For what values of a and b will [img=793x239]1803474d8c6aa22.png[/img]be differentiable for all values of x?
A: a=-1/2, b=1
B: a=1/2, b=1
C: a=-1/2, b=-1
D: a=1/2, b=-1
A: a=-1/2, b=1
B: a=1/2, b=1
C: a=-1/2, b=-1
D: a=1/2, b=-1
举一反三
- 求函数[img=148x49]17da6537a5eee98.png[/img]的导数； ( ) A: 1/(x^2*(2/x^2 + 1)) B: -1/(x^2*(2/x^2 + 1)) C: (x^2*(2/x^2 + 1)) D: -1/(x^2*(2/x^2 + 1))+2/x^2 + 1
- 积分[img=136x52]1803d6afd4e6f95.png[/img]的计算程序和结果是 A: clearsyms xy=1/x^2/sqrt(x^2-1)int(y,x,-2,-1)3^(1/2)/2 B: clearsyms xint(1/x^2/sqrt(x^2-1),x,-2,-1)3^(1/2)/2 C: clearsyms xint(1/x/sqrt(x^2-1),x,-2,-1)-pi/3 D: clearsyms xint(1/x/sqrt(x^2-1),x,-2,-1)3^(1/2)/2 E: clearsyms xint(1/x^2*sqrt(x^2-1),x,-2,-1)log(3^(1/2) + 2) - 3^(1/2)/2
- 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)
- 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)
- 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)