1.

If \[y={{\sin }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\], then \[\frac{dy}{dx}\] equals [EAMCET 1991; RPET 1996]

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


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