2012级郑州大学工学院高数课后习题答案
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习题2.1
1. 已知质点做直线运动的运动方程为s?t2?3,求该质点在t?5时的瞬时速度. 解:因为 s??t??2t,所以,v?5??s??5??2?5?10. 2. 若limf?x??f?a?x?ax?a,试判断下列命题是否正确. ?A(A为常数)
(1)f?x?在x?a处可导; (2)f?x?在x?a处连续. 解:
(1)因为f??a??limf?x??f?a?x?ax?a?A,所以,f?x?在x?a处可导,命题正确;
(2)因为可导必连续,故f?x?在x?a处连续,命题正确. 3.利用导数定义求下列函数在指定点的导数.
3(1)y?x,x0?1; (2)y?2x?1,x0?1;
(3)y?cosx,x0?解:
(1)f??1??lim(2)f??1??lim?2; (4)y?1x,x0?2.
f?x??f?1?x?1f?x??f?1?x?1x?1?limx?1x?13x?1?lim?x?1??x2?x?1x?1x?1?lim2?x?1?x?1??lim?xx?12?x?1?3;
?x?1?lim?2x?1??3x?1x?1x?1?2;
???(3)f????lim?2?x?????f?x??f???2?x??2?limx?cosxx??2?2?limx????sin??x??2?x???lim2x??x?2?2?22x??2??1;
1(4)f??2??lim
f?x??f?2?x?2x?2112?x??lim??. ?limx2?limx?22xx?2x?2x?22x?x?2?4?1?x2,x?1,4.讨论f?x???在点x?1处的连续性与可导性.
?2x,x?1,解:
(一) limf?x??limx?1;
2x?1?x?1?1
limf?x??lim2x?2.
x?1?x?1? 因为 limf?x??limf?x? ,因此f?x?点x?1处无极限,从而也不连续.
x?1?x?1?(二) 因为连续是可导的必要条件,所以,f?x?点x?1处不可导. ?x,x?0,1?5.讨论f?x???在点x?0处的连续性与可导性.
1?ex?x?0,?0,解:
x?(一)limf?x??limx?0?x?01?01?0?0;lim?f?x??lim?x.x?0e?1x?1x?0?01?0?0,
1?ex??x?01?e因为limf?x??limf?x??f?0?,故f?x?点x?0处连续.
x?0x?0(二)f???0??limf?x??f?0??x?0x?0?lim?x?011?11?01x??1;
1?ex f???0??limf?x??f?0??x?0x?0?lim?x?011?e?1x?01?0?0.
1?ex1?e因为f???0??f???0?,所以,f?x?点x?0处不可导. 6.已知f??3??2,求极限lim解:limf?3?h??f?3?2h??f?3?h??f?3?.
2hf?3???h???f?3??hh?012h?0limh?0??12f??3???1.
?2,x?1,?7.设f?x???1?x2已知函数点x?1处可导,试确定a,b的值.
?ax?b,x?1,?解:(一)因为可导必连续,所以f?x?在x?1处连续,即
limf?x??limf?x??f?1? ①
x?1?x?1?其中 f?1??1;
lim?f?x??lim?x?1x?1??21?x2?1;
limf?x??lim?ax?b??a?b.
x?1x?12
所以,有
a?b?1. ② (二)因为以f?x?在x?1处可导,所以,应满足f???1??f???1?.
2其中,f???1??limf?x??f?1??x?1x?1?lim?1?xx?1x?1??lim?x?12?1
??1;
?lim?x?11?x21?x1?x2?x?1??1?x2?f?x??f?1?x?1x?1f???1??lim?x?1?lim?x?1?ax?b??1x?1(由②)
?a?lim?x?1a?x??1?a???1??lim?x?1a?x?1?1?x2.
所以,a??1,b?2.
8.证明:
(1)可导的奇函数的导数为偶函数; (2)可导的偶函数的导数为奇函数;
(3)可导的周期函数的导数为具有相同周期的周期函数. 证明:
(1)设奇函数f?x?处处可导.在点x处由导数定义有 f??x??lim ?limf?x??x??f?x??x?xf??x??x??f??x??xf???x?????x???f??x???x?x?0(因为f?x?为奇函数)
??f??x??x?????f??x??
?x?0 ?lim??x?0 ?lim?x?0
?f???x?
即证明了对于任意点x,有 f???x??f??x?,所以,f??x?为偶函数. (2)设偶函数f?x?处处可导.在点x处由导数定义有 f??x??lim ?limf?x??x??f?x??x?0?xf??x??x??f??x??xf??x(因为f?x?为偶函数)
?x?0 ??lim???x?????x???f??x??x?0 ??f???x?
3
即证明了对于任意点x,有 f???x???f??x?,所以,f??x?为奇函数. (3)设以T为周期的函数f?x?处处可导.在点x处由导数定义有 f??x?T??lim ?limf?x?T??x??f?x?T?x?x??x?0(因为f?x?为周期函数)
f?x??x??f?x??x?0
??f??x?
