A) \[\frac{1}{2}\]
B) \[-\frac{1}{2}\]
C) 1
D) None of these
Correct Answer: A
Solution :
Let \[{{y}_{1}}={{\tan }^{-1}}\left( \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right)\] and \[{{y}_{2}}={{\tan }^{-1}}x\] Now \[\frac{d{{y}_{1}}}{dx}=\frac{d}{dx}\left[ {{\tan }^{-1}}\tan \frac{\theta }{2} \right],\] [By putting \[x=\tan \theta ]\] Þ\[\frac{d{{y}_{1}}}{dx}=\frac{d}{dx}\left[ {{\tan }^{-1}}\tan \frac{\theta }{2} \right]=\frac{1}{2(1+{{x}^{2}})}\]& \[\frac{d{{y}_{2}}}{dx}=\frac{1}{1+{{x}^{2}}}\] Hence \[\frac{d{{y}_{1}}}{d{{y}_{2}}}=\frac{1}{2}\].
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