数学分析讲义第四版刘玉琏傅沛仁著高等教育出版社课后答案第五单元.pdf

8 111?1 ?SSS1.1(1)?SSS1.2(5) ?SSS1.3(9) 111?444? 13 ?SSS2.1(13)?SSS2.2(21) ?SSS2.3(36)?SSS2.4(40) ?SSS2.5(51) 111nnn?YYY 57 ?SSS3.1(57)?SSS3.2(64) 111ooo?YYY555 75 ?SSS4.1(75)?SSS4.2(83) 111? 95 ?SSS5.1(95)?SSS5.2(100) ?SSS5.3(109)?SSS5.4(113) ?SSS5.5(114) 111888?nnn444?AAA.125 ?SSS6.1(125)?SSS6.2(136) ?SSS6.3(142)?SSS6.4(148) 111? . 170 ?SSS7.1(170)?SSS7.2(172) ?SSS7.3(179)?SSS7.4(187) 111lll? . 194 ?SSS8.2(194)?SSS8.3(204) ?SSS8.4(210)?SSS8.5(232) 111? . 244 ?SSS9.1(?)(244)?SSS9.1(?)(252) ?SSS9.1(nnn)(266)?SSS9.2(?)(273) ?SSS9.2(?)(290)?SSS9.3(298) ?SSS9.4(309) 1 111?.323 ?SSS10.1(323)?SSS10.2(132) ?SSS10.3(334)?SSS10.4(352) 111? . 366 ?SSS11.1(366)?SSS11.2(375) ?SSS11.3(378)?SSS11.4(389) 111?222?CCC? 393 ?SSS12.1(393)?SSS12.2(400) ?SSS12.3(406) 111?nnn? . 425 ?SSS13.1(?)(425)?SSS13.1(?)(430) ?SSS13.2(493) 111?ooo?.465 2 1? ? S K5.1 (56?1 157 ?) 3. ?y = x2?y = 2 x23?:?(?)?. ) ?Jn)?:?I(1,1)?(1,1). ?y = x2?y = 2 x23:(1,1)?Ok1?k2,k k1= lim x0 y x = lim x0 (1 + x)2 12 x = lim x0(2 + x) = 2. k2= lim x0 y x = lim x0 2 (1 + x)2 2 + 12 x = lim x0(2 x) = 2. u,tg1= k2 k1 1 + k1k2 = 3 4, =1= arctg 3 4. ?3:(1,1)?2= arctg 3 4. 3. ?,e?3:x?: (3) f(x) = x + 1,x 1. ) f 0(x) = lim x0 p(x + x) + 1 x + 1 x = lim x0 1 p(x + x) + 1 +x + 1= 1 2x + 1 (4) f(x) = sin3x. ) f 0(x) = lim x0 sin3(x + x) sin3x x = lim x0 2sin 3x 2 cos 6x + 3x 2 x = lim x0 3 sin 3x 2 3x 2 cos 6x + 3x 2 = 3cos3x. 6. f(x) = x2,x c, ax + b,x 0,K 0,x (a,a + ),kf(a) 0,?4?5, 0,x : a a,x a 0.u,f(x) f(a) 0f(x) a,=x a 0?,x a?,k f(x) f(a) x a g(x) g(a) x a h(x) h(a) x a . fi ? lim xa+ f(x) f(a) x a = f 0 +(a)? lim xa+ h(x) h(a) x a = h 0 +(a),?f 0 +(a) = h 0 +(a).?Y n, lim xa+ g(x) g(a) x a ?3,=g(x)3am?,? f 0 +(a) = g 0 +(a) = h 0 +(a). ?yf 0 (a) = g 0 (a) = h 0 (a). 13. ?f(x)3a?,f(a) 6= 0,4?: lim n “f?a +1 n ? f(a) #n ) lim n “f?a +1 n ? f(a) #n = lim n “f(a) + f?a +1 n ? f(a) f(a) #n = lim n “ 1 + f ? a + 1 n ? f(a) f(a) #n = lim n (“ 1 + f ? a + 1 n ? f(a) f(a) # f(a) f ? a+1 n ? f(a) )1 f(a) f(a+1 n)f(a) 1 n =e 1 f(a)f 0(a) = e f0(a) f(a) 14. ?f(x)3a?,4?: (2) lim h0 f(a) f(a h) h . ) lim h0 f(a) f(a h) h = lim h0 f(a h) f(a) h = f 0(a). (4) lim t0 f(a + 2t) f(a + t) 2t . ) lim t0 f(a + 2t) f(a + t) 2t = lim t0 f(a + 2t) f(a) f(a + t) f(a) 2t 5 =lim t0 f(a + 2t) f(a) 2t 1 2 lim t0 f(a + t) f(a) t = f 0(a) 1 2f 0(a) =1 2f 0(a). 