第二章 导数计算及应用
n??n?????1?x2?n?n?1?cos?x???2nxsin?x??
2?2?????11?1?11?e?xe??2?1?ex?ex?x??x,x?0 37. 解：f??x??1212?????1?ex??1?ex?????????1x1xhf???0??lim?n?0f?h??f?0?1?e?limh?0?hhf?h??f?0??limh?0?h11?e1h?0，
?1。
f???0??lim?h?01h所以f??0? 不存在。 38、解f??x??arctan1??x?2?x?21?1?1????x?2?2
?x?2?，x?2 1?arctan?x?2?x?2?2?131harctanf?2?h??f?2?h?? f???2??lim?limh?0?h?0?h2hf?2?h??f?2?1?f???2??lim?limarctan??
h?0?h?0?hh2因
f???2??f???2? 故f??2?不存在
23???x?1??x?1?,x??139、解：f?x???
22????x?1??x?1?,x??13222?2x?1x?1?x?1?3x?1?x?1x?1????????????3x?1?,x??1?? f??x???2????x?1??x?1??3x?1?,x??1f??1?h??f??1???2?h?h3?0f????1??lim?lim?0
h?0?h?0?hh2 58 / 49
第二章 导数计算及应用
f??1?h??f??1???2?h?h3?0f????1??lim?lim?0
h?0?h?0?hh2?f???1??0
?2?x?3??x?1?,x??1???3?x??x?1?,?1?x?3 ?2??x?3??x?1?,x?3??240、解：y?2?x?3??x?1??2x?2x?3ln2??2x?2?,x??1?2?y???2?x?2x?3ln2???2x?2?,?1?x?3
?x2?2x?32ln2??2x?2?,x?3??f??1?h??f??1?e?4?h?h?1?4 f????1??lim?limh??1?0h??1?0hhf??1?h??f??1?e??4?h?h?1??4 f????1??lim?limh??1?0h??1?0hh故x??1，y不可导;
f?3?h??f?3?eh?4?h??1?4 f???3??lim?limh?3?0h?3?0hhf?3?h??f?3?e?h?4?h??1??4 f???3??lim?limh?3?0h?3?0hh?x?3时，y?x?不可导
211?2xsin?cos,x?0?41、解（1）f??x??? xx??2x,x?0f???0??limf?h??f?0??limh?0?hh2sinhh?0?1h?0
f?h??f?0?h2?0，故f??0??0 f???0??lim?limh?0?h?0?hh（2）limf??x??lim?2xsinx?0?x?0?故f??x?在x?0处不连续
42、解ylny?x?0，方程两边对x求导
??11??cos?不存在 xx?lny
dyy??y?1?0 dxy 59 / 49
第二章 导数计算及应用
?lny?1?y??1?0, ???
y???1
lny?11?y??2?0，得 y对（*）两端再次对x求导，
y???lny?1??y????y?lny?1??y??2??1y?1?lny?3
?2?x2,?2?x?2??2x,?2?x?243、解：f(x)??，f?(x)??
?0,x?2或x??2?2,x?2或x??2函数f?x?在x??2时间断，故x??2时，f?x?不可导。
x(g?(x)?e?x)?g(x)?e?x44、解：（1）f?(x)?，x?0
x2g(h)?eh?0f(h)?f(0)g(h)?e?hh?limf?(0)?lim?lim
h?0h?0h?0hhh2g?(h)?e?hg??(h)?e?hg??(0)?1??lim ?lim
h?0h?022h2（2）当x?0时，由g?(x)的连续性知f?(x)连续 g?(x)?e?x?x(g??(x)?e?x)?g?(x)?e?xlimf?(x)?lim x?0x?02xg??(x)?e?xg??(0)?1??lim?f?(0)
x?022f?(x)在x?0处连续。
综合得f?(x)在(??,??)上处连续。
45．证明：设切点为(x0,y0)且满足x0?y0?a232323
2?32?3dyxx?yy??0???()3 33dxy111yyk??(0)3，切线方程为y?y0??(0)3
x0x0yx令x?0得y?y0?(0)3x0，令y?0得x?x0?(0)3y0。
x0y0切线于坐标轴之间的长度：
1111 60 / 49
第二章 导数计算及应用
l?(y0?32323y0xx0)2?(x0?30y0)2 x0y02322302323?(y0(y0?x0)?x(x0?y0)2
?(y0?x0)?(a)?a
2323322332xy'?y2xy'?yx?2yy'12x?2yy'x46．解：，， ??2222y22x2?y2x?yx?y1?()xx?y?2y'?y?x?2yy'，y'?
