若 (x)=ln (x+sqrt (1+{x)^2}) 则 '(x)=dfrac (1)(sqrt {1+{x)^2}}-|||-5-|||-○ C.若 (x)=x(x-1)(x-2)... (x-100), 则 '(0)=1001!-|||-D. 若 (x)=(e)^dfrac (1{sqrt {x-1)}}, 则 '(0)=dfrac (e)(2)

(2) (x)=ln (x+sqrt (1+{x)^2});

(11) =ln (x+sqrt (1+{x)^2}) ;

求极限 lim _(xarrow +infty )dfrac (ln (x+sqrt {{x)^2+1})-ln (x+sqrt ({x)^2-1})}( l

[题目]求导数: =ln (x+sqrt (1+{x)^2})

求下列极限-|||-lim _(xarrow 0)[ dfrac (1)(ln (x+sqrt {1+{x)^2})}-dfrac (1)(ln (1+x))]

求lim _(xarrow infty )dfrac (ln (x+sqrt {{x)^2+1)}-ln (x+sqrt ({x)^2-1})}({({e)^d

(14)若函数 (x)=dfrac (1+sqrt {1+{x)^2}}(x), 则 (dfrac (1)(x))= __ (2019年理]

求极限 lim _(xarrow +infty )((x+sqrt {1+{x)^2})}^dfrac (1{x)}

(19) =ln ((e)^x+sqrt (1+{e)^2x}).

设 gt 0, 则 =(int )_(-a)^asqrt ({a)^2-(x)^2}ln dfrac (x+sqrt {1+{x)^2}}(3)dx= __