16. ef(x)3aY,?f(a) 6= 0,?f(x)23a?,Kf(x)3a? y fi ?f(x)3aY,K lim x0 f(a + x) = f(a). qfi ?f(x)23a?,=4? lim x0 = f(a + x)2 f(a)2 x ?3. ? lim x0 = f(a + x)2 f(a)2 x = A.u,4? lim x0 = f(a + x) f(a) x = lim x0 = f(a + x) f(a)f(a + x) + f(a) xf(a + x) + f(a) = lim x0 = f(a + x)2 f(a)2 x 1 f(a + x) + f(a) = A 2f(a), =f(x)3a?. ? S K5.2 (56?1 175 ?) 1. e?: (1) y = x4+ 3x2 6 ) y 0 = 4x3+ 6x (3) y = (1 + 4x3)(1 + 2x2) ) y 0 = 12x2(1 + 2x2) + 4x(1 + 4x3) = 40x4+ 12x2+ 4x = 4x(10x3+ 3x + 1). (5) y = x3+ 1 x2 x 2. ) y 0 = 3x2(x2 x 2) (x3+ 1)(2x 1) (x2 x 2)2 = x4 2x3 6x2 2x + 1 (x2 x 2)2 (7) y = xtgx ctgx. 6 ) y 0 = tgx + x cos2x + 1 sin2x = tgx + xsecx+csc2x. (9) y = 1 lnx 1 + lnx. ) y 0 = 1 x(1 + lnx) (1 lnx) 1 x (1 + lnx)2 = 2 x(1 + lnx)2. (11) y = arctgx x . ) y 0 = ?1 xarctgx ?0 = 1 x2 arctgx + 1 x 1 1 + x2 = 1 x2 ? x 1 + x2 arctgx ? . (13) y = x2arccosx. ) y 0 = 2xarccosx x2 1 x2 (15) xsinxlnx. ) y 0 = sinxlnx + cosxlnx + xsinx 1 x = sinx(lnx + 1) + xlnxlnx 2. e?: (1) y = (2x2 3)2. ) y 0 = 2(2x2 3) 4x = 8x(2x2 3). (3) s 1 + x 1 x. ) y 0 = 1 2 s 1 + x 1 x 1 x (1 + x)(1) (1 x)2 = 1 (1 x)1 x2. (5) q x + px +x . ) y 0 = 1 q x + px +x h 1 + 2px + x ? 1 + 1 22 ?i = 1 + 2x + 4xpx + x xpx +xqx +px +x. 7 (7) y = sin2xcos3x. ) y 0 = 2cos2xcos3x 3sin2xsin3x. (9) y = asin3 x 3. ) y 0 = 3asin2 x 3 cos x 3 1 3 = asin2 x 3 cos x 3. (11) y = lntgx. ) y 0 = 1 tgx 1 cos2x = 1 sinxcosx = 2 sin2x. (13) y = (xctgx)2. ) y 0 = 2xctgx ? ctgx x sin2x ? = 2xctgx(ctgx xcsc2x). (15) y = log3(x2 sinx) ) y 0 = 1 (x2 sinx)ln3 (2x cosx) = 2x cosx (x2 sinx)ln3. (17) y = xln(x + 1 + x2) 1 + x2. ) y 0 = ln(x + p 1 + x2) + x ? 1 + 2x 21 + x2 ? x + 1 + x2 2x 21 + x2 = ln(x + p 1 + x2) (19) y = ln 1 + x2 x 1 + x2 + x. )? ) y 0 = 1 1 + x2 x 1 + x2 + x (x2+ 1 + x) ? x 1 + x2 1 ? (x2+ 1 x) ? x 1 + x2+ 1 ? (x2+ 1 + x)2 = 2 x2 + 1 )? ,y = ln 1 + x2 x 1 + x2 + x = ln(1 + x2 x) (1 + x2+ x). ) y 0 = 1 1 + x2 x ? x 1 + x2 1 ? 1 1 + x2 + x ? x 1 + x2+ 1 ? 8 = 2 x2 + 1 5 l?)?)?w?,?,kA?5?z ?.,?2?. (21)y = a2 + x2 aln a + a2 + x2 x . ) y = p a2+ x2 alna + p a2+ x2 lnx. y 0 = x a2 + x2 a ? 