即证明了对于任意点x,有 f??x?T??f??x?,所以,f??x?也是以T为周期的函数. 9.设f?0??0,f??0?存在,求lim. 2tanxf?1?cosx?f?0??1?cosx???f?0?1?cosx?lim.解:lim 22x?0x?0tanx1?cosxtanxx?0f?1?cosx? ?limf?0??1?cosx???f?0?1?cosx11 ?f??0???f??0?.
22x?0.lim1?cosxtanx2x?0
1其中 lim
f?0??1?cosx???f?0?1?cosxx?0?f??0? ;lim1?cosxtanx2x?01(等价替换)?lim22?.
x?02xx2习题2.2
1. 求下列函数的导数.
2(1).y?4x?3x?1; (2). y?2x?51x?sinx;
(3). y?x4lnx; (4). y?sinx?x?1;
(5).y?2cosx?3x; (6). y?2?3;
xx(7). y?log2x?x2; (8). y?x?lnx?1;
x2(9).y?e?x?1?cosx; (10). y?1?x1?x.
解:
(1).y??4?x2???3?x????1???8x?3;
5(2). y??2?x????1??4?????sinx??2?5x???x?1????2???cosx? ?x?4
?10x?41x2??cosx?; ?1(3).y???x4?lnx?x4?lnx???4x3lnx?x4.?4x3?lnx?1?;
x(4).y???sinx????x????1???cosx?1?0?cosx?1;
(5).y??2?cosx??3?x???2sinx?3;
??(6). y???2x???3x??2xln2?3xln3; (7). y???log2x???x2????1xln2?2x??;
(8). y???x????lnx????1??;
???(9).y???ex??x2?1?cosx?ex?x2?1?cosx?ex?x2?1??cosx?
?e?ex?x?x2?1cosx?2xecosx?sinxe?1?cosx?sinx??2xecosxx?xx?x2?1?
x2?;
(10).
??1?x???1?x???1?x??1?x??2?1?x?y?????.; ??22?1?x??1?x??1?x?2. 求下列函数的导数.
(1).y?4?x?1???3x?1?; (2). y??1?2x?;
227(3). y?lnlnx; (4). y?lnsinex;
(5).y?2ln1x; (6). y?arctan2x;
?x?2?(7). y?e?3xsin2x; (8). y?lntan?(9).y?sinln2x?12???;
4?22; (10). y?lnx??x?a.; 1?x;
2?(11).y?arcsin?sin??x?; (12).
1??; x?y?arccosx(13). y?ln?arccos2 (14). y?a?x1?x1?x2322;
(15).y?sinx2sinx; (16). y?3;
5
(17). y?arctan(19).y?x221?x1?x2; (18). y?arctan?tan2x?;
a2x?a?2ln(x?x?a)
22(20). y?lnx21?x?1?x?221?x1?x;
xa(21).y?a?x?a22arcsin;
(22).y?sin?sin(sinx)?. 解:
(1).y??8?x?1??.x?1??2?3x?1??.3x?1??8?x?1??6?3x?1??26x?14;
??(2). y??7.?1?2x?6?1?2x????14(1?2x)6; (3). y??1lnx1sine1xx?lnx???x1xlnx;
1x (4). y??(5).y??2ln?sine???sine.cose.ex????exxcotex;
?1ln?1??ln2?ln??2xln2??lnx?; ?x?1(6). y??1??2x?2?2x???21?4x2;
??(7). y???e?3x??3x??sin2x?e?3x?cos2x?2x???e?3x??3sin2x?2cos2x?;
????????(8).
?x??cos???????1x?x?24?x??????????2y??tan?????sec???.???? ??x???x????24???24??24??tan????sin?????2424??????x??cos???11?24??..??2?x??2?xsin???cos????24??24?
?1?x?2sin???24??x??cos??????24??1?sin?x??2?????1cosx?secx.
6
(9).y??cosln?1?1?2x?1?ln?2x?1???cosln2?2?cosln2x?1.
?1?2x?1???2x?1???2x?1?
?12x?1(10). y??x?1x?a1x?x?a2222?x?x?a22??x??1x?a22?1221?x?a?2????x?212?a2????
? ???11?.2x???22x?2x?a??1x?a22????x?a2?x?? 22x?a??2 ??1x?a22.
(11).y??11?sin?2x?2?sin2?x ??11?sin4?.2sinx?sinx? x ?11?sinx4.2sinxcosx?sin2x1?sinx4;
??12?1?x?? 2?21?x?(12). y???1?1?1?x?22?1?x????21x2??
?x1?1??????2x??2x?21?x2x1?x?;
????1??x??11x?1arccosx(13). y??1arccos??arccos1?x?11???x?arccos???1?1????? 2?1??x??1?????x?? ????1arccos?x1?11??x?1?2xxx???????2x;
(14). y????12222?1.a?x?x.?a?x?22?2a?x????a?x22?2
???1??2x??1.a?x?x.?22?2a?x?22?a?x22?2?a2?a?x22?3
7
(15).y???2sinx?sinx???.sinx2?sin????sin22?22x.cosx.x????????x2
??2sinxcosx?.sinx2?sin2x.?2xcosx2?sinx22222
?sin2x.sinx?2xsinx.cosxsinx2?2322;
1?1?x??(16). y???3?3?1?x????1?x2?1????1?x3?3??3?2x.1?x3?1?x2.?3x2?32221?x??1?x?????1?x3???1?????????
