1、limln(1+x+2x^2)+ln(1-x+x^2)/secx-cosx
举一反三
- 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
- 已知\( y = \ln (1 + {x^2}) \),则\( y' \)为( ). A: \( { { 2x} \over {1 + {x^2}}} \) B: \( {x \over {1 + {x^2}}} \) C: \( {1 \over {1 + {x^2}}} \) D: \( { { {x^2}} \over {1 + {x^2}}} \)
- 已知\( y = {x^x} \),则\( y' \)为( ). A: \( {x^x} \) B: \( {x \over {1 + {x^2}}} \) C: \( {1 \over {1 + {x^2}}} \) D: \( {x^x}(1 + \ln x) \)
- 下列函数相等的是( )。 A: \( f(x) = \ln {x^2},g(x) = 2\ln x \) B: \( f(x) = x,g(x) = \sqrt { { x^2}} \) C: \( f(x) = \sqrt { { x^2}} ,g(x) = \left| x \right| \) D: \( f(x) = { { {x^2} - 1} \over {x - 1}},g(x) = x + 1 \)
- \( \int {({1 \over x} - {2 \over {\sqrt {1 - {x^2}} }})dx} = \)( ) A: \( \ln \left| x \right| + 2\arcsin x + C \) B: \( \ln \left| x \right| - 2\arcsin x + C \) C: \(- \ln \left| x \right| - 2\arcsin x + C \) D: \(- \ln \left| x \right| +2\arcsin x + C \)