If \[u={{\tan }^{-1}}\left\{ \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right\}\] and \[v=2{{\tan }^{-1}}x\], then \[\frac{du}{dv}\] is equal to [RPET 1997]

If \[u={{\tan }^{-1}}\left\{ \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right\}

1.	If \[u={{\tan }^{-1}}\left\{ \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right\}\] and \[v=2{{\tan }^{-1}}x\], then \[\frac{du}{dv}\] is equal to [RPET 1997]
A.	4
B.	1
C.	¼
D.	?1/4
Answer» D. ?1/4

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If \[u={{\tan }^{-1}}\left\{ \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right\}\] and \[v=2{{\tan }^{-1}}x\], then \[\frac{du}{dv}\] is equal to [RPET 1997]

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