A) \[\frac{1}{1+{{x}^{2}}}\]
B) \[\frac{1}{2(1+{{x}^{2}})}\]
C) \[\frac{{{x}^{2}}}{2\sqrt{1+{{x}^{2}}}(\sqrt{1+{{x}^{2}}}-1)}\]
D) \[\frac{2}{1+{{x}^{2}}}\]
Correct Answer: B
Solution :
Let \[y={{\tan }^{-1}}\frac{\sqrt{1+{{x}^{2}}}-1}{x}\] Put \[x=\tan \theta ,\] then \[y={{\tan }^{-1}}\left( \frac{\sqrt{1+{{\tan }^{2}}\theta }-1}{\tan \theta } \right)\] \[y={{\tan }^{-1}}\left( \frac{\sec \theta -1}{\tan \theta } \right)={{\tan }^{-1}}\left( \frac{1-\cos \theta }{\sin \theta } \right)\] \[y={{\tan }^{-1}}\left( \frac{2{{\sin }^{2}}\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}} \right)={{\tan }^{-1}}\tan \frac{\theta }{2}\] \[y=\frac{\theta }{2}=\frac{1}{2}{{\tan }^{-1}}x\],\[(\because \theta ={{\tan }^{-1}}x)\]. Hence \[\frac{dy}{dx}=\frac{1}{2(1+{{x}^{2}})}\].
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