If \[y=\sec ({{\tan }^{-1}}x),\]then \[\frac{dy}{dx}\] is [DCE 2002; Kurukshetra CEE 2001]

\[\frac{-x}{\sqrt{1+{{x}^{2}}}}\]

\[\frac{x}{\sqrt{1+{{x}^{2}}}}\]

\[\frac{x}{\sqrt{1-{{x}^{2}}}}\]

If \[y=\sec ({{\tan }^{-1}}x),\]then \[\frac{dy}{dx}\] is [DCE 2002; K

1.	If \[y=\sec ({{\tan }^{-1}}x),\]then \[\frac{dy}{dx}\] is [DCE 2002; Kurukshetra CEE 2001]
A.	\[\frac{x}{\sqrt{1+{{x}^{2}}}}\]
B.	\[\frac{-x}{\sqrt{1+{{x}^{2}}}}\]
C.	\[\frac{x}{\sqrt{1-{{x}^{2}}}}\]
D.	None of these
Answer» B. \[\frac{-x}{\sqrt{1+{{x}^{2}}}}\]

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If \[y=\sec ({{\tan }^{-1}}x),\]then \[\frac{dy}{dx}\] is [DCE 2002; Kurukshetra CEE 2001]

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