第二章：一元函数微分.pdf

【解解】)(xf1 x0 x在在和和不不可可导导,故故选选（C C）xnnxnnexxxf,1maxe|1lim)(0e0111xxxxx,maxlim121iminnmnnnaaaa ),2,1(0miai 其其中中.1)(212 dtexfxt21 x1 y224416)(,2)(xxxexfexf )(1)(xfy dydxxfdxdy )(1)(【解解】知知，时时 233188)21()21()1(eeeff )(1)()(2xfxfxf 由由【解解】)1)(2(11211)(2 xxxxxf)2(1)1(13!)1()(11)(nnnnxxnxf)2111(311 xxaxxxxxf 432)(234)12)(2)(1(24664)(23 xxxxxxxf0)(xf.1,21,2 xxx【解解】令令,则则令令,得得,4)1()2(aff4 aaf 1617)21(axxxxxf 24683)(234)1)(2)(1(1224122412)(23 xxxxxxxf0)(xf.2,1,1 xxx【解解】令令,则则令令,得得,8)2(,13)1(,19)1(afafaf ,0)2(,0)1(,0)1(fff.813 a .utx xxxduuufduufxduufux000)()()()(【解解】（令令)xdttxtfxF0)()(kxxxkxbxxduuufduufxbxxxF 00202021)()(lim21)(lim100)()()(lim kxxbkxxxxfxxfduuf20)1(1)(lim kxxkbkxf 20)1ln()(limxxxxfx 2)(2()(lim2220 xxxxxxfx 由由0 x1 y0ee11 yyyxy10 xy【解解】令令得得.0)1(e1 yyyy.0e)2(12 yyyyy,cos)sin(lndd xyyxyfxz.0dd0 xxz,sin)()sin(lncos)sin(lndd22222 xyyyyxyfxyyxyfxz1)12)(0(dd022 fxzx.20 xy【解解】0|)6()0(,2|)23()0(002 tttxtx tyuut10de2t,1e)0(y;2)0(2ey 等等式式 两两端端对对求求导导可可得得 ,210 edxdyt2edd2022 txy则则utx 2duufxduufxxxxxx sin02sin0222)(1)(1)()0(x.0)0(【解解】令令，则则 由由已已知知得得0 x xxxxxxxxfxduufxxsin022232)cossin2()sin(1)(2)(xxxxxxxfduufxsin0232)cossin2)(sin()(2（1 1）当当时时，有有xxx)0()(lim)0(0 xxxduufxsin0302)(1lim)(sinlim320 fxxxx.2)0(f .0,2;0),cossin2)(sin()(2)(sin0232xxxxxxxfduufxxxx xxxxxxxxxfduufxxsin023002)cossin2)(sin()(2lim)(lim)0(2)0(3)0(2 ff【解解】,0232 xyyxyyyy0 yxy 01223 xx1 x,012)13(2)23(22 yyyyxyy021)1,1(y【解解】令令，得得，将将此此代代入入原原方方程程有有.,12dd txy3221ddtxy 【解解】（I I）由由于于（I II I）),1(12)4(200200 txttty.0,1 yx02020 tt10 t20 t令令，得得解解得得或或（舍舍去去）.切切线线方方程程为为1 xy.2121dd)1(xyxxS.37d)4(2291032 ttt 1022)1d()4(29ttt)0(,sin)(3 xxxxxf【解解】令令xxxffcos13)(,0)0(2 )(lim,02)0(xffx0sin6)(xxxf),0(x由由于于 ),0(0 x0)(0 xf则则存存在在唯唯一一的的使使，且且),0(0 xx)(,0)(xfxf ;0)(xf 当当时时,单单调调减减,),(0 xx)(,0)(xfxf 当当时时,单单调调增增,又又0)(0 xf )(limxfx,de)(302xxtxfxt )(xf),(【解解】令令则则是是上上的的奇奇)0,(),0(),0(函函数数,从从而而,原原方方程程在在区区间间和和上上实实根根个个数数相相同同，上上实实根根个个数数。因因此此,只只需需讨讨论论13)(,0)0(22 xexffx )(lim,02)0(xffx062)(2 xxexfx),0(x又又 ),0(0 x0)(0 xf则则存存在在唯唯一一的的使使,且且),0(0 xx 0)(xf 当当时时，),(0 xx0)(xf0)(0 xf )(limxfx当当时时，),0(0 x),(0 x则则原原方方程程在在区区间间上上无无实实根根，在在区区间间上上有有唯唯一一实实根根。故故原原方方程程共共有有三三个个实实根根.,1e2axx .1e)(2axxfx 【解解】原原方方程程变变形形得得 令令0)2(e)2(e)(2 xxxxxfxx2,0 xx令令得得)0,(x)(,0)(xfxf 当当 时时，单单调调减减.)2,0(x)(,0)(xfxf 当当 时时，单单调调增增.),2(x)(,0)(xfxf 当当 时时，单单调调减减.)1e(lim)(lim2axxfxxxafaf1e4)2(,01)0(2 01)1e(lim)(lim2 aaxxfxxx1 1）,4e02 a一一个个根根；2 2）,4e2 a两两个个根根；3 3）,4e2 a三三个个根根.xxxxf e)1ln()(xxxxxxxxfe111ee11)(0e)1(1e2 xxxx)0(x【证证】令令 ,0)0(f)0(,0)(xxf 则则)(xf.4sin2xeexxx .4cos2)(xxeexxf.sin2)(xxeexxf .0cos2)(xxeexf.0)0()0()0(fff【证证】令令则则axaxaaln)()ln()ln(ln)()(xaaaxaxf ,0ln)(xaaaxf【证证】只只需需证证设设 ，则则baabalnln2.lnln2bbaab bbaalnln.lnlnbbaa xxxgxxxfln)(,ln)(【证证】只只要要证证和和即即和和分分别别证证明明单单调调增增即即可可.【证证】由由 13210)(3)(dxxfdxxf得得，132132320)(3)()(dxxfdxxfdxxf则则 132320)(2)(dxxfdxxf，由由积积分分中中值值定定理理得得)(32)(32 fcf)1,32(),32,0(c 从从而而)()(fcf 令令)()()()(fxfexFxg ),(c 0)(F则则存存在在,使使 ).()()(22xfxfxF )0,2()()2(0)2()0(aafff【证证】令令 由由拉拉格格朗朗日日定定理理知知：)2,0()(02)0()2(bbfff,1|)2(|)0(|21|)(|ffaf1|)2(|)0(|21|)(|ffbf.2)()()(.2)()()(2222 bfbfbFafafaF而而4)0()0()0(22 ffF)(),(xFba ,ba0)(F则则在在点点取取到到它它在在上上的的最最大大值值,则则2)(!2)()()()(cxfcxcfcfxf 214)(cf ),0(1c 22)1(4)(cf )1,(2c 【证证】21)0(!2)()0)()()0(xfxxfxff （1 1）22)2(!2)()2)()()2(xfxxfxff （2 2）（2 2）式式减减（1 1）式式得得)()2)(21)(2)0()2(2122xfxfxfff )|)(|)2(|)(|412|)2(|)0(|)(|2122xfxfffxf 24411)2(41122 xx