If \({\rm{y}} = {\cos ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right)\), then \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) is equal to

\(- \frac{2}{{1 + {{\rm{x}}^2}}}\) for all |x| > 1

\(- \frac{2}{{1 + {{\rm{x}}^2}}}\) for all |x| < 1

\(\frac{2}{{1 + {{\rm{x}}^2}}}\) for all |x| < 1

If \({\rm{y}} = {\cos ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}

1.	If \({\rm{y}} = {\cos ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right)\), then \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) is equal to
A.	\(- \frac{2}{{1 + {{\rm{x}}^2}}}\) for all \|x\| < 1
B.	\(- \frac{2}{{1 + {{\rm{x}}^2}}}\) for all \|x\| > 1
C.	\(\frac{2}{{1 + {{\rm{x}}^2}}}\) for all \|x\| < 1
D.	None of the above
Answer» B. \(- \frac{2}{{1 + {{\rm{x}}^2}}}\) for all \|x\| > 1

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If \({\rm{y}} = {\cos ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right)\), then \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) is equal to

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