设(x)=dfrac (1-x)(1+x), 则(x)=dfrac (1-x)(1+x)

=dfrac (sqrt {1+x)-sqrt (1-x)}(sqrt {1+x)+sqrt (1-x)}

(9) =dfrac (sqrt {1+x)-sqrt (1-x)}(sqrt {1+x)+sqrt (1-x)} ;

[题目]-|||-6.设函数 (dfrac (1-x)(1+x))=x 则 f(x)= __

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

设(x)=dfrac (1)(1-x), 求(x)=dfrac (1)(1-x)和(x)=dfrac (1)(1-x)设, 求和

设y=dfrac(x{(1+x))^2}({(1-x))^3}，则y=A. $\dfrac{6+x-{x}^{2}}{x-{x}^{3}}$B. $\dfrac

计算lim _(xarrow 0)dfrac (ln (dfrac {1+x)(1-x))}((1+cos x)sin x)计算

14)设 (x)=arctan dfrac (1+x)(1-x), 整数 geqslant 0, 则 ^(2n+1)(0)= __

求下列函数的n阶导数:-|||-(1) =sqrt [m](1+x) ;-|||-(2) =dfrac (1-x)(1+x)

设函数f（x）=(1-x)/(1+x)，则f[f（x）]= ____ ．设函数f（x）=$\frac{1-x}{1+x}$，则f[f（x）]= ____ ．