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

Let $x=\mathrm{tan}\theta$x=tanθ

$${y}_{1}={\mathrm{tan}}^{-1}(\frac{\mathrm{sec}\theta -1}{\mathrm{tan}\theta })={\mathrm{tan}}^{-1}(\mathrm{tan}\frac{\theta }{2})=\frac{\theta }{2}=\frac{1}{2}{\mathrm{tan}}^{-1}x$$y1​=tan−1(tanθsecθ−1​)=tan−1(tan2θ​)=2θ​=21​tan−1x

$$x=\mathrm{sin}\phi ,{y}_{2}={\mathrm{tan}}^{-1}(\frac{2\mathrm{sin}\phi \mathrm{cos}\phi }{\mathrm{cos}2\phi })={\mathrm{tan}}^{-1}(\mathrm{tan}2\phi )=2\phi =2{\mathrm{sin}}^{-1}x$$x=sinϕ,y2​=tan−1(cos2ϕ2sinϕcosϕ​)=tan−1(tan2ϕ)=2ϕ=2sin−1x

$$\frac{d{y}_{1}}{d{y}_{2}}=\frac{d{y}_{1}/dx}{d{y}_{2}/dx}=\frac{\frac{1}{(1+{x}^{2})}\cdot \frac{1}{2}}{2\cdot \frac{1}{\sqrt{1-{x}^{2}}}}$$dy2​dy1​​=dy2​/dxdy1​/dx​=2⋅1−x2​1​(1+x2)1​⋅21​​

$$=\frac{\sqrt{1-{x}^{2}}}{4(1+{x}^{2})}=\frac{\sqrt{1-\frac{1}{4}}}{4(1+\frac{1}{4})}=\frac{\sqrt{3}}{10}$$=4(1+x2)1−x2​​=4(1+41​)1−41​​​=103​​