?f?(?x)?limf(?x??x)?f(?x)f(x??x)?f(x)＝lim?x?0?x?0?x?x
f(x??x)?f(x)＝-lim?－f?(x)??x?0??x ?f?(x)为奇函数.
若
f(x)为奇函数时, f(?x)??f(x)
?f?(?x)?limf(?x??x)?f(?x)?f(x??x)?f(x)＝lim?x?0?x?0?x?x
f(x??x)?f(x)＝lim?f?(x)??x?0??x?f?(x)为偶函数. 习题2-2
★ 1. 计算下列函数的导数：
知识点：基本初等函数的导数和导数的四则运算法则
思路：利用基本初等函数的导数和导数的四则运算法则求导数
(1)
y?3x?5x;
52x
解: y??(3x?5x)??(3x)??(5x)??3?(2)
y?5x2?3x?3ex;
2xx2xxxx解: y??(5x?3?3e)??(5x)??(3)??(3e)??10x?3ln3?3e
(3)
y?2tanx?secx?1;
2解: y??(2tanx?secx?1)??(2tanx)??(secx)??(1)??2secx?secxtanx
(4)
y?sinx?cosx;
22解: y??(sinx?cosx)??(sinx)?cosx?sinx(cosx)??cosx?sinx?cos2x
(5)
y?x3lnx;
1?x2(3lnx?1) x33323解: y??(xlnx)??(x)?lnx?x(lnx)??3xlnx?x(6)
y?excosx;
xxxxx解: y??(ecosx)??(e)?cosx?e(cosx)??ecosx?esinx
(7)
y?lnx; x1x?lnx(lnx)?x?x?lnxx1?lnx解:y?? ??222xxx(8)y?(x?1)(x?2)(x?3);
解:y??(x?1)?(x?2)(x?3)?(x?1)(x?2)?(x?3)?(x?1)(x?2)(x?3)?
?(x?2)(x?3)?(x?1)(x?3)?(x?1)(x?2)
(9)s?1?sint;
1?cost解:s??(1?sint)?(1?cost)?(1?sint)(1?cost)?cost(1?cost)?(1?sint)(?sint)? 22(1?cost)(1?cost)1?sint?cost
(1?cost)2 ?(10)
y?3xsinx?axex;
解：y??(3xsinx)??(axex)??(3x)?sinx?3x(sinx)??(ax)?ex?ax(ex)?
11?2xxxx ?x3sinx?x3cosx?aelna?ae
3(11)
y?xlog2x?ln2;
1 ln2解:y??(xlog2x)??(ln2)??x?log2x?x(log2x)??0?log2x?5x2?3x?4(12)y?. 2x?1(5x2?3x?4)?(x2?1)?(5x2?3x?4)(x2?1)?解:y'?
(x2?1)2(10?3x)(x2?1)?(5x2?3x?4)(2x)3(x2?6x?1) ? ?2222(x?1)(x?1)★ 2.计算下列函数在指定点处的导数:
知识点：基本初等函数的导数和导数的四则运算法则
思路：利用基本初等函数的导数和导数的四则运算法则求导数
3x3?(1)y?3?x3,求
y?(0);
13x332??y(0)?解:y??( )??()???x33?x3(3?x)2(2)
y?ex(x2?3x?1),求y?(0).
x2x2xx2解:y????e(x?3x?1)???e(x?3x?1)?e(2x?3)?e(x?x?2)
??y?(0)?ex(x2?x?2)★ 3.求曲线
x?0?1(1?1?2)??2
y?2sinx?x2上横坐标为x?0的点处的切线方程与法线方程.
知识点：导数的几何意义，基本初等函数的导数和导数的四则运算法则 思路：利用基本初等函数的导数和导数的四则运算法则求导数得切线的斜率
解:y??2cosx?2x ?在x?0的点处切线的斜率k?y?|x?0?2cos0?20?2
又当x?0时,y?0 ?在x?0的点处切线方程为y?2x，法线方程为y??1x 2★ 4.写出曲线
y?x?1与x轴交点处的切线方程. x知识点：导数的几何意义，基本初等函数的导数和导数的四则运算法则 思路：利用基本初等函数的导数和导数的四则运算法则求导数得切线的斜率 解:y??(x?)??1?1 x21当y?0时，即x??0 解得x?1或?1 ?曲线与x轴的交点为(1,0),(?1,0)
x1x ?点(1,0)处的切线的斜率为k1?y?|x?1?2 ?切线方程为y?2(x?1)，即y?2x?2 ?y?|x??1?2 ?切线方程为y?2(x?1)，即y?2x?2
?点(?1,0)处的切线的斜率为k2★ 5.求下列函数的导数:
知识点：基本初等函数的导数以及复合函数的求导法则 思路：利用链式法则求复合函数的导数
(1)
y?cos(4?3x);
解:y???cos(4?3x)???(4?3x)???sin(4?3x)(?3)?3sin(4?3x)
(2)
y?e?3x22;
22解:y??(e?3x)??e?3x?(?3x2)???6xe?3x(3)
y?a2?x2(a2?x2)?2a?x22;
解:y???12a?x22(?2x)??xa?x22
(4)
y?tan(x2);
解:y??sec2(x2)?(x2)??2xsec2(x2)
(5)
y?arctan(ex);
(ex)?ex解:y'??1?(ex)21?e2x(6)
y?arcsin(1?2x);
解:y??(1?2x)?1?(1?2x)2??1x?x2
(7)
y?arccos1; x11()?21x解:y??? ?x?2111?()21?2|x|x?1xx(8)
y?ln(secx?tanx);
11(secx?tanx)??(secxtanx?sec2x)?secx
secx?tanxsecx?tanx解:y??(9)
y?ln(cscx?cotx).
11(cscx?cotx)???(?cscxcotx?csc2x)?cscx
cscx?cotxcscx?cotx解:y??★ 6.求下列函数的导数:
知识点：导数的四则运算法则和复合函数的求导法则
思路：利用导数的四则运算法则和复合函数的求导法则求导数
(1)
y?(2?3x2)1?5x2; 2222解:y??(2?3x)?1?5x?(2?3x)?(1?5x)??x(16?45x2)1?5x2
(2)
y?lnx?lnx; 1(lnx)?11?(x)????x2lnx2x2xlnx
解:y??(3)
y?ln1?x1?x; 11?x1?x1?x2x解:y???()???1?x1?x1?x(4)
?(1?x)?12x(1?x)2?(1?x)?1(1?x)?x
y?lntanx; 2解:y??x1x11?(tan)???sec2???cscx xx222sinxtantan221(5)
y?lnlnx;
11?(lnx)?? lnxxlnx解:y??(6)
y?x1?x2?arcsinx;
22解:y??1?x?x?(1?x)??11?x2?1?x2?x??2x21?x2?11?x2?21?x2
(7)
xy?(arcsin)2;
2解:y??2arcsinxxx1x?(arcsin)??2arcsin??()??2221?(x2)222arcsin4?xx22
(8)
y?1?ln2x; (1?ln2x)?21?lnxx2解:y???2lnx(lnx)?21?lnx2?2lnx(1x)21?lnx2?lnxx1?lnx2
(9)
y?earctan 解:y??earctanx?(arctanx)??earctanx(x)?arctan??e1?(x)2x11earctanx ???21?x2x2x(1?x)(10)
y?10xtan2x;
xtan2x解:y??10 ?10?ln10?(xtan2x)??10xtan2xln10[tan2x?xsec22x?(2x)?] ln10(tan2x?2xsec22x)
xtan2xe4x(11)y?ln; 4xe?1