7 - 198 C CHAPTERS DIFFERENTIATION RULES 67. (a) If y : f...

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Unformatted text preview: 198 C CHAPTERS DIFFERENTIATION RULES 67. (a) If y : f _1(m), then f : m. Differentiating implicitly with respect to a: and remembering that y is a function , dz; d 1 7' 0f$,Weg€tf(y)E=1,SOEZ:m => (f (b) f<4> : 5 : f"1<5> = 4-Bypart<a>v(f*1)’<5>= 1/f’(f’1(5))= 1/f’(4) : 1/<%) = 68. (a) f(:c) : 25c + cossc => f’(x) : 2 — since > 0 for all ac. Thus, f is increasing for all a: and is therefore one-to-one. (b) Since f is one—to—one, f_1(1) : k <=> f(k) = 1. By inspection, we see that f(0) = 2(0) + c030 : 1, so k : f’1(1) : 0. (c) (f’1)'(1)= 1/f’(f’1(1)) : 1/f’(0) : 1/(2 — sine) =% 69. x2 + 4:112 2 5 => 2x + 4(2yy') = 0 => 3/ = —4£y. Now let h be the height of the lamp. and let (a, b) be the point of tangency of the line passing through the points (3, h) and (—5, 0). This line has slope (h r 0)/[3 * (—5)] = éh. But the slope of the tangent line through the point ((1,1)) can be expressed as y’ : — b — 0 _ b . . a b a ; (—5) — a + 5 [smce the line passes through (*5, 0) and (a, b)], so —4—b = a + 5 4b2 = 7(12 ~ 5a <=> a2 + 4b2 : —5a. But a2 + 4b2 : 5 [since (a,b) is on the ellipse], so 5 = #511 <=> or as a = —1. Then 4b2 — (12 5a : 1 5( 1) ‘ 4 -> b — 1. since the point is on the top half of the ellipse. h b 1 8 , So e h i 2. So the lamp is located 2 units above the w—axis. 3.7 Higher Derivatives fig—J!!!“— 1. a : f, b = f’, c = f". We can see this because where a has a horizontal tangent, b : 0, and where b has a horizontal tangent, c : 0. We can immediately see that c can be neither f nor f’ . since at the points where c has a horizontal tangent, neither (1 nor b is equal to 0. 2. Where d has horizontal tangents, only c is 0, so (1' : c. c has negative tangents for :c < 0 and b is the only graph that is negative for x < 0, so c' = b. b has positive tangents on IR (except at a: 2 0), and the only graph that is positive on the same domain is a, so 17' : a. We conclude that d : f, c : f’, b : f", and a : f’”. 3. We can immediately see that a is the graph of the acceleration function, since at the points where a has a horizontal tangent, neither c nor b is equal to 0. Next, we note that a : O at the point where b has a horizontal tangent, so b must be the graph of the velocity function, and hence, b’ : a. We conclude that c is the graph of the position function. 