The derivative of ^ - 1 ( 1 + x ^ 2 - 1 x ) with respect to ^ - 1 (...

Let $x = tan\theta$
${y}_{1} = \tan^{- 1}\left(\dfrac{sec\theta - 1}{tan\theta}\right) = \tan^{- 1}\left(tan\dfrac{\theta}{2}\right) = \dfrac{\theta}{2} = \dfrac{1}{2}\tan^{- 1}x$
$x = sinϕ, {y}_{2} = \tan^{- 1}\left(\dfrac{2sinϕcosϕ}{\cos 2ϕ}\right) = \tan^{- 1} ( \tan 2ϕ ) = 2ϕ = 2\sin^{- 1}x$

$\dfrac{d{y}_{1}}{d{y}_{2}} = \dfrac{d{y}_{1} / dx}{d{y}_{2} / dx} = \dfrac{\dfrac{1}{\left(1 + {x}^{2}\right)} \cdot \dfrac{1}{2}}{2 \cdot \dfrac{1}{\sqrt{1 - {x}^{2}}}}$
$= \dfrac{\sqrt{1 - {x}^{2}}}{4\left(1 + {x}^{2}\right)} = \dfrac{\sqrt{1 - \dfrac{1}{4}}}{4\left(1 + \dfrac{1}{4}\right)} = \dfrac{\sqrt{3}}{10}$