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1、1P61 习题 3-1 1、根据定义求导数：(1)cosyx00000cos()coslim2sinsin22limsin()sin22lim2sin2lim sin() lim2 2 sinxxxxxxxxyx xxxxxxx xxxxx xxxx 1 2(2)yx11 220001 2()lim()lim 1lim11 22xxxxxxyx xxx xxxxxxxxx (3)yx x033022302202()lim()lim()33lim()33lim()3 2 3 2xxxxxxxxx xyx xxx x xxxxx xxxx xx x xxxxx xxx xx xxxxx xx x
2、xx (4)xya001limlimxxxx xxxaaayaxx 设，则tx 01limt xtayat 再设，则，于是tsalogats111 1110 11limlog1lim log1lim log 1 (1)1 loglnxsaxs s axs s axaxsyasa sa saeaa 2、20000000()()(1) lim()()lim()xxf xxf x x fxxf x x fx 00000000000000000000000()()(2) lim()()()()lim()()()()limlim()()()()limlim()()2()xxxxxxf xxf xx x
3、f xxf xf xf xx x f xxf xf xf xx xx f xxf xf xxf x xx fxfxfx 000( )(3)lim()lim(0)(0)lim(0)xxxf x x fx x fxf x f 00001001(4)lim ()()1()() lim1()nnn f xf xnf xf xnn fx 3、证：为偶函数且，则( )f xQ(0)0f00000(0)(0)(0)lim()(0)lim()(0)lim()(0)lim()(0)lim(0)xxxxxfxffx fxf x fxf x fxf x fxf x f 又在处可导，则( )f x0x (0)(0)f
4、f即(0)(0)ff 所以(0)0f故。(0)0f 4、证：（1）设为可导的奇函数，则：( )f x0000()()()lim()( )lim()( )lim()( )lim( )xxxxfxxfxfxx f xxf x x f xxf x x f xxf x x fx 所以为偶函数。( )fx（2）设为可导的偶函数，则：( )f x30000()()()lim()( )lim()( )lim()( )lim( )xxxxfxxfxfxx f xxf x x f xxf x x f xxf x x fx 所以为奇函数。( )fx（3）设为可导的周期函数且其周期为( )f x，则：T00()()
5、()lim()( )lim( )xxf xTxf xTfxTx f xxf x x fx 所以仍为以为周期的周期函数。( )fxT5、解：00|1x xxyeQ在点处的切线斜率为 1，法线xye(0,1)斜率为了-1，故所求的切线和法线方程分别 为：1,1yx yx 即。1,1yxyx 6、解：001lim( )lim sin0(0) xxf xxfxQ所以在处连续；( )f x0x 000000011()sinsin lim( )lim111sinsinsin lim11sinsin1limlim sin11(sinsin)1limsin11112 cossin12sinlimxxxxxxx
6、xxxxxxf xxxxxxxxxx xxxxxx xxxxxxx xxxxxxxxxx 0022sin112 ()sin2 coslim1sin112 ()2 ()sin2 coslim2 () 111sin2 cos()2 111sincosxxx x x xxxxxx x x xxx xxxxxx x xxxxxxxxx 即：111( )cossinfxxxx于是在处不可导。( )f x0x 7、解：在处连续，( )f xQ1x 112lim( )lim()(1)11xxf xaxbabf即1ab又在处可导，( )f xQ1x 400(1)(1)(1)lim(1)()limxxfxffx
7、 axbab x a Q022020(1)(1)(1)lim(1)1lim2lim2xxxfxffx x x xx x 又在处可导，( )f xQ1x 2a由得。1ab1b 8、解：由已知，产品的产量并不随劳动力数量N的增加而均匀增加，当时劳动生产x0xx率应为： 。00 0 0( )()lim() xxN xN xN xxxP67 习题 3-2 1、求下列函数的导数：52524(1)(31)()(3)( )1561yxxxxxxxx2222(2)(cos )() cos(cos )2 cossinyxxxxxxxxxx221(3)()sin (1) sin(1)(sin ) sin sin(
8、1)cos sinxyx xxxx x xxx x2(4)(tansin )(tan )(sin )seccosyxxxxxx2222222(5)()(2)4xxxyeexxe2(6)(arctan) 1()1 () 111 12 1 2(1)yxxxxxxx 2222222(7)(cos3 )() cos3(cos3 )2cos3( 3sin3 )2cos33sin3xxxxxxxyexexexexxexxexex(8)( ln(cos )ln(cos )(ln(cos )ln(cos )(cos )cos sinln(cos )cos ln(cos )tanyxxxxxxxxxx xxxx
9、 xxx2、解：2212122211( )()21 11()()21 (2) (1) 12 (2)(1)fxxxxx xx x xx Q522212 01(0)(02)(01)4f 22212 ( 1)1( 1)( 12)( 1)1)2f 22212 111(1)(12)(11)18f 3、解： 11( )1111( )1xfxx xf xxQ211( )()1(1)fxxx 4、解：当时，1x (1)1yx 当时，12x(1)(2)23yxxx当时，2x (2) 11yx 1,1 23,12 1,2x yxx x P69 习题 3-3 1、求下列函数的二阶导数：32 535527 553(1
10、)()()5 36()525yxxxyxx Q22222222222(2)()( 2 )2( 2)( 2 )( 2 )()22 ( 2)24xxxxxxxxxxyeexxeyxex ex eexxeex e Q22(3)(sincos)cossin( cossin)sincosyaxbxaaxbbxyaaxbbxaaxbbx Q(4)(sin )sincos(sincos )(sin )(cos )sincoscossin2cosxxxxxxxxxxxxyexexexyexexexexexexexexexQ21(5)(ln2 )11( )yxxyxx Q22(6) ln(2)ln(2)2ln(
11、2)2 12 2(2) 4 (2)xyxxxx xyxx xx xx x x Q2、求下列函数的阶导数：n3222(1)(1)321(321)62 (62)6yxxxxxyxxx yx 当时3n ( )0ny6112(4)23(5)34( )1(2)( ln )ln1(ln1)()()2(2)6( ) (2)!,(2)nnnyxxxyxxyxxyxxyxxynxn L L2(4)(5)(6)(3)(sin)2sin cossin2(sin2 )2cos22sin(2 )2 2(2cos2 )4sin24sin(2 )2 3( 4sin2 )8cos28sin(2 )2 4( 8cos2 )16
12、sin216sin(2 )2 5(16sin2 )32cos232sin(yxxxxyxxxyxxxyxxxyxxxyxx ( )12 )212sin(2 )2nnxnyxL L( )(4)()(1)(1)(1)(2)()xxxxxxxxnxyxeexexeyxeexexeyxn eL LP73 习题 3-4 1、解：由两边对求导得：0xyxyeex0xyxyyxyeeyeyyex2、解：由两边对求导得：1 ln()yyxye x11 1yyyyyeyxyyxeyexy 3、解：由两边对求导得：23xyexyx()202 1xyxyxyeyxyyyeyxe把代入解得，0y 23xyexy1x 12|11xyxxyyeyxe 即所给曲线在点处的切线斜率( 1,0)1k 故所求切线方
