函数f(x)=sinx在x=π3处的切线方程是( ) A.y−12=32(x−π3) B.y−32=12(x−π3) C.y−32=32(x−π3) D.y−12=12(x−π3)
问题描述:
函数f(x)=sinx在x=
处的切线方程是( )π 3
A. y−
=1 2
(x−
3
2
)π 3
B. y−
=
3
2
(x−1 2
)π 3
C. y−
=
3
2
(x−
3
2
)π 3
D. y−
=1 2
(x−1 2
) π 3
答
求导函数可得y′=cosx
∴x=
时,y′=π 3
,y=1 2
3
2
∴所求切线方程为y−
=
3
2
(x−1 2
)π 3
故选B.