Derivative of $\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}$ with respect to $\tan^{-1}x$
Method 1
$$
w=\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}\quad\&\quad z=\tan^{-1}x
$$
Put $\theta=\tan^{-1}x\implies\tan\theta=x$
$$
w=\tan^{-1}\frac{|\sec\theta|-1}{x}=\tan^{-1}\frac{|\sec\theta|-1}{\tan\theta}=\tan^{-1}\frac{\dfrac{1}{|\cos\theta|}-1}{\dfrac{\sin\theta}{\cos\theta}}=\tan^{-1}\frac{1-|\cos\theta|}{|\cos\theta|}.\frac{\cos\theta}{\sin\theta}\\
=\tan^{-1}\frac{1\mp\cos\theta}{\pm\cos\theta}.\frac{\cos\theta}{\sin\theta}=\pm\tan^{-1}\frac{1\mp\cos\theta}{\sin\theta}\\
w=\begin{cases}\tan^{-1}\dfrac{1-\cos\theta}{\sin\theta}=\tan^{-1}\tan(\theta/2)\quad;\quad\theta\in(-\pi/2,\pi/2)\\
-\tan^{-1}\dfrac{1+\cos\theta}{\sin\theta}=-\cot^{-1}\cot(\theta/2)\quad\&\quad\text{elsewhere}
\end{cases}\\
=\begin{cases}n\pi+\dfrac{\tan^{-1}x}{2}\quad;\quad\theta\in(-\pi/2,\pi/2)\\
-n\pi-\dfrac{\tan^{-1}x}{2}\quad\&\quad\text{elsewhere}
\end{cases}\\
\frac{dw}{dx}=\dfrac{\pm1}{2(1+x^2)}\\
\frac{dz}{dx}=\frac{1}{1+x^2}\implies\boxed{\frac{dw}{dz}=\pm\frac{1}{2}}
$$
Method 2
$$
\frac{dw}{dx}=\frac{1}{1+\Big(\frac{\sqrt{1+x^2}-1}{x}\Big)^2}.\frac{x.\dfrac{x}{\sqrt{1+x^2}}-(\sqrt{1+x^2}-1)}{x^2}\\
=\frac{x^2}{2(x^2+1-\sqrt{1+x^2})}.\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}}=\frac{1}{2(1+x^2)}\\
\frac{dz}{dx}=\frac{1}{1+x^2}\implies\boxed{\frac{dw}{dz}=\frac{1}{2}}
$$
Why do I seem to get slightly different solutions in methods 1 and 2 ?
Best Answer
In the first method for $\theta\in(-\pi/2,\pi/2)$ we have $|\cos\theta|=\cos\theta$ therefore only plus sign holds.
Second method seems more effective and clear to me.