x1x22costdt?1costdt?x??0f(x)?f(0)0x解. f?'(0)?lim ?lim?lim2??x?0?x?0x?0xxx1?x22cosx?12 ?lim?lim?0 x?0?x?0?2x2x2(1?cosx)?12f(x)?f(0)2(1?cosx)?x2x f?'(0)?lim ?lim?lim3???x?0x?0x?0xxx2sinx?2x2(cosx?1)?lim?0 ?lim2?x?0?x?06x3x 所以 f'(0)?0, f(x)在x?0处连续可导.
12. 设f(x)在x = 0的某领域内二阶可导, 且lim??sin3xf(x)??2??0, 求3x?0x??xf(0),f'(0),f''(0)及limx?0f(x)?3. 2xsin3x?f(x)sin3x?xf(x)?sin3xf(x)?x?2??lim?lim?0. 所以 解. lim?332x?0x?0x?0xxxx?? lim?3x?sin??f(x)??0. f(x)在x = 0的某领域内二阶可导, 所以f(x),f'(x)在x = 0
x?0?x?连续. 所以f(0) = －3. 因为
sin3xsin3x?f(x)?3?f(x)?3 limx2?0, 所以limx?0, 所以 2x?0x?0xxsin3x3?f(x)?3sxx?lim3x?sin3x?lim3?3co3 lim ?limx?0x?0x?0x?0x2x2x33x23sin3x9? =limx?02x2f(x)?f(0)f(x)?3f(x)?39?lim?limx??0??0 f'(0)?lim2x?0x?0x?0x?0xx2f(x)?39?, 将f(x)台劳展开, 得 由lim2x?0x21f(0)?f'(0)x?f''(0)x2?0(x2)?31992!f''(0)? lim, 所以, 于是 ?x?022x22f''(0)?9.
第二章 导数与微分
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一. 填空题 1. f(x)? f(k?1)1?x, 则f(n)(x)= _______. 1?x(?1)k?12?(k?1)!(?1)n2?n!(n), 所以f ??k?1?1n?1(1?x)(1?x)?x?1?t2d2y
2. 设? , 则2?______.
dx?y?costd2y?sint?dt2tcost?2sint1sint?tcostdy?sint??????解. , ??223dx4t2t4tdx2t?2t?tdx3. 设函数y = y(x)由方程ex?y'?cos(xy)?0确定, 则
dy?______. dx解. ex?y(1?y')?(y?xy')sinxy?0, 所以
ysinxy?ex?y y'?x?y
e?xsinxy4. 已知f(－x) =－f(x), 且f'(?x0)?k, 则f'(x0)?______. 解. 所以 f'(x0)?f'(?x0)?k
f(x0?m?x)?f(x0?n?x)?____(m?n)f'(x0)___.
?x?0?xf(x0?k?x)?f(x0)11?f'(x0), 则____k?____. 6. 设lim?x?0?x335. 设f(x)可导, 则lim7. 已知
d??1??1?1?, 则f'f????_______. ??2??dx??x??x?2?x2?1?21?1??1?解. ?f'?2??3?, 所以f'?2???. 令x2 = 2, 所以f'?2???1
x2?x?x?x??x?8. 设f为可导函数, y?sin{f[sinf(x)]}, 则解.
dy?_______. dxdy?f'(x)cosf(x)f'[sinf(x)]cos{f[sinf(x)]} dx2x?y9. 设y = f(x)由方程e为_______. y?1?
二. 单项选择题
?cos(xy)?e?1所确定, 则曲线y = f(x)在点(0, 1)处的法线方程
1x, 即 x－2y + 2 = 0. 2 7
1. 已知函数f(x)具有任意阶导数, 且f'(x)?[f(x)]2, 则当n为大于2的正整数时, f(x)的n阶导数是
(a) n![f(x)]n?1 (b) n[f(x)]n?1 (c) [f(x)]2n (d) n![f(x)]2n 解.. (a)是答案.
2. 设函数对任意x均满足f(1 + x) = af(x), 且f'(0)?b, 其中a, b为非零常数, 则 (a) f(x)在x = 1处不可导 (b) f(x)在x = 1处可导, 且f'(1)?a (c) f(x)在x = 1处可导, 且f'(1)?b (d) f(x)在x = 1处可导, 且f'(1)?ab 解., 所以. (d)是答案
注: 因为没有假设f(x)可导, 不能对于f(1?x)?af(x)二边求导. 3. 设f(x)?3x3?x2|x|, 则使f(n)(0)存在的最高阶导数n为
(a) 0 (b) 1 (c) 2 (d) 3 所以n = 2, (c)是答案.