1 a + a2 + x2 x a2 + x2 1 x ? = x a2 + x2 + a2 xa2+ x2 = a2 + x2 x (23) y = 7x 2+2x. ) y 0 = 7x 2+2x ln7 (2x + 2) = 2(x + 1)7x 2+2x ln7. (25) y = ln ex 1 + ex. ) y = lnex ln1 + ex= x ln(1 + ex). y 0 = 1 ex 1 + ex = 1 1 + ex (27) (arcsinx)2. ) y 0 = 2arcsinx 1 1 x2= 2arcsinx 1 x2. (29) arctg 2x 1 x2. ) y 0 = 1 1 + ? 2x 1 x2 ?2 2(1 x2) 2x(2x) (1 x2)2 = 2 1 + x2. (31) y = xa2 x2+ a2arcsin x a. ) y 0 = a2 x2+ x2 a2 x2 + a2 1 s 1 ?x a ?2 = 2a2 x2. (33) y = arcsin(sinx). ) y 0 = 1 p 1 sin2x cosx = cosx |cosx| 9 = 1,x ? 2k 2,2k + 2 ? 1x ? 2(k + 1) 2,2(k + 1) + 2 ? . k Z (35) y = arctg a x + ln s x a x + a . ) y = arctg a x + 1 2ln(x a) ln(x + a) y 0 = 1 1 + ?a x ?2 ? a x2 ? + 1 2 ? 1 x a 1 x + a ? = 2a3 x4 a4. (37) y = x 1 x. ) y = e 1 x lnx. y 0 = e 1 x lnx? 1 x2 lnx x2 ? = x 1 x2(1 lnx). (39) y = (sinx)x ) y = exlnsinx y 0 = exlnsinx ? lnsinx + x cosx sinx ? = (sinx)x(lnsinx + ctgx) 4. y:ef1(x),f2(x), ,fn(x)3x?,?3x?0,?g(x) = f1(x)f2(x)fn(x),K g(x)3x?,? g 0(x) = g(x)hf 0 1(x) f1(x) + f 0 2(x) f2(x) + + f 0 n(x) fn(x) i y fi ?f1(x),f2(x), ,fn(x)3x?,Kf1(x)f2(x)fn(x),3x?,d? ?“,k g 0(x) = f 1(x)f2(x)fn(x) 0 = n X k=1 f1(x)f2(x)f 0 k(x)fn(x) = n X k=1 g(x) f1(x)f2(x)f 0 k(x)fk(x) f1(x)f2(x)fk(x)fn(x) = g(x) hf0 1(x) f1(x) + f 0 2(x) f2(x) + + f 0 n(x) fn(x) i 5. y:?;?,?fl ? A?. y ?f(x)?,=f(x) = f(x),k f 0(x) = lim x0 f(x + x) f(x) x = lim x0 f(x x) f(x) x = lim x0 f(x x) f(x) x = f 0(x) 10 =?. ?y,?. A?:?uy?,?z?:?3?,3 :(x,y)?(x,y)?=?K?,=?. 7. y:3?y = x2+ x + 1?I?x1= 0,x2= 1,x3= 1 2?n:?u? :. y y 0 = 2x + 1.?y = x2+ x + 13n:? ?:y 0 x1=0 = 1,y 0 x2=1 = 1,y 0 x3=1 2 = 0 ?: 1 y 0 x1=0 = 1, 1 y 0 x2=1 = 1, 1 y 0 x3=1 2 = . ?n:?I(0,1),(1,1), ? 1 2, 3 4 ? ,3?Ln:?n? x + y 1 = 0, x y + 2 = 0, x + 1 2 = 0. d)?A?,? ? ? ? ? ? ? ? 111 112 10 1 2 ? ? ? ? ? ? ? = 0 n?u?:. 8. y:e?y = xa?RR 0,Ky 0 = axa1 y 35.210,fi y:x 0,k(xa) 0 = axa1.e?y = xa?R 0,= R,?x 6= 0,d?,k y 0 = (xa) 0 = lim x0 (x + x)a xa x = lim x0 xa h? 1 + x x ? 1 i x = lim x0 xa x ? 1 + x x ? x x (d3.29) e?y = xa?RK:0uy = xa?, d?ka 1?,? y = xa3:?,? (xa)|x=0= lim x0 (0 + x)a 0a x = lim x0(x) a1 = 0a 1, 1,a = 1. w,?y = xa,(a 1)?“y 0 = axa13“:. 11 9. y:efij(x)?(i,j = 1,2, ,n,)K ? ? ? ? ? ? ? ? ? ? ? f11(x)f12(x)f1n(x) . . . . . . . . . fk1(x)fk2(x)fkn(x) . . . . . . . . . fn1(x)fn2(x)fnn(x) ? ? ? ? ? ? ? ? ? ? ? 