4. a must be the jerk since none of the graphs are 0 at its high and low points. a is 0 where b has a maximum, so b’ = a. b is 0 where c has a maximum, so 0’ = b. We conclude that d is the position function, 0 is the velocity, b is the acceleration, and a is the jerk. 5.: x5 + 6302 7 7:16 :> f’(:£) : 5x4 +1256 — 7 => f”($) : 209:3 + 12 6. f(t) : t8 — 71:6 + 22:4 :> f’(t) : 8t7 7 425’ + 85’ => f”(t) : 56t6 7 210:4L + 241:2 7. y = 00320 => y' : *2sin29 :> y” : —4cos26 8. y : 6sin0 => y' : 0c0s6 +sin0 => y" : 6(~sin0) +cos6 - 1 +cos0 : 2cos€ ~ 65in6 SECTION 3.7 HIGHER DERIVATIVES D 9. F(t) = (1 — 71¢)“ => F'(t) = 6(1 7t)5( 7) 2 42(1 71)5 _> F”(t) : 742 - 5(1 — 7t)4(—7) : 1470(1 — 71:)4 199 2$+1 ($_1)(2)_(2x+1)(1) 2x—2—2m~1_ —3 or _3$_1—2 ‘° 9‘9”) = $71 :> 9'“) ($71? (ac—112 (ac—1)? ( ) 6 7 73 => g”(w) : ~3(—2)(w — 1) 3 : 6(at — 1) or (a; _1)3 1—4u I _ (1+3u><—4>—<1—4u><3> 2 1M 2 i 11. h(u) = 1+3” => h(u)— “ma—— (14.3102 (1+3u)2 7 42 _7(1 +3u)‘2 :> h”(u) : —7(—2)(1 + 3u)_3(3) = 42(1+ 3“) 3 or (1 + 3103 12. H(s) : a‘/§ + % : 6131/2 +bs—1/2 :» H'(s) : 6163—1” +b(~%s‘3/2) 2%(15—1/2 — %bs_3/2 => H"(s) _;a gym) _ M gym) gays/2 + 2134/2 I 1 2 —1/2 x : 2 : — 1 2 : i 13 h(:c) m+1 :> h($) 2($+) m2+1 h. _ 962 +1 H [W +1>‘1/2<296>] _ (x2 + 1) W W + 1) 421 Z 1 (m) ‘ ( $2 +1)2 _ (:c2 +1)1 (3:2 +1)3/2 14. y :zzceCm :> y 2x e” c+e”-1=e”(cx+1) => (3102) = 2562(133 -+-1)_1/3 => 4.27 16. 2 => y Vz+1 ,_\/:t+1-4—4x-%(:C+1)_1/2_4\/;L'+1—2z/\/x+1_4($+1)~2w_ 2x+4 y — ( /—$+1)2 x+1 (w+1)3/2 (m+1)3/2 y" _ (x+1)3/2.2~ (2;c+4).§(ac+1)1/2 _ (m+1)1/2[2(x+1)—3(x+2)] “Hus/2]? <w+1>3 _2:c+2—3:c~6_ —:I:—4 E (:1:+1)5/2 ‘(ar:+1)5/2 17. H(t) = tan3t :> H’(t) : 3sec2 3t :> H”(t) : 2 ~ 3sec 3t diff (sec 3t) 2 Gsec 3t (3 sec 3t tan 3t) 2 18 sec2 3t tan 3t 18. 9(3) 2 32 0053 => g'(s) : 2sc0ss — 32 sins :> 9"(3) : 2coss — 2ssins — ZSSins — 32 0053 = (2 — s2) coss — 4ssins 19. g(t) : t3e5‘ => g’(t) = 75%“ ~ 5 + e“ - 3t2 2 t2e5‘(51 + 3) => g”(t) = (2t)e5‘(5t + 3) + 12 (5e5t) (5t + 3) + t2e5‘(5) 2 t6“ [2(5t + 3) + 5t(5t + 3) + St] = t65‘(25t2 + 302: + 6) 200 3 CHAPTER3 DIFFERENTIATIONRULES 71 2 ' 1 2a: 20 h<$>=tan => h(3'3):1+(ac2)2 2m=1+$4 => h//(I) _ (1+;124)(2)_ (2x)(4m3) F 2_6x4 (1 “‘92 _ (1 + 9:4)2 21. (a) = 2cosx + sin2 a: => f’(;v) = 2(—sinm) + 2sinac (cosx) : sin 2% — 2sinx : f”(m) : 2e0s 2:0 — 2eosac = 2(eos 2m — cos cc) We can see that our answers are plausible, since f has horizontal tangents where f'(x) = 0, and f’ has horizontal tangents where f”(x) = 0- The graphs seem reasonable since f has horizontal tangents where f’ is zero, f I is positive where f is increasing, and f ' is negative where f is decreasing; and the same relationships exist between f ' and f". 23. y : «276+ 3 = (22: + 3)”2 :> y’ = %(2:c + 3)-1/2.2 = (2x + 3r“? :» y" = —l(2x + 3)_3/2 - 2 = ——(2:c + 3)_3/2 => 7/" = %(2m + 3YS/2 ~ 2 : 3(2x + 3)_5/2 93 I (296— 1)(1) "30(2) '1 -2 — 1 2 1 ”—> 23571 9 y (2m—1)2 (zcc—i)2 or (m ) y" = —1(—2)(2z ! 1)’3(2) = 4(250 v1)—3 :> y'" : 4(—3)(2x — 1)—4(2) = —24(2a: — 1)’4 or —24/(2$ — 1)4 25. f(t) : teost => f’(t) : t(— sint) +cost-1 => f"(t) — t( cost) sint - 1 sint —> f’"(t) = tsint — cost - 1 — cost - cost = tsint — 3eost, so f’"(0) : 0 — 3 = —3. 