1.

If \[y={{\tan }^{-1}}\left( \frac{x}{\sqrt{1-{{x}^{2}}}} \right)\], then \[\frac{dy}{dx}=\]      [MP PET 1999]

A.   \[-\frac{1}{\sqrt{1-{{x}^{2}}}}\]
B.  \[\frac{x}{\sqrt{1-{{x}^{2}}}}\]
C.  \[\frac{1}{\sqrt{1-{{x}^{2}}}}\]
D.   \[\frac{\sqrt{1-{{x}^{2}}}}{x}\]
Answer» D.   \[\frac{\sqrt{1-{{x}^{2}}}}{x}\]


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