0 = n X k=1 ? ? ? ? ? ? ? ? ? ? ? f11(x)f12(x)f1n(x) . . . . . . . . . f 0 k1(x) f 0 k2(x) f 0 kn(x) . . . . . . . . . fn1(x)fn2(x)fnn(x) ? ? ? ? ? ? ? ? ? ? ? y dn?1?“?, ? ? ? ? ? ? ? ? ? ? ? f11(x)f12(x)f1n(x) . . . . . . . . . fk1(x)fk2(x)fkn(x) . . . . . . . . . fn1(x)fn2(x)fnn(x) ? ? ? ? ? ? ? ? ? ? ? = X i1,i2,in (1)f1i1(x)fkik(x)f1n(x)0 ?i1,i2, ,in1,2, ,n?,?;?(i1,i2, ,in)?_S, P i1,i2,inL (i1,i2, ,in)?k?,?kn!?.u ? ? ? ? ? ? ? ? ? ? ? f11(x)f12(x)f1n(x) . . . . . . . . . fk1(x)fk2(x)fkn(x) . . . . . . . . . fn1(x)fn2(x)fnn(x) ? ? ? ? ? ? ? ? ? ? ? = X i1,i2,in (1)f1i1(x)fkik(x)f1n(x)0 = X i1,i2,in (1) n X k=1 f1i1(x)f 0 kik(x)f1n(x) = n X k=1 X i1,i2,in (1)f1i1(x)f 0 kik(x)f1n(x) = n X k=1 ? ? ? ? ? ? ? ? ? ? ? f11(x)f12(x)f1n(x) . . . . . . . . . f 0 k1(x) f 0 k2(x) f 0 kn(x) . . . . . . . . . fn1(x)fn2(x)fnn(x) ? ? ? ? ? ? ? ? ? ? ? 10. ef(x)?g(x)3x?,e?: (3) y = g(x) pf(x) g(x) 6= 0,f(x) 0. ) y = e ln f(x) g(x) y 0 = e ln f(x) g(x) ? lnf(x) g(x) ? = g(x) pf(x)?f 0(x) g(x)f(x) g 0(x)lnf(x) g(x)2 ? 12 (4) y = logf(x)g(x)g(x) 0,f(x) 0. ) y = logf(x)g(x) = lng(x) lnf(x). y 0 = g 0(x) g(x) lnf(x) f 0(x) f(x) lnf(x) lnf(x)2 = g 0(x) g(x)lnf(x) f 0(x)lng(x) f(x)lnf(x)2 ? S K5.3 (56?1 183 ?) 1. e?(? dy dx (2) b2x2+ a2y2= a2b2. ) 2b2x + 2a2y y 0 = 0,y 0 = b2x a2y (4) x 2 3+ y 2 3= a 2 3. ) 2 3x 1 3+ 2 3y 1 3 y 0 = 0,y 0 = 3 s y x . (6) y = cos(x + y). ) y 0 = sin(x + y) (1 + y 0), y 0 = sin(x + y) 1 + sin(x + y) (8) sin(xy) = x. ) cos(xy)(y + xy 0) = 1, y 0 = 1 y cos(xy) xcos(xy) 2. A?,e?: (2) y = x2 1 x 3 s 3 x (3 + x)2 ) ln|y| = ln ? ? ? ? ? x2 1 x 3 s 3 x (3 + x)2 ? ? ? ? ? = 2ln|x| ln|1 x| + 1 3 ln|3 x| 2 3 ln|3 + x|. y 0 y = 2 x + 1 1 x 1 3(3 x) 2 3(3 + x), = 2 x x(1 x) + 3(9 x2) =y 0 = x2 1 x 3 s 3 x (3 + x)2 ? 2 x x(1 x) + x 9 3(9 x2) ? 13 (4) y = (x a1)1(x a2)2(x an)n,?a1,a2,an,1,2, ,n?. ) ln|y| = 1ln|x a1| + 2ln|x a2| + + nln|x an| y 0 y = 1 x a1 + 2 x a2 + + n x an = n X k=1 k x ak =y 0 = (x a1)1(x a2)2(x an)n n P k=1 k x ak 3. e?3?:? (1) x2+ 3xy + y2+ 1 = 0,3(2,1). ) 2x + 3(y + xy 0) + 2yy0 = 0,y 0 = 2x + 3y 3x + 2y. 3(2,1),?y 0 = |x=2 y=1 = 1 4. (3) 3 2x 8 y = 13(4,1). ) 3 2 3 x 2 3 1 8y 7 8 y 0 = 0,y 0 = 8 3 2y7 8 3x 2 3 . 