??? ?133?1?x3?2x?3x2?x4??; ?1?x3?.321?x??2??(17).
?21?x?1?x?2??y???.??2?1?x?2??1?x?2?1?x?2?1?x??1?x?1???1?x??111?x2
?;
(18). y?arctan?tan2x?;
y??11?tan?2x?22?tan;
2x???11?tan4?2tanx?tanx??????x?
?2tanx.secx1?tanxx24(19).y?x?a?22a22ln(x?x?a);
22?xy????2?2a?x?a??ln(x?2?22?x?a)
2????a?????2????x?22??
??1????2x?a22x?122??x?a2?2x2?a2??1x?a22(x??x?a)???22
?1???2x?a22?a2?????2222x?a???x?x2???1?1?.2x???22?22x?a?2x?a???1
8
?x2?a2?x2?a2????222?2x?a???1x?a1?x1?x22?2x?a2?22?2?x?a.
222x?a(20). y?ln1?x?1?x?1?x?1?x?1?x;
y?ln??ln?1?x???1?x?1?x2x?22?1?1?x??ln
x?? ?ln1?1?x2?lnx.
y??ln1?1?x????2????lnx????11?1?x2?1?1?x2??1 x? ?11?1?x2?121?x?0?221?x??????1
?x2? ?1?11?x2.x1?x2?1x?1?1?xx2.x1?x2?1x
?1?1?xx1?x22?1x?1x1?x22;
xa(21).y??xy????2x2a?x?2a22arcsin;
??2ax???22a?x???arcsin?
2a?????a2x?12222??a?x??.a?x???2?2a2?x22???
??1????2???x???2a?x???1????a?1?
?1212a?x?22x222?a22aa?x22a?x222.1a
?a?x?aa?x222;
2a?x(22).y?sin?sin(sinx)?.
解:y??cos?sin(sinx)?(sin(sinx))??cos?sin(sinx)?.cos(sinx).(sinx)?
?sin(sixn)?. ?cosx.cos(sinx).cos9
3. 设y?f?sinx2?,其中f?x?可导,求y?.
?2?22222解:y??f??sinx2??.sinx??f??sinx?.cosx.?x??2xcosxf??sinx?.
??????4. 设函数u?x?,v?x?可微,求下列函数的导数.
(1)y?u2?x??v2?x?,?u2?x??v2?x??0?; (2)y?arccot解: (1)y??1212u?x?v?x?,?v?c??0?.
?u.2?x??v?x??2?12.u?2?x??v2?x??
? ?1u2?x??v?x?2.?2u?x?.u??x??2v?x?.v??x??;
(2)
?2?u??x?.v?x??u?x?.v?x??v?x??u?x??y???? 2???u2?x??v2?x??2???vx??vx????u?x???1????v?x??1 ??u??x?v?x??u?x?.v?x?u2?x??v2?x?.
5.设y????ln??x???,其中??x?为正值可导函数,求y?. ???x????1?????????????x.?x?ln?x.?x??????ln??x??????x??ln??x???ln??x???? 解:y??????.??.??????2???x????x?????x?????x?????? ????x??1?ln??x???2?x??ln??x??.????. ???x??6.设y?f??2x?3?2?,f??x??arctanx?2x?3? ① ,求y??0?.
?12?2x?3??2x?3??2x?3???f.解:因为y??f??. ② .?????2?2x?3??2x?3??2x?3??2x?3?10
所以,y??0??f???1?.x129?arctan??1??2129??3.
7.设y?arctane?ln解:y?arctanex? y??11??exe2x2x1?e,求y??0?.
2x12x?2x?ln?1?e??
11?e2x?2?e???1??2?2? ?1?e????2x??ex2x1?e2xx?1?2ee?1??2??. 2x?2x2?1?e?1?e所以,y??0??1.
习题2.3
5. 求下列方程所确定的隐函数的导数.
(1).exy?sin?x?y??0; (2)y3?y2?2x2;
(3). y?1?xey; (4)arctanyx?lnx?y.
22解:(1)方程exy?sin?x?y??0两边对自变量x求导,有 exy?y?xy???cos?x?y?.?1?y???0,即
?xexy?cos?x?y?.y???ye?xy?cos?x?y?
所以 y???yexexyxy?x?y??cos?x?y??cos.
(2)方程y3?y2?2x2两边对自变量x求导,有 3y2y??2yy??4x,即
3y?2yy??4x.
?2? 所以 y??y4x3y?2y2.
(3)解:方程y?1?xe两边对自变量x求导,有
y??0?ey?xeyy?,即?1?xey?y??ey.
11
所以 y??(4)先将方程arctanarctanyx2eyy1?xe2?ey2?y2.
yx??ln122x?y.化简为
lnx?y.
