已知函数f(X)=负根号3sin平方x+sinxcosx，求函数f(X)的最小正周期

f(X)=负根号3sin平方x+sinxcosx
=根号3/2-根号3/2 cos2x+1/2sin2x
=sin(2x-π/3)+根号3/2
所以，周期为π

f(x) = -3^(1/2)[sin(x)]^2 + sin(x)cos(x)

= -3^(1/2)[1-cos(2x)]/2 + sin(2x)/2

= sin(π/3)cos(2x) + cos(π/3)sin(2x) - 3^(1/2)/2

= sin(2x + π/3) - 3^(1/2)/2

2x + 2π + π/3 = 2(x + π) + π/3,

所以，
函数f(x) = -3^(1/2)[sin(x)]^2 + sin(x)cos(x)的最小正周期为π。

另外，
y(x) = x^2 + 2x + 1,

y'(x) = 2x + 2 = 2(x+1)

y'(1) = 2(1+1) = 4.

所以，
y(x) = x^2 + 2x + 1 在 x = 1处的导数等于4.

1.f(x)=-根号3(1-cos2x)/2+1/2sin2x=sin(2x+π/3)-根号3/2
T=2π/2=π
2.
y'=2x+2,所以结果是4

f(x)=sin(2x+∏/3)-根号3/2
所以T=2∏/2=∏

2.1/4

f(x)=-√3/2(1-cos2x)+1/2sin2x=sin(2x+∏/3)-√3/2
T=2∏/2=∏