3(4,1),?y 0 = |x=4 y=1 = 4 3. 4. e? dy dx (2) x = 3at 1 + t3, y = 3at2 1 + t3. ) dy dx = d dt ? 3at2 1 + t3 ? d dt ? 3at 1 + t3 ? = t(2 t3) 1 2t3 (3) ? x = acos3t, y = bsin3t. ) dy dx = d dt(bsin 3t) d dt(acos 3t) = 3bsin2tcost 3acos2tsint = b atgx. 7. e?d?I? = f()L,K?z?9? ? x = cos = f()cos, y = sin = f()sin. dy dx. ) dy dx = d d(f()sin) d d(f()cos) = f 0()sin + f()cos f 0()cos f()sin 8. A17Ky:?%9? = a(1 + cos)? = a(1 cos)3?:?R?. y ?J?,?%9?k?: ? a, 2 ? ?%9? = f() = a(1 + cos), = g() = a(1 cos). 14 ?%9?k?: ? a, 2 ? ?Ok1?k2,d17?“,k k1= dy dx ? ? ? ?= 2 = f 0()sin + f()cos f 0()cos f()sin ? ? ? ?= 2 = asinsin + a(1 + cos)cos asincos a(1 + cos)sin ? ? ? ?= 2 = 1, k2= dy dx ? ? ? ?= 2 = g 0()sin + g()cos g 0()cos g()sin ? ? ? ?= 2 = asinsin + a(1 + cos)cos asincos a(1 + cos)sin ? ? ? ?= 2 = 1. u,k1 k2= (1)(1) = 1, =?%9?3z?:?R?. ? S K5.4 (56?1 192 ?) 1. e?: (5) eaxcosbx. ) y 0 = aeaxcosbx beaxsinbx = eax(acosbx + bsinx). u dy = y 0dx = eax(acosbx + bsinx)dx. (6) y = arcsin 1 x2. ) y 0 = 1 q 1 (1 x2)2 2x 21 x2 = x |x| 1 1 x2 = 1 1 x20 0. ?4?S5,x 0,?|y x| 0. 20 ?y x?,kf(y) f(x); ?y x?,kf(y) f(x).u,x (x0 ,x0+ )?3+ ?(x x,x + x),f(x)3(x x,x + x)?O. x,y (x0 ,x0+ ),?x y.?k?CXn,?3?k?mm(xi xi,xi+ xi) ,i 1,2, ,n,?x1= x,xn= y?x1 x2 xn,?CX4mx,y,= x,y n i=1 (xi xi,xi+ xi). ?ai (xi,xi+1),i = 1,2, ,n 1,?ai (xi xi,xi+ xi) T(x i+1 xi+1,xi+1+ xi+1).u ,k f(x) f(a1) f(a2) f(an1) f(y), =f(x)3(x0 ,x0+ )?O. ?3.2n7,y = f(x)3(x0,x0+)?3?x = (y),?3(f(x0),f(x0+ ) Y,?O. ?gy,?x = (y),3y0= f(x0)?3?. ?5.2n4,y (f(x0 ),f(x0+ ),k 0(x) = 1 f 0(x). ?y 6= 0,y0+ y (f(x0 ),f(x0+ ),l?,kx = (y0+ y) (y0) 6= 0. x0+ x (x0 ,x0+ ),f(x0+ x) = y0+ y,?y 0 y 0.u lim y0 (y0+ y) 0(y 0) y = lim x0 1 f 0(x0 + x) 1 f 0(x0) f(x0+ x) f(x0) = lim x0 f 0(x 0+ x) f 0(x 0) x f 0(x0)f0(x0 + x) f(x0+ x) f(x0) x = f 00(x 0) f 0(x0)3, = 00(y 0) = f 00(x 0) f 0(x0)3. 12. y:f(x)ng?“,a?f(x) = 0?k(k n)? f(a) = f 0(a) = = f(k1)(a) = 0,f(k1)(a) 6= 0. y =ea?f(x) = 0?k(k n),Kf(x) = (x ak)g(x),?g(x)n kg? “,?g(a) 6= 0.d4?Z“,m = 1,2, ,k 1,k,k f(m)(x) = C0 m(x a) k(m)g(x) +C1 m(x a) k(m1)g0(x) + +C(l1) m (x a)k(ml+1)g(l1)(x) +C(l) m(x a) k(ml)g(l)(x) + + C(m) m (x a)kg(m)(x).(1) ?m k 1,?,(1)“?m?z?kx a?“,u f(a) = f(0a) = = f(k1)(a) =