=dfrac (x)(sqrt {1-{x)^2}}，则=dfrac (x)(sqrt {1-{x)^2}}=_________.

设=dfrac (arcsin x)(sqrt {1-{x)^2}}（1）证明：=dfrac (arcsin x)(sqrt {1-{x)^2}}（2）求=df

=dfrac (arcsin x)(x)+dfrac (1)(2)ln dfrac (1-sqrt {1-{x)^2}}(1+sqrt {1-{x)^2}}

设=ln sqrt (dfrac {1-x)(1-{x)^2}}则 dy|=ln sqrt (dfrac {1-x)(1-{x)^2}}设则dy|

(6) () =dfrac (1)(sqrt {1-{x)^2}} int dfrac (1)(sqrt {1-{x)^2}}dx=() .

.int dfrac (sqrt {1+{x)^2}+sqrt (1-{x)^2}}(sqrt {1-{x)^4}}dx.

(4) =dfrac (x)(sqrt {1-{x)^2}}

已知f(x)=(2x)/(sqrt (1-{x)^2)}，则(df(sqrt (1-{x)^2)})/(dx)=(,,,,,)A、-2；B、-dfrac(2x)

函数(x)=dfrac (ln |x|)(sqrt {1-{x)^2}}的定义域是 A(x)=dfrac (ln |x|)(sqrt {1-{x)^2}} B

、证明:当 -1lt xlt 0 时, arcsin sqrt (1-{x)^2}-arctan dfrac (x)(sqrt {1-{x)^2}}=dfrac

函数 =arcsin sqrt (1-{x)^2}+dfrac (1)(sqrt {1-{x)^2}} 的定义域为 __ 。