26. g(av) = \/5' — 23: => g'(z) = as — 2m)_1/2(—2) = —(5 — 235V“? 2 9"(33) = as — 2x)r3/2(—2) : —(5 — 290*” => g'"(ac) = 3(5 - 2x)*5/2(—2) = —3(5 — 2x)—5/2, so g'"<2> = 4(1)!” = —3. 27. f(t9) = cot0 :> f’(0) : —cse20 => f”(9) = —2cse0(— escficote) : 2ese20eot6 => f”'(9) = 2 (#2 ese2 6001: 0) cot6 + 2esc2 0 (# ese2 6) : —2esc2 t9 (2 cot2 6 + ese2 0) :> Ms) = 4(2)? [2(\/§)2 + (2)2] : -80 28. g(x) : seem => g'(x) = secactana: :> g”(1‘) : see as sec2 :3 + tans: (sec :3 tan at) : sec3 x + see :6 tan 2 2m :see3cc+seca:(see a: — 1) = 28603.13 — seem => "'(w) : 6 $802 a: (seecc tan x) — seen: tanac : seem tan at (6 see2 an — 1) a g gn/(g) : fi(1)(6.2 _ 1): 11\/2_ SECTION3J HIGHER DERIVATlVES C 201 29. 9m2+y229 => 18$+2yy’—0 —> 2yy’— 1893 —> y'——93:/y : 2 2 ll — 9 — _9 EM : —9 . fl -_— —9 . £3 [sincexandy must Satisfy y _ _ 212 _ y2 313 y the original equation, 9:132 —|— 312 = 9] Thus, y” : 781/743. 1 I 30.\/E+\/yj=1 : 2fi+2y__\/y:0 : y/:_\/g/fi :> ” x/E [1/(2 y' — x/E7l1/(2 x/EH x/E (1/x/37) (—x/g/fi) — fiU/fi) : 1+ \2/17/x/5 2m :0 y = ——————?———— : _ = M = 1 since :0 and y must satisfy the original equation. fi + fl 2 1. 253 fi 2:3 fl at2 31.$3+y3=1 :> 312+3y2y'20 => y'=~? :> u 142(293) — $2 ‘ 2yy' _ 2551/2 — 2x29(—$2/y2) _ 296314 + 2:64:11 _ “293M113 + x3) _ 2m y = _ (y2)2 — _ y“ y6 y6 315’ since :1: and y must satisfy the original equation, $3 + y3 : 1. 32. m4 + y4 = a4 => 42:3 + 4y3y' = 0 => 4y3y’ = ~4$3 => 31’ = fins/ya => / 4 4 4 _ 4 2 y” I _ 313 ~3962 — $3 - 331231 : _3x2y2 . y ~$(~$3/y3) : _3x2 . 31 2 _3w2 . i7 : 3‘? (313)2 y6 y y y y 33- 10(15): :5” => f’(m) = nxn‘l => f”(m) = n(n —1)$"‘2 => -_—> f(")(a:) = n(n —1)(n ~ 2) - - ~2- 1m"‘” 2 n! 34. f(:t) = 5 1 1 z (51: —1)—1 :> f'($) : —1(5$ — 1)‘2 . 5 => f"(a:) = (—1)(—2)(5x —1)-3-52 :9 w a We) = <—1>(—2>(«3><5w —1>-4 ~53 :» => Mo) = (—1>"n!5"<5z — 1>-<"+1> 35. f($) : 62’” => f’(ac) : 2621 => f"(;c) = 2 - 262m : 22e2GE => Jen/(1:) : 22 _ 262$ : 2382w :> U ' : f(n)($) : 2n621 36 : fl: 1.1/2 :> : éx~l/2 z} fI/(aj) = %(—%)m-3/2 : Jul/(x) 2 %(i%)(_%)m—5/2 : 1 _1-3~5 We):§<—%)<—%><~2>z-"2— 24 W 2 a 13-57 _ fem:a—at—gM—gx—ax 9/2=2—5w 9/2 2 é a V _ 1-305~---(2n—3) a , n 2n 1 2 n 1 2n 1 2 f( we):a—gx—ame—nm < V =<—1) 2n 96‘ V 37- f(96)=1/(3333)=%f3 => f'(96)=§(-3)$‘4 => f"($)=%(~3)(—4)$75 => f’”($)‘§( 3)( 4)( 5)“?6 —> —> (n) fi;_ _ m_ 4M3);(*1)"'3'4‘5-~'~(n+2),2 (—1)"(n+2)’ f (x)—3( 3)( 4) [ (n+2)]m 3mn+s 2-‘GWT 38.Dsinx=cosx 2 Dzsinscz—sinm :> Dssinx2—cosx : D4sinx=sin$.Thederivativesof sina: occur in a cycle of four. Since 74 : 4(18) + 2, we have D74 sinx = D2 sin a: = — sin w. 39. Let f(:c) : cos ac. Then Df(2x) : 2f'(2a:), D2f(2$) = 22f"(2:c), D3f(2x) : 23f’”(2m), ..., D(")f(2:1:) = 2"f("> Since the derivatives of cosz' occur in a cycle of four, and since 103 2 4(25) + 3, we have f(103)(ac) = fa) : sinus and D103 cos 2:1: 2 2103f(1°3)(2m) = 2103 sin 23. 