4. 设函数y = f(x)在点x0处可导, 当自变量x由x0增加到x0 + ?x时, 记?y为f(x)的增量, dy为f(x)的微分, lim?y?dy等于
?x?0?x?y?dyo(?x)?lim?0. (b)是答案.
?x?0x?0?x?x(a) －1 (b) 0 (c) 1 (d) ? 解. 由微分定义?y = dy + o(?x), 所以lim1?2x?0?xsin5. 设f(x)?? 在x = 0处可导, 则 x
x?0??ax?b(a) a = 1, b = 0 (b) a = 0, b为任意常数 (c) a = 0, b = 0 (d) a = 1, b为任意常数
解. 所以 0 = a. (c)是答案. 三. 计算题
1. y?ln[cos(10?3x)]，求y'
2?sin(10?3x2)?6x解. y'???6xtan(10?3x2) 2cos(10?3x)2. 已知f(u)可导, y?f[ln(x?a?x2)]，求y'
2解. y'?f'[ln(x?a?x)]??2x?1??x?a?x2?2a?x21?? ?? =
f'[ln(x?a?x2)]a?x2
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3. 设y为x的函数是由方程lnx2?y2?arctany确定的, 求y'. x解.
y'x?y22x?2yy'x ?22222yx?y2x?y1?2xx?y x?y x?yy'?y'x?y, 所以y'??x?etsintd2y4. 已知?, 求. 2tdx?y?ecostdyetcost?etsintcost?sint?t?解. , tdxecost?esintcost?sintd2yd?cost?sint?dt?(cost?sint)2?(cost?sint)21???? ??dxdx2dt?cost?sint?dx(cost?sint)2dtd2y2 ??dx2et(cots?sint)35. 设x?y?y，u?(x?x)解. dx?(2y?1)dy, du?
223/2, 求
dy du(2y?1)dy ?32dux?x(2x?1)dx2dy2 ?2du3(2y?1)x?x(2x?1)132(x?x)2(2x?1)dx 2dx
?x?f(t)??dyd2y6. 设函数f(x)二阶可导, f'(0)?0, 且?, 求, . 23tt?0t?0dxdx?y?f(e?1)dyf'(e3t?1)3e3tdy解. , 所以=3. ?dxf'(t)dxt?0d2y[f''(e3t?1)3(e3t)2?3e3tf'(e3t?1)]f'(t)?e3tf'(e3t?1)f''(t) ?3dx2[f'(t)]3所以
d2y[3f''(0)?3f'(0)]f'(0)?f'(0)f''(0)9f'(0)?6f''(0) ?3?232dxt?0[f'(0)][f'(0)]
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?x?tet7. 设曲线x = x(t), y = y(t)由方程组?t确定. 求该曲线在t = 1处的曲率. y?e?e?2eetetetdyyt'et?2e1解. yt'??y?t. 所以 ??t?ttee?2edxxt'e?te(1?t)(e?2e) 所以
dy1??.
dxt?12e?dtd2yd?12et?2e?tet ?????2t3t2t??dxdt?(1?t)(e?2e)?dx(1?t)(e?2e)ed2y1 所以 . 在t = 1的曲率为 ??22dxt?18e|y''|(1?y'2)t?132 k??18e21???1?2??4e?x?0 x?032?e(1?4e)
?2?32四. 已知当x ? 0时, f(x)有定义且二阶可导, 问a, b, c为何值时 F(x)??二阶可导.
?f(x)2?ax?bx?c
F(x)?limF(x), 所以c = f(－0) = f(0); 解. F(x)连续, 所以lim??x?0x?0因为F(x)二阶可导, 所以F'(x)连续, 所以b = f?'(0)?f'(0), 且 F'(x)??x?0?f'(x)
2ax?f'(0)x?0??F''(0)存在, 所以F?''(0)?F?''(0), 所以
2ax?f?'(0)?f'(0)f'(x)?f'(0)?lim?2a, 所以 ?x?0x?0xx1 a?f''(0)
2 lim?x2(n)，求f(0). 五. 已知f(x)?21?x解. f(x)??1?1111??? 21?x21?x10