What is \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\sin }^{ - 1}}\frac{{2{\rm{x}}}}{{1 + {\rm{\;}}{{\rm{x}}^2}}}} \right)\) equal to?

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{-1}{{1 + {{\rm{x}}^2}}}\)

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{1}{{1 + {{\rm{x}}^2}}}\)

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{-2}{{1 + {{\rm{x}}^2}}}\)

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{2}{{1 + {{\rm{x}}^2}}}\)

What is \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\sin }^{ - 1}}\frac{{

1.	What is \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\sin }^{ - 1}}\frac{{2{\rm{x}}}}{{1 + {\rm{\;}}{{\rm{x}}^2}}}} \right)\) equal to?
A.	\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{1}{{1 + {{\rm{x}}^2}}}\)
B.	\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{-2}{{1 + {{\rm{x}}^2}}}\)
C.	\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{2}{{1 + {{\rm{x}}^2}}}\)
D.	\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{-1}{{1 + {{\rm{x}}^2}}}\)
Answer» D. \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{-1}{{1 + {{\rm{x}}^2}}}\)