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If \(y=\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x}\), then \(y^{\prime}(1)=\)
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\(1 / 4\)
Hints: \(y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \quad\) Put \(x=\tan \theta\)
\(\begin{aligned} & =\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \\ & =\tan ^{-1}\left(\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)=\tan ^{-1} \tan \frac{\theta}{2} \\ & =\frac{\theta}{2}=\frac{1}{2} \cdot \tan ^{-1} x, y^{\prime}=\frac{1}{2\left(1+x^2\right)} \\ & y^{\prime}(1)=\frac{1}{2.2}=\frac{1}{4}\end{aligned}\)
\(\begin{aligned} & =\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \\ & =\tan ^{-1}\left(\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)=\tan ^{-1} \tan \frac{\theta}{2} \\ & =\frac{\theta}{2}=\frac{1}{2} \cdot \tan ^{-1} x, y^{\prime}=\frac{1}{2\left(1+x^2\right)} \\ & y^{\prime}(1)=\frac{1}{2.2}=\frac{1}{4}\end{aligned}\)
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