2(1?y)g(x)?cosxg(x)?sinx?lim?g?(0) 47．解：（1）limf(x)?limx?0x?0x?0x1由x?0处连续，可知a?f(0)?limf(x)?g?(0)
x?0x(g?(x)?sinx)?g(x)?cosx
x2f(h)?f(0)g(h)?cosh?g?(0)hg?(h)?sinh?g?(0)f'(0)?lim?lim?lim 2h?0h?0h?0hh2hg??(h)?coshg??(0)?1?lim? h?022xf'(x)?g(x),g??x?在x?0处连续 48．解：当x?0时，g?(x)?x2（2）当x?0时，f'(x)?f(h)?f?(0)g(h)?g(0)f(h)?f?(0)hh g?(0)?lim?lim?lim2h?0h?0h?0hhhf?(h)?f?(0)f??(h)f??(0)?lim?lim? h?0h?02h22xf?(x)?f(x)f??xf??(x)?f?(x)f??(x)f??(0)?lim?lim?因limg?(x)?lim 2h?0x?0h?0x?0x2x22f??(0)??所以limg(x)?g(0)?，故g?(x)在=0处连续。 x?02综上所述g(x)有一阶连续导数。
11?21?x2xx49．(1).原式=lim?lim?lim?1. 2x???arccotxx???x???1x?21?x （2）原式=ex?0?limsinx?lnx?ex?0?limxlnx?elnxx?0?1/xlim?ex?0??1/x2lim1/x?e0?1
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第二章 导数计算及应用
(3) 原式= lim1?x(lnx?1)?limx?1x?11?1?xx?xx(lnx?1)2?xx.?1x21x
1??xx?(lnx?1)2??x??lim??2 x?11x2xlnx?1?xxlnx?1?x?lim (4) 原式=lim
x?1(1?x)lnxx?1(1?x)ln(1?x?1)1xlnx?1?xlnx?1?1x??1。 ?lim?lim?lim2x?1x?1?2(x?1)x?1?22??x?1?x(ex?1)?2(ex?1)ex?1?xex?2ex?lim (5)原式=lim
x?0x?0x33x2ex?ex?xex?2exex1?lim?lim?。 x?0x?066x6xx1xcosx?x?(1?cos)1?cossin2?lim2?lim22?1。 (6) 原式= limx?0x?01sinx(1?cosx)?cosxx?0sinx450、(1) f(x)?lnx在区间[1,1+x]上满足拉格朗日中值定理条件, 故存在??(1,1?x),使得:
1ln(1?x)?ln1?f????(1?x?1）?x
?所以，
x1?ln(1?x)?x?x。 1?x?3(2) f(x)?x在区间[a,b]上满足拉格朗日定理条件,故存在??(a,b),使
b3?a3?f????(b?a)?3?2(b?a)
所以,3a(b?a)?b?a?3?(b?a)?3b(b?a) （3）令F(x)?1?xln(x?1?x2)?1?x2，F?0??0
23322F?(x)?ln(x?1?x2)?x1?x2?x1?x2?ln(x?1?x2)?ln1?0(x?0)。
故F(x)在x?0上严格单调上升，F(x)?F(0)?0即原命题得证。 （4）令F(x)?2x?3?111?2?0(x?1)，所以F(x)，F(1)?21?3?1?0，F?(x)?xxx在x?1时严格单调上升，可知F(x)?F(0)?0，即原命题得证。
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