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

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

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

lim _(xarrow 0)dfrac (sqrt {1+x)+sqrt (1-x)-2}(sqrt {1+{x)^2}-1}

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

求下列函数的导数:-|||-(1) =arcsin (sin x);-|||-(2) =arctan dfrac (1+x)(1-x);-|||-(3) =ln

求下列函数的导数:-|||-(1) =arcsin (sin x);-|||-(2) =arctan dfrac (1+x)(1-x);-|||-(3) =ln

设 lim _(xarrow 0)dfrac (ln (1+x)-(ax+b{x)^2)}({x)^2}=2, 则设 lim _(xarrow 0)dfrac

(int )_(0)^1arctan dfrac (x)(2)dx=()(int )_(0)^1arctan dfrac (x)(2)dx=()(int )_(

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

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