22??方程arctan
yx?lnx?y两边对自变量x求导,有
1?y?x?y?1.?..?2x?2yy??,即 ??2222?2x?y?y??x1????x?11?y?x?y?1.?..?2x?2yy?? ??22222x?y?x?2x?yx2 化简得 y?x?y?x?yy? 即 ?x?y?y??x?y 所以 y??2.求下列函数的导数(1) 解:
dydtx?yx?y.
.
dydx?x?a?t?sint?, ???y?a1?cost.??asintdydx?dydt;
dxdt??a?1?cost?. 1?costsint.
所以
dxdt?x?etcost,(2)? t?y?esint.解:
dydt?e?sint?cost?;
tdydt?e?cost?sint?.
t所以
dydx?dydtdxdt?sint?costcost?sint.
?x?3e?t,(3)? t?y?2e.解:
12
dydt?2et;
dydt??3e?t.
所以
dydx?dydtdxdt??23e.
2t3at?x?,2?1?t(4)? 23at?y?.21?t?解:
dydtdxdt?3a1?t???3at.2t?1?t?2222?3a?1?t2?1?t?6at2222?;
?6at?1?t??3at?1?t?222.2t?2t?1?t?2;.
所以
dydx?dydtdxdt?1?t.
?x?2cos3?,(5)? 3y?4sin?.?解:
dxd?dyd??6.cos?(?sin?);
2?12sin?.cos?.
2所以
dydx?dyd?dxd???2tan?.
2??x?3t?2t,(6)?y
??esint?y?1?0.解:(将方程中x,y均视为t的函数),对所给方程两边关于t求导,得:
?dx?dt?6t?2?2?1?3t?,? ?dyydy?y??t??0?e.?.sint?ecos?dtdt????1?
?2?.由(2)式,得: 所以
dydx?dydtdxdt?ecost2?1?3t?1?esintyydydt?ecost1?esintyy (3)
??.
13
3.求下列函数的导数
dydx.
(1)y?xx?x?0?; 解:
dydx?e?xlnx???exlnx?xlnx?1?lnx??xx(1?lnx).?xlnx??e ①.
(2)y?x?xx?xx?x?0?; 解:?xxxx???e?xlnxx???exlnxx?xxlnx??
?exxlnx???x?x.?x?lnx?x?lnx?(由①) ?????xx1?.?x.?1?lnx?.lnx?x.?
x?? ?exxlnx ?xx.?xx.?1?lnx?.lnx?xx?1?; ②
xdydxx??x?x(将①、②代入) ??x???x??x?? ?1?xx?1?lnx??xx.?xx.?1?lnx?.lnx?xx?1?.
x(3)
1??y??1??x??x
解:lny?x?ln?1?x??lnx?. 上式两边关于x求导,得
11??1y???ln?1?x??lnx??x???y?1?xx?
?ln?1???1?1??x?1?x .
所以
??1?1??1???1?1?y??y?ln?1????1?.ln1?????????x?1?x??x???x?1?x???2x.
??x?1??x?2??x?3??3(4)y??? 3x?x?4???解:lny?14
23?ln?x?1??ln?x?2??ln?x?3??3lnx?ln?x?4??.
上式两边关于x求导,得
1yy??2?11131??????3??x?1x?2x?3xx?4?2
所以
2??x?1??x?2??x?3??3y??.??33?x?x?4??x1131??1.???????x?1x?2x?3xx?4?.
(5)y?x?x?0?.
解:lny?1xlnx.
上式两边关于x求导,得
1yy??1?lnxx2x ,所以, y??1?e.
14xx2x.?1?lnx?. .
(6). y?解:lny?12?xsinx?lnx?12ln?sinx??ln1?e?x?.
两边关于x求导,得
1y1y??111111x.?..cosx?..?e. 即 x2x2sinx41?e??1111e y??.?.cotx?.. xy2x241?ex所以
x?111?1e1y??y?.?.cotx?.?x?241?e??2x2?xsinx?1?exx?11e??cotx?.x21?e?x??. ?4.求曲线?解:故
dydtdydx?x?sint,?y?cos2t.在参数所对应的点处的切线方程与法线方程.
?cost??2sin2t?dydtdxdt;
?dxdt.
?2sin2tcost??4sint.
曲线对应t???4处点的切线斜率为k?2dydx|t??4??4sin?4??22.
当t??2?,y?0,故切点坐标为?时,x?,0?. ??42?2?15
所以曲线对应t??4处点的切线方程为
?2??,即 y??22x?2. y??22?x??2???曲线对应t??4处点的法线方程为
?2??x??,即 ?2??22??1 y?? y?122x?14.
5.求曲线x2?xy?y2?4 ①在点?2,?2?处的切线方程与法线方程. 解:方程①两边对自变量x求导,有 2x??y?xy???2yy??0,即
?x?2y?y??所以 y????2x?y.
.
