The derivative of tan - 1 1 + x 2 - 1 x with respect to tan - 1 2 x 1 - x 2 1 - 2 x 2 at x = 1 2 is :

Let $x=\tan \theta$
$y_1={\tan }^{-1}\left(\frac{\sec \theta -1}{\tan \theta }\right)={\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}=\frac{1}{2}{\tan }^{-1}x$
$x=\sin \varphi , y_2={\tan }^{-1}\left(\frac{2\sin \varphi \cos \varphi }{\cos 2\varphi }\right)={\tan }^{-1}\left(\tan 2\varphi \right)=2\varphi =2{\sin }^{-1}x$

$\frac{d{y}_1}{d{y}_2}=\frac{d{y}_1/dx}{d{y}_2/dx}=\frac{\frac{1}{\left(1+x^2\right)}·\frac{1}{2}}{2·\frac{1}{\sqrt{1-x^2}}}$
$=\frac{\sqrt{1-x^2}}{4\left(1+x^2\right)}=\frac{\sqrt{1-\frac{1}{4}}}{4\left(1+\frac{1}{4}\right)}=\frac{\sqrt{3}}{10}$