​$$\text { The derivative of } \tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \text { with respect to } \tan ^{-1} x \text { is: }$$​​
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Correct option is B

​$$\text{We are given:} \\f(x) = \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right), \quad g(x) = \tan^{-1}x \\[10pt]\text{We need to compute:} \\\frac{d}{d(\tan^{-1}x)} \left[ \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right) \right] \\[10pt]{{{Use substitution:}}} \\[5pt]\text{Let } x = \tan\theta \Rightarrow \\f(x) = \tan^{-1} \left( \frac{\sqrt{1 + \tan^2\theta} - 1}{\tan\theta} \right) = \tan^{-1} \left( \frac{\sec\theta - 1}{\tan\theta} \right) \\[10pt]= \tan^{-1} \left( \frac{1 - \cos\theta}{\sin\theta} \right) = \tan^{-1} \left( \frac{2\sin^2 \frac{\theta}{2}}{2\sin\frac{\theta}{2} \cos\frac{\theta}{2}} \right)= \tan^{-1}(\tan \frac{\theta}{2}) \\[10pt]= \frac{\theta}{2} \quad \text{and since } \theta = \tan^{-1}x, \text{ we get:} \\f(x) = \frac{1}{2} \tan^{-1}x \\[10pt]\boxed{{\text{Now differentiate:}}} \\\frac{d}{d(\tan^{-1}x)} f(x) = \frac{d}{d(\tan^{-1}x)} \left( \frac{1}{2} \tan^{-1}x \right) = {\frac{1}{2}}$$​​