2x?yx?2y曲线在点?2,?2?的切线斜率为k?y?|?2,?2??1. 所以,曲线在点?2,?2?的切线方程为 y?2?1.?x?2?.即 y?x?4. 曲线在点?2,?2?的法线方程为
y?2??1.?x?2?.即 y??x. 6.求下列函数的导数. (1)y?x?3;
?3?x,x?3,?解:(1)y??0,x?3,
?x?3,x?3.?当x?3时,y??(3?x)???1; 当x?3时,y??(x?3)??1; 当x?3时,
??3??limy??x?3f?x??f?3?x?3?lim?x?3?3?x??0x?3??1;
16
??3??lim y?f?x??f?3??x?3x?3?lim?x?3?x?3??0x?3?1.
??3??y???3?,故y??3?不存在. 因为y?x?3,??1,?所以,y??不存在,x?3,
?1,x?3.??x3?2x2,x?0,?(2) y??2x2?x3,0?x?2,;
?32x?2x,x?2.?当x?0时,y??(x3?2x2)??3x2?4x; 当0?x?2时,y??(2x2?x3)??4x?3x2; 当x?2时,y??(x3?2x2)??3x2?4x.
??0??limy??x?0f?x??f?0?x?0f?x??f?0?x?0?lim?x?0x?2xx2x?xx232?0;
3??0??limy??x?0?lim?x?0?0.
??0??y???0??0,故y??0??0. 因为y???2??limy??x?2f?x??f?2?x?2f?x??f?2?x?2?lim?x?22x?xx?223?lim?x?2x2?2?x?x?2??4;
??2??limy??x?2?lim?x?2x?2xx?232?lim?x?2x2?x?2?x?2?4
??2?,故y??2?不存在. ??2??y?因为y??3x2?4x,x?0,?2?4x?3x,0?x?2所以,y???
?不存在,x?2,?3x2?4x,x?2.?1?xarctan,x?0,? f?x???x?0,x?0.?(3)
17
解:当x?0时
??11y??(xarctan)??arctan?x??xx?1???arctan??0??limy??x?0??1?1??.?2? 2??1??x??????x??1x1?xf?x??f?0?x?0?x2;
1x1x????lim?arctanx?0?2;
??0??limy??x?0f?x??f?0?x?0?lim?arctanx?0?2??0??y???0?,故y??0?不存在. ,因为y?所以,
1x??,x?0,?arctan2? f?x???x1?x?不存在,x?0.??xe?x,x?0,(4)f?x???
?ln(1?x),x?0.解:当x?0时,y??(xe?x)??e?x?xe?x??1?x?e?x;
当x?0时,y??(ln?1?x?)????0??limy??x?011?x.
?1;
f?x??f?0?x?0f?x??f?0?x?0?lim?x?0xe?xx??0??limy??x?0?lim?x?0ln?1?x?x?1.
??0??y???0??1,故y??0??1.所以 因为y????1?x?e?x,x?0,?x?0, f??x???1,?1?,x?0.?1?x习题2.4
6. 设k为常数,求下列函数的n阶导数.
(1)y?e; (2)y?sinkx; (3)y?coskx;(4)y?a. 解:(1)y???ekxkxkx???e?kx???kekxkxkxkx; ekxy???ke18
????k?ke??k2;
归纳可得 y?n??knekx.
(2)y???sinkx??coskx?kx??kcoskx;
??2y???k?coskx??k???sinkx??kx????ksinkx;
????y????ky?4?2???sin3??23kx???k?coskx?kx????kcoskx;
??????34kx???k??sinkx?kx???ksinkx.
??????k?cos 归纳可得 y?n??knsin?kx???n???. 2???n???. 2?(3)类似于(2)的求法,y?n??kncos?kx???(4)y???akx??akxlna.?kx??klna.akx;
y???klnaa????klna.?klna.a??kkxkx2.lna.a2kx;
归纳可得 y?n??knlnna.akx.
2.设y?a0?a1?x?x0??a2?x?x0??a3?x?x0????an?1?x?x0?23n?1?an?x?x0?
n,求函数在点x0处的各阶导数. 解:
y??0?a1?2a2?x?x0??3a3?x?x0?????n?1?an?1?x?x0?2n?2?nan?x?x0?n?1
所以,y??x0??a1;
y???2.1a2?3.2a3?x?x0?????n?1??n?2?an?1?x?x0?所以,y???x0??2!a2;
y????3.2.1a3????n?1??n?2??n?3?an?1?x?x0?n?4n?3?n?n?1?an?x?x0?n?2;
?n?n?1??n?2?an?x?x0?n?3;
所以,y????x0??3!a3; 归纳可得 yy?k??x0??k!ak(k?1,...n,)
?n).
?k??x0??0.(k23.已知y?ln?1?x19
?,求y???0?.
解:y??11?x2?1?x???22x1?x2;
?2221?x?2x.2x21?x?2x? y????; ???22222?1?x?1?x1?x???????? 所以,y???0??2.
1?x?t?,2?dy?t4.求由方程??t?0?确定的函数的二阶导数2. 2dx?y?t?lnt,?2?解: (一)
dydt?t?1t?1?tt?2;
dydtdxdtdxdt?1?1t2?1?tt22.
