y=-1/x$ and $y=\frac{1}{2}\arccos(-1/x)$. Thus
$$
\frac{dy}{dx}=-\frac{1}{2}\frac{-1}{\sqrt{1-\dfrac{1}{x^2}}}\frac{-1}{x^2}=
=-\frac{1}{2}\frac{-\sqrt{x^2}}{\sqrt{x^2-1}}\frac{-1}{x^2}=
-\frac{1}{2|x|\sqrt{x^2-1}}
$$
because
$$
\frac{\sqrt{x^2}}{x^2}=\frac{|x|}{|x|^2}=\frac{1}{|x|}
$$
A: Let $x = \sec \theta, \ 0 \le \theta \le \pi, \ \theta \ne \frac{\pi}{2}$
Then $\sqrt {\dfrac {x+1}{x-1}} =\sqrt {\dfrac {\sec \theta+1}{\sec \theta-1}} = \sqrt {\dfrac {1+\cos \theta}{1-\cos \theta}} = \tan \dfrac{\theta}{2} $ (positive root as $\dfrac{\theta}{2}$ is in the 1st Quadrant.
$\therefore y = \tan^{-1} \tan \dfrac{\theta}{2} = \dfrac{\theta}{2}$
Hence $\dfrac{dy}{dx} = \dfrac{1}{2 \sec \theta \tan \theta}= \dfrac{1}{2|x| \sqrt{x^2-1}}$ (as $\sec \theta \tan \theta \ge 0$ in the first two quadrants)
A: Your calculations are likely to be correct when $x>1$. When $x <-1$ write $\sqrt {\frac {x+1} {x-1}}$ as  $\sqrt {\frac {-x-1} {-x+1}}=\frac {\sqrt {-x-1}}{\sqrt {-x+1}}$ and now compute the derivative as before. You will get the correct answer now. 
A: By the chain rule,
$$\left(\arctan\sqrt{\frac{x+1}{x-1}}\right)'=-\frac2{(x-1)^22\sqrt{\dfrac{x+1}{x-1}}\left(1+\dfrac{x+1}{x-1}\right)}=-\frac1{2x(x-1)\sqrt{\dfrac{x+1}{x-1}}}
\\=-\frac1{2x\text{ sgn}(x-1)\sqrt{x^2-1}}.$$
The claim follows from
$$\text{ sgn}(x-1)=\text{ sgn}(x)$$ (when $|x|>1$).
A: Notice that with
$$f(x):=\frac{1+x}{1-x}$$ we have
$$f(-x)=\frac1{f(x)}$$ so that
$$\arctan\sqrt{f(-x)}=\text{arccot}\sqrt{f(x)}=\frac\pi2-\arctan\sqrt{f(x)}.$$
This shows that to an additive constant, the initial function is odd and its derivative must be even.

