例10]证明 (arcsin x)'=dfrac (1)(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}}

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

(7) int dfrac ({10)^2arcsin x}(sqrt {1-{x)^2}}dx;

(6) int dfrac ({10)^2arcsin x}(sqrt {1-{x)^2}}dx

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

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

(5)下列函数的导数计算正确的有 () .-|||-① (arcsin x)=-dfrac (1)(sqrt {1-{x)^2}}; ② ((log )_(a)

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

证明等式 arctan x=arcsin dfrac (x)(sqrt {1+{x)^2}} , in (-infty ,+infty