所以
dydx22dydx?t.
2(二)
d?dy?dt1t???. ???2dxdx?dx?dx1?tdt125.求由方程x?y?siny?0 ①所确定的隐函数的二阶导数
dydx22.
解:(一)①式两边关于x求导,得 1???dydx?12cosy.dydx?0
即 ?1?dydx21?dycosy?.?1 2?dx所以 ?1?112cosy?22?cosy. ②
(二)
dydx2?d?dy?d2() ???dx?dx?dx2?cosy ?ddy2?cosy4siny(2).dydx??2siny?2?cosy?2.22?cosy
???2?cosy?y3.
6.设函数y?y?x?由方程e?xe20
f?y? ①所确定,其中f二阶可导,且f??1,求y??.
解:①两边取对数,得
y?lnx?f?y? ② ②式两边关于x求导,得 y??1x?f??y?.y?
即 xy??1?xf??y?.y? ③ 由③式解得
y??1x?1?f??y?? 所以
y?????1???x?1?f??y?????x?1?f??y???????x?1?f??y???2 ???1?f??y???x?0?f???y?y??1?f??y??xy?f???y?x2?1?f??y??2??x2?1?f??y??2(将④代入)1?f??y??x.1x?1?f??y??.f???y? ??x2?1?f??y??2
???1?f??y??2?f???y?x2?1?f??y??3.
7.设y?1?n?x2?2x?3,求y.
解:y?1?x?1??x?3??1?11?14???x?1x?3????x?1??1??x?3?4??1?. y??14???1?.?x?1??2???1??x?3??2?;
y???14???1?.??2?.?x?1??3???1?.??2?.?x?3??3?;
归纳可得 y?n??1???1?.??2????n??x?1??n?1???1?.??2?.???n??x?3??n?14?
???1?nn!?14?n??1???x?1?1?x?3?n?1?. ?8.设y??1?x2?cosx,求y?n?.
9.设f?x?具有任意阶导数,且f??x???f?x??2.,求f?n??x??n?2?.
21
④
证明:因为f??x???f?x??2,
所以 f???x??2f?x?.f??x??2.1.?f?x??3; f????x??3.2.1?f?x??2.f??x??3!?f?x??4;
??归纳可得:f?n??x??n!?f?x??n?1.
复习题2
1. 设??x?在x?a处连续,试讨论f?x???x?a???x?与g?x??x?a??x?在x?a处的可导性. 解:
(一)因为limf?x??f?a?x?ax?a?lim?x?a???x?x?ax?a?lim??x????a?,所以,f?x?在x?a处
x?a可导,且f??a????a?.
??a??lim(二)g?g?x??g?a??x?ax?ag?x??g?a?x?a?lim??x?a???x?x?ax?a??lim??x?????a?;
x?a??a??lim g?x?a??lim?x?a???x?x?ax?a?lim??x?????a?.
x?a??a??0,所以,g?x?在x?a处可导,且g??a??0;??a??g?(1)当??a??0时,因为g? ??a??g???a?,所以,g?x?在x?a处不可导. (2)当??a??0时,因为g?2.设f?0??0,f??0?存在,求limf?1?cosx?2tanxf?1?cosx?f?0??1?cosx???f?0?1?cosx?lim.解:lim 22x?0x?01?cosxtanxtanxx?0.
?limf?0??1?cosx???f?0?1?cosx12?12f??0?.
x?0.lim1?cosxtanx2x?0
?f??0??1其中 limf?0??1?cosx???f?0?1?cosxx?0?f??0? ;limf?x?x1?cosxtanx2x?01(等价替换)?lim22?.
x?02xx23.若函数f?x?在点x?0处连续,且lim证明: 设limf?x?xx?0存在,证明f?x?在点x?0处可导.
x?0?A ①
.limf?x??A?0?0. ② 则 limf?x??lim?.x??limx?0x?0x?0x?0x?x?22
?f?x??f?x?
又因为f?x?在点x?0处连续,故 f?0??limf?x??0. ③
x?0所以 f??0??limf?x??f?0?x?0x?0 ?limf?x?xx?0?0.
4.设曲线y?f?x?在原点与曲线y?sinx相切,求limx????2?xf??. ?x?解:因为曲线y?f?x?在原点与曲线y?sinx相切,故
f?0??0 ① 且
?f??0???sinx?|x?0?cosx|x?0?1 ②
?2?f???f?0??x?.2?2xx???lim?2?xf???limx????x??2?f???f?0??x?2?2xx???lim2f??0??2.
5.设函数f?x?在???,???内有定义,f?x??0,f??0??1,且对任意x,y????,???,恒有f?x?y??f?x?.f?y?①成立,证明f?x?在???,???内可导,且f??x??f?x?. 证明:
(一)①中,取x?y?0,得 f?0??f所以f?0??1.
