(tan x)^prime=1/(cos x)^2=(sec x)^2=( )

题型说明:共15题,每题2分。36.(tan x)^prime=1/(cos x)^2=(sec x)^2=( )题型说明:共15题,每题2分。36.(2.0分

判断题：tan x)=(sec )^2xtan x)=(sec )^2x 正确 tan x)=(sec )^2x 错误判断题：正确错误

lim _(xarrow 0)dfrac (ln (1+{x)^2)}(sec x-cos x)..

(5)lim_(xto(pi)/(2))(1+cos x)^2sec x(5)$\lim_{x\to\frac{\pi}{2}}(1+\cos x)^{2\se

11 lim_(xto0)(ln(1+x+x^2)+ln(1-x+x^2))/(sec x-cos x)=_.11 $\lim_{x\to0}\frac{\ln

【11】lim_(xto0)(ln(1+x+x^2)+ln(1-x+x^2))/(sec x-cos x)=【11】$\lim_{x\to0}\frac{\ln

10) lim _(xarrow 0)dfrac (ln (1+{x)^2)}(sec x-cos x)

lim _(xarrow 0)dfrac ({(1+2{x)^2)}^dfrac (3{2)}-1}(tan 2x(cos sqrt {x)-1)}。。

lim _(xarrow dfrac {pi )(2)}((1+cos x))^3sec x;

设函数 (x,y)=1-dfrac (cos sqrt {{x)^2+(y)^2}}(tan ({x)^2+(y)^2)} ,则当定设函数 (x,y)=1-df