The derivative of tan−1⁡1+x2−1x with respect to tan−1⁡2x1−x21−2x2 at x=12

y=tan−1⁡1+x2−1x Put x=tan⁡α ⇒y=tan−1⁡sec⁡α−1tan⁡α⇒y=tan−1⁡1−cos⁡αsin⁡α=tan−1⁡tan⁡α2=α2=12tan−1⁡xLet z=tan−1⁡2x1−x21−2x2 Put x=sin⁡ϕ ⇒z=tan−1⁡sin⁡2ϕcos⁡2ϕ=2ϕ=2sin−1⁡xdydz=dydxdzdx=12⋅11+x22⋅11−x2=1−x241+x2 dydzx=12 =1−1441+14=325=310

The derivative of tan−1⁡1+x2−1x with respect to tan−1⁡2x1−x21−2x2 at x=12

y=tan−1⁡1+x2−1x Put x=tan⁡α ⇒y=tan−1⁡sec⁡α−1tan⁡α⇒y=tan−1⁡1−cos⁡αsin⁡α=tan−1⁡tan⁡α2=α2=12tan−1⁡xLet z=tan−1⁡2x1−x21−2x2 Put x=sin⁡ϕ ⇒z=tan−1⁡sin⁡2ϕcos⁡2ϕ=2ϕ=2sin−1⁡xdydz=dydxdzdx=12⋅11+x22⋅11−x2=1−x241+x2 dydzx=12 =1−1441+14=325=310

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The derivative of $\tan^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x})$ with respect to $\tan^{- 1} (\frac{2 x \sqrt{1 - x^{2}}}{1 - 2 x^{2}})$ at $x = \frac{1}{2}$

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$\frac{2 \sqrt{3}}{5}$

$\frac{\sqrt{3}}{10}$

$\frac{2 \sqrt{3}}{3}$

$\frac{\sqrt{3}}{2}$

answer is B.

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Detailed Solution

$y = \tan^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x}) Put x = \tan α \Rightarrow y = \tan^{- 1} (\frac{\sec α - 1}{\tan α})$

$\begin{array}{l} \Rightarrow y = \tan^{- 1} (\frac{1 - \cos α}{\sin α}) = \tan^{- 1} (\tan \frac{α}{2}) = \frac{α}{2} = \frac{1}{2} \tan^{- 1} x \\ Let z = \tan^{- 1} (\frac{2 x \sqrt{1 - x^{2}}}{1 - 2 x^{2}}) Put x = \sin ϕ \Rightarrow z = \tan^{- 1} (\frac{\sin 2 ϕ}{\cos 2 ϕ}) = 2 ϕ = 2 \sin^{- 1} x \\ \frac{dy}{dz} = \frac{\frac{dy}{dx}}{\frac{dz}{dx}} = \frac{\frac{1}{2} \cdot \frac{1}{1 + x^{2}}}{2 \cdot \frac{1}{\sqrt{1 - x^{2}}}} = \frac{\sqrt{1 - x^{2}}}{4 (1 + x^{2})} {(\frac{dy}{dz})}_{x = \frac{1}{2}} = \frac{\sqrt{1 - \frac{1}{4}}}{4 (1 + \frac{1}{4})} = \frac{\frac{\sqrt{3}}{2}}{5} = \frac{\sqrt{3}}{10} \end{array}$