函数\(y = \sqrt {1{\rm{ - }}x} \)的导数为（ ）. A: \({\rm{ - }}{1 \over {2\sqrt {1{\rm{ - }}x} }}\) B: \({1 \over {2\sqrt {1{\rm{ - }}x} }}\) C: \({1 \over {\sqrt {1{\rm{ - }}x} }}\) D: \( - {1 \over {\sqrt {1{\rm{ - }}x} }}\)

函数\(y = {\left( {\arcsin x} \right)^2}\)的导数为（ ）. A: \(2\arcsin x{1 \over {\sqrt {1 - {x^2}} }}\) B: \( - 2\arcsin x{1 \over {\sqrt {1 - {x^2}} }}\) C: \(2\arcsin x{1 \over {\sqrt {1 + {x^2}} }}\) D: \( - 2\arcsin x{1 \over {\sqrt {1 + {x^2}} }}\)

求函数$y = \root 3 \of {x + \sqrt x } $的导数$y' = $( ) A: ${{1 + 2\sqrt x } \over {\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ B: $ {{1 + 2\sqrt x } \over {6\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ C: $ {{1 + 2\sqrt x } \over {6\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ D: $ {{1 + 2\sqrt x } \over {\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$

\( d(\arcsin x) \)=（ ）. A: \( {1 \over {\sqrt {1 - {x^2}} }}dx \) B: \( - {1 \over {\sqrt {1 - {x^2}} }}dx \) C: 0 D: 1

设幂级数\(\sum\limits_{n = 0}^\infty { { a_n}} {x^n}\)与\(\sum\limits_{n = 1}^\infty { { b_n}{x^n}} \)的收敛半径分别为\( { { \sqrt 5 } \over 3}\)与\({1 \over 3}\)，则幂级数\(\sum\limits_{n = 1}^\infty { { {a_n^2} \over {b_n^2}}} {x^n}\)的收敛半径为( )。 A: 5 B: \( { { \sqrt 5 } \over 3}\) C: \({1 \over 3}\) D: \({1 \over 5}\)