202 CHAPTER 3 DIFFERENTIATION RULES 40. f(x) : ace’m => f’(m) 2 x(—e_z) +e‘I = (1 — we" => f"(;c) : (1 * $)(‘€_I) + e_x(—1)=(sc — 2)e_ac :> f”'(x) : (m * 2)(ie’m) +e’m : (3 — (3)671 => f(4)($) = (3 - HEX-e”) + e_z(*1) = (m - 4k?—z => "' => f‘")(w) = (-1)"($ - me“- So DmooacrfCE : (:L‘ — 1000)e‘1. 41. By measuring the slope of the graph of s = f(t) att : 0. 1, 2, 3, 4. and 5, and using the method of Example 1 in Section 2.9, we plot the graph of the velocity function 1; = f’(t) in the first figure. The acceleration when t = 2 s is a = f "(2), the slope of the tangent line to the graph of f’ when t = 2. We estimate the slope of this tangent line to be (1(2) 2 f”(2) = v’(2) 3 ¥ 2 9 ft/sz. Similar measurements enable us to graph the acceleration function in the second figure. 50 25 0 10 20 t (b) Drawing a tangent line at t = 10 on the graph of a, (1 appears to decrease by 10 ft/s2 over a period of 20 s. So at t : 10 s, the jerk is approximately —10/20 = #05 (ft/s2)/s or ft/s3. 43. (a) s = 2t3 — 151:2 + 36t + 2 2 11(25): s'(t) = 6t2 # 30t + 36 => a(t) = v’(t) = 12t — 30 (b) a(1) = 12 ~ 1 — 30 = —18 m/s2 (0) v(t) = 6(t2 — 525 + 6) : 6(t — 2)(t — 3): 0 whent = 2 or3 and (1(2) : 24 — 30 = —6 m/s2, (1(3) 2 36 — 30 : 6 m/sz. 44. (a) s a 22:3 3t2 124‘. —> v(t) — s’(t) ~ 6t2 — 6t # 12 => a(t) = v’(t) : 12t — 6 (b) a(1):12~1—6:6m/s2 (c) v(t) : 6(t2 —t — 2) = 6(t+1)(t— 2) = 0whent : —10r2. Sincet 2 0,t;é —1 and a(2) : 24 — 6 : 18 m/s2. 45. (a) s = sin(%t) +608 (%t),0 s t s 2. W) — 3’05) ! cos(2§t) - % sin(’ét) ' % — %Icos(%t) - Sin(%t)l :» an) = wt) : gI—sin(%t> - % ~cos(%t) - a = ~§- Isin(%t) +c°s(%t>l cal: (b) a(1) = —;r—; [sin(% - 1) +cos( .1)] : —7T— [1 + fl] = -77r— (1 + t/fi) x —0.3745 m/s2 SECTION3J HIGHER DERIVATIVES E 4 . 4 sin(%t) (c)v(t):0for0§tg2 => cos(gt)2sm(gt) => :Cos(%t) => tan(§t)=1 => %t2tan_11 :> :23 §2§=1.5S.Thus. 2 _ 7r 7r2 fl fl _ F2 N 2 «1(3) : {—6 [sm(% ~ g) +cos(—6— . g)] = 2% [7 7] _ Ewe 2038771114 46. (a) s = 213 — 72:2 +4t +1 2 v(t) = s’(t) : 6t2 —14t+ 4 : a(t) = v’(t) = 121‘. — 14 (b) a(1) : 12 — 14 : 22 m/s2 ' (c) v(t) = 2(3t2 — 7t + 2) : 2(3t — 1)(t — 2) : Owhen t = g 0r2 and (1%) (1(2) : 12(2) — 14 : 10 m/s2. 12g) — 14 : —10m/s2, 47. (a) s(t) = t4 — 4153 + 2 —> v(t) ~ s’(t) ~ 4t3 12112 > a(t) v'(t) 122:2 24: _12t(t 2 2) = 0 when t 2 O or 2. (b) 3(0) 2 2 m, 11(0) 2 0 m/s, 3(2) 2 —14 m. 11(2) 2 —16 m/s 48. (a) 3(1) : 2t3 — 9t2 —> 41(1) — s'(t) ~ 6t2 1823 _> a(t) 2 2/ (t) _ 12t — 18 : 0 when t : 1.5. (b) 50.5) : —13.5 m, v(1.5) = 213.5 m/s ll 49. (a) s = f(t t3 — 12112 + 36t, t 2 0 2 v(t) = f’(t) : 3t2 — 241: + 36. a(t) = v’(t) = 623 — 24. a(3) = 6(3) 2 24 : —6 