举一反三
- 为什么当x趋于0时,(1+x^2)^1/3-1与1/3x^2是等价无穷小?
- 17e0b849d3a4a3b.jpg,计算[img=19x34]17e0ab14a855463.jpg[/img]的实验命令为（ ）. A: syms x; f=diff((1+sin(x)^2)/cos(x),1)f=2*sin(x) + (sin(x)*(sin(x)^2 + 1))/cos(x)^2 B: f=diff((1+sinx^2)/cosx,1)f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;f=diff((1+sinx^2)/cosx,1)f=2*sin(x) + (sin(x)*(sin(x)^2 + 1))/cos(x)^2
- limxx趋于负无穷（1-x）^（1/x）
- 求函数[img=107x38]17da6537b12a2e0.png[/img]的导数； ( ) A: 2*x*sin(1/x) - sin(1/x) B: 2xsin(1/x) - cos(1/x) C: 2*x*sin(1/x) - cos(1/x) D: 2*x*cos(1/x) - cos(1/x)
- x趋于零(1+ax^2)^1/3-1和cosx-1是等价无穷小求a的值.
内容
- 0
X趋于正无穷时 括号sin X 1 加cos影X 1 括号X方
- 1
17e0b849b7d64bd.jpg,计算[img=19x34]17e0ab14a855463.jpg[/img]实验命令为(). A: syms x;f=diff(asinsqrt(x))f=1/2/x^(1/2)/(1-x)^(1/2) B: f=diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2)
- 2
17da42840675a6d.jpg,计算[img=19x34]17da4275482315f.jpg[/img]实验命令为(). A: syms x;f=diff(asinsqrt(x))f=1/2/x^(1/2)/(1-x)^(1/2) B: f=diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2) C: syms x;diff(asin(sqrt(x)))f=1/2/x^(1/2)/(1-x)^(1/2)
- 3
$\int {{{x\cos x} \over {{{\sin }^3}x}}} dx = \left( {} \right)$ A: $ - {x \over {2{{\sin }^2}x}} - {1 \over 2}\tan x + C$ B: $ - {x \over {2{{\sin }^2}x}} - {1 \over 2}\cot x + C$ C: $ - {x \over {2{{\cos }^2}x}} - {1 \over 2}\cot x + C$ D: $ - {x \over {2{{\cos }^2}x}} - {1 \over 2}\tan x + C$
- 4
函数\(y = \sin {1 \over x}\)的导数为（ ）. A: \({1 \over { { x^2}}}\sin {1 \over x}\) B: \( - {1 \over { { x^2}}}\sin {1 \over x}\) C: \( - {1 \over { { x^2}}}\cos {1 \over x}\) D: \({1 \over { { x^2}}}\cos {1 \over x}\)