2019-2020学年北师大版选修1-1 基本初等函数和导数运算法则 教案
2019-2020学年北师大版选修1-1   基本初等函数和导数运算法则   教案第3页

(12)作业布置:教科书P13探究二;P18A组4(1)-(5),6,7 练习与测试:

1. 求下列函数的导数:(1) (2) (3) y = tanx (4)

2.求函数的导数.

(1)y=2x3+3x2-5x+4 (2)y=sinx-x+1 (3)y=(3x2+1)(2-x) (4)y=(1+x2)cosx

3.填空:

(1)[(3x2+1)( 4x2-3)]′=( )(4x2-3)+(3x2+1)( )

(2)(x3sinx)′=( )x2sinx+x3( )

4.判断下列求导是否正确,如果不正确,加以改正.

[(3+x2)(2-x3)]′=2x(2-x3)+3x2·(3+x2)

5.y=3x2+xcosx,求导数y′.

6.y=5x10sinx-2cosx-9,求y′.

参考答案:

1.(1)y′′;

(2)y′′;

(3)y′= (tanx)′=()′;

(4)y′′=.

2.(1)(2x3+3x2-5x+4)′=(2x3)′+(3x2)′-(5x)′+4′=2·3x2+3·2x-5=6x2+6x-5

(2)y′=(sinx-x+1)′=(sinx)′-x′+1′=cosx-1

(3)y′=[(3x2+1)(2-x)]′=(3x2+1)′(2-x)+(3x2+1)(2-x)′

  =3·2x(2-x)+(3x2+1)(-1)=-9x2+12x-1

(4)y′=[(1+x2)cosx]′=(1+x2)′cosx+(1+x2)(cosx)′

  =2xcosx+(1+x2)(-sinx)=2xcosx-(1+x2)sinx

3.(1)[(3x2+1)(4x2-3)]′=(3x2+1)′(4x2-3)+(3x2+1)(4x2-3)′

  =3·2x(4x2-3)+(3x2+1)(4·2x)=(6x)(4x2-3)+(3x2+1)(8x)

(2) (x3sinx)′=(x3)′sinx+x3(sinx)′=(3)x2sinx+x2(cosx)

4.不正确.[(3+x)2(2-x3)]′=(3+x2)′(2-x3)+(3+x2)(2-x3)′

=2x(2-x3)+(3+x2)(-3x2)=2x(2-x3)-3x2(3+x2)

5.y′=(3x2+xcosx)′=(3x2)′+(xcosx)′

=3·2x+x′cosx+x(cosx)′=6x+cosx+xsinx

6.y′=(5x10sinx-2cosx-9)′=(5x10sinx)′-(2cosx)′-9′

  =5·10x9sinx+5x10cosx-(·cosx-2sinx)

  =50x9sinx+5x10cosx-cosx+2sinx

  =(50x9+2)sinx+(5x10-)cosx