\[\frac{d}{dx}({{\sin }^{-1}}x)\] is equal to [RPET 1995]

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

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

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

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

\[\frac{d}{dx}({{\sin }^{-1}}x)\] is equal to

1.	\[\frac{d}{dx}({{\sin }^{-1}}x)\] is equal to [RPET 1995]
A.	\[\frac{1}{\sqrt{1-{{x}^{2}}}}\]
B.	\[-\frac{1}{\sqrt{1-{{x}^{2}}}}\]
C.	\[\frac{1}{\sqrt{1+{{x}^{2}}}}\]
D.	\[\frac{-1}{\sqrt{1+{{x}^{2}}}}\]
Answer» B. \[-\frac{1}{\sqrt{1-{{x}^{2}}}}\]

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\[\frac{d}{dx}({{\sin }^{-1}}x)\] is equal to [RPET 1995]

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