2?0?,故 f?0??1 或f?0??0.又因为f?x??0,
(二)由导数定义,对x????,???,
f??x??lim?limf?x??x??f?x??xf?x?.?f??x??1??x?0
?x?0 ?f?x?.lim?xf?0??x??f?1??x
?x?0?f?x?.f??0??f?x?.
?1?cosax,x?0,?x??x?0,在在???,???内处处可导,并求f??x?. 6.求a,b的值,使函数f?x???0,?2lnb?x?,x?0.?x???解:(一)因为可导必连续,所以f?x?在x?0处连续,即
23
limf?x??limf?x??f?0?.----------------------(1)
x?0?x?0?1 其中,limf?x??limx?0?1?cosaxxln?b?xx2x?0??lim2x?0?ax?x2?0;
所以,limf?x??limx?0??x?0??lim?fx?0?x??0.
因此,必有,limln?b?x2??lnb?0?b?1;
x?0(二)因为以f?x?在x?0处可导,所以,应满足: f???0??f???0?.
1?cosax其中,f???0??limx?0f1?x??f?0?x?0??lim?x?0xx?0
?lim1?cosxx2x?0?lim2x?0?ax?x22?12a;
2f???0??lim?x?0f?x??f?0?x?0?limln?b?xx22?x?0?limln?1?xx22?x?0?1.
所以,
a22?1?a??2. ??xsin2x?1?cos2x,x?0,2?x?且:f??x???1,x?0,.
?2222x?1?xln1?x?????,x?022?x?1?x??7.设f?x??3x?xx,试求f323??2x,x?0,解: f?x???3??4x,x?0.?n??0?存在的最高阶数n.
(一)f??0??limf?x??f?0?x?0x?0 ?lim3x?xxx32?lim3x?xx?0;
x?0x?0?2?2??6x,x?0, f??x???2??12x,x?0.
24
(二)f????0??limf??x??f??0??x?0x?0 ?limx?06x?0?2x?02?lim?6x?0;
x?0 f????0??limf??x??f??0??x?0x?0 ?limx?012x?0?x?0?lim?12x?0.
x?0 故 f???0??0. f???x????12x,x?0,?24x,x?0.
12x?0?(三)f?????0??lim f????0??limf???x??f???0??x?0x?0f???x??f???0?x?0?n? ?limx?0x?0? ?limx?0x?024x?0x?0?12;
??24..
故 f????0?不存在.所以求f?0?存在的最高阶数n?2.
22?dy?x?3t?2t?3,8.设y?y?x?由方程?所确定,求. 2|t?0ydx??esint?y?1?0,解:(将方程中x,y均视为t的函数),对所给方程两边关于t求导,得:
?dx?6t?2?21t?3??????????dt ?dyydyy?sinte?ecot?s???0??dtdt?(1)------------(*)
(2).由(2)式,得:
dydt?ecost1?eyysitn?eycots????(3) ?2y由(1)、(3)式得:
y??dydxy?dydt
?ecostdx?dt2?1?3?t??2cots2?1?t3?y???(4)y?2?ye
y?????????????(5)(5)式两端关于x求导,得:
?y?e?ey??2?y?yye2yy??2?yeyy????1?sitn??1t?323tcdtosdx?1?3t?32,
即:?3?yeyy??22?yeyy????1?1?3t?sint?3cost4?1?3t???????(6).
25
又当t?0时,y?1?y?|所以,
dydx22t?0?e2,代入(6)式??e2?1ey??|t?0??34
|t?0?y??|t?0?e3?e?e?????2e?3?. ?24?4y9.设y?y?x?由方程y?1?xe ①所确定,求解:(一)①关于x求导得
dy?ydy?y?0?1.e?x?e.? dxdx??ydydx33.
即 ?1?xe由②式解得
2?dydx?e ②
ydydx?eyy1?xe(因为①)?ey1??y?1??ey2?y. ③
(二)
dydx2yd?dy?d?e?????dx?dx?dx??2?yyy?d?e????dy?2?y??2y?dy?.?dx ? ?e?2?y??ey??1?ey.22?y?2?y??e?3?y?. ④ 3?2?y?(三)
dydx332d?dy??2dx?dx?2y2y2y?d?e?3?y??d?e?3?y??dy?????.?? 3?3??dx?????2?y??dy??2?y??dx?2e ? ?e3y?3?y??e2y?.?2?y?3??e2y?3?y??.?3?2?y?2.??1??ey.2?y?2?y?62
19?12y?2y?2?y?5.
10.设雨点为球状物,若雨点体积对时间的变化率与表面积成正比,证明雨点半径增加的速
率为一常数.
解:设半径为r的球体的体积及表面积分别为V和S. 则有 V?43?r ① , S?4?r2 ②
3 ① ①式两边关于t求导得
dVdt?4?r.2drdt?S.drdt ③
由题意,雨点体积对时间的变化率与表面积成正比,故可设
26
将④代入③得
dVdt?kS(k为正常数) ④
kS?S.drdt
所以,有
1xdrdt?k,即证明了雨点半径增加的速率为常数k.
11.曲线y?的切线与x轴、y轴围成一个图形.记切点的横坐标为a,试求切线方程
和这个图形的面积.当切点沿曲线趋于无穷远时,该面积的变化趋势如何? 解:(一)y?1x在点M?a,??1??处的切线斜率为 a??1? k????x??|x?a??112xx|x?a??12aa.1.
故y?1x在点M?a,??1??处的切线方程为 a? y?即
x3a1a??12aa.1?x?a?
?y32a?1 ①
(二)点M?a,??1??处的切线与x轴、y轴围成一个图形面积为 a? S?a??12.3a.32a94?94a.
94(三)limS?a??lima???a???a???;lim?S?a??lim?a?0a?0a?0.
12.一电线杆高AB?5m,顶端B处有一盏灯,一人高CD?1.8m,且以1m/s的速率前进.求当人与电线杆的距离AC?10m时(1)人影头顶E处前进的速率;(2)人影CE 的伸长速率.
解:(一)由于?CDE~?ABE,故有
即
27
CDAB1.85??CEAE
?1?ACAEAE?ACAE
所以 AC?①两边关于t求导得
将AC?10m,
d?ACdt1625AE ①
d?ACdt??16d?AE? ② .25dt??1m/s代入②,得
d?AEdt
?|AC?10?2516(m/s).
(二)由于?CDE~?ABE,故有
即
CDAB1.85??CEAECEAE925
也就是 CE?③两边关于t求导得
将AC?10m时,
d?AEdtAE ③
d?CEdt2516dt??925.d?AEdt? ④
?|AC?10?m/s代入④,得
13.设y?xa?xx,求y?. 解:y?ealnxxxad?CE??92516.25?916(m/s).
?exlnxa,所以
ay??e?alnxx????exlnx???e1alnxx?axlnx????exlnxa??xa?lnx
? ?eaxlnx?x?aln???2x?ax?xlnxa?1a?1 a??eaxlnx?xx????? ?xa.ax?lna.lnx?2xa1?x?a?1?alnx?1?. ??xx?14.求抛物线y?x?ax与y?x?bx(b?a?0)的公切线的方程.
解:所谓两条曲线的公切线是指同时与两条曲线相切的那条直线,比如中学时学过的两个圆周曲线的内、外公切线.
由y?x?ax得y??2x?a.抛物线y?x?ax在点M1x1,x1?ax1处的切线方程 为 y?x1?ax1??2x1?a??x?x1?,即
2222?2??? y??2x1?a?x?x1 ②
228
同理,抛物线y?x2?bx在点M2?x2,x2?bx2处的切线方程
2?为 y?x2?bx2??2x2?b??x?x2?,即
2?? y??2x2?b?x?x2 ③
2由切线方程②,③是完全相同的,故有
?x1??x2 ④ 且 2x1?a?2x2?b ⑤ 由④式得 x1?x2(舍去,否则与⑤式矛盾!)或x1??x2.将x1??x2代入⑤,得 x1?222b?a4
b?a4b?3a?2ab1622从而 x1?ax1?所以公切线斜率为
k?2x1?a?2.?b?a?216?a.?
b?a4?a?a?b2
所以公切线方程为 y?b?3a?2ab1622?a?b?b?a??x?? 2?4?2即 y?a?b2a?b2x?b?3a?2ab162?b?a822
也就是 y?x??b?a?216.
15.设函数g?x?在x?0处的某邻域内有定义,且g?x?在x?0处可导,
1???gxarctan2,x?0,? f?x???x?0,x?0.?
其中g?0??g??0??0,求f??0?.
解:f??0??limf?x??f?0?x?0g?x??g?0?x?0x?0?limg?x?xx?0arctan1x2
?limx?01?.limarctan2?g??0???0. x?02x29
16.设函数f?x??limxe4n?x?1??ax?b3??,其中a,b为常数,则a,b取何值时,函数f?x?n??enx?1?1在其定义域内处处可导. 解:(一) 1. 当x?1时,f?x??1?a?b2;
2. 当x?1时,f?x??ax3?b; 433. 当x?1时,f?x??limxen?x?1??ax?b43e?n(x?1)?be?n(x?1)n??en?x?1??1?limx?axn??1?e?n?x?1? ?x4. 即
?ax3?b,x?1,? f?x????1?a?b,x?1, ?2?x4?,x?1.(二)要使得f?x?在其定义域内处处可导,只须保证f?x?在x?0处可导. 1.要使f?x?在x?0处可导,首先要保证f?x?在x?0处连续,为此有 limf?x??limfx?1??x??f?1?
x?1?因此有 a?b?1 2.要使f?x?在x?0处可导,必须要保证f???1??f???1? ff?x??f?1?ax3?b?f?1????1??limx?1?x?1?limx?1?x?1(因为①)
32 ?lima?x?1??alim?x?1??x?x?1?x?1?x?1x?1?x?1?3a;
?1??limf?x??f?1?x4?1x??1??x3?x2 f?x?1???.
x?1?x?1?limx?1?x?1?limx?1?x?1?4因此,有 3a?4?a?43.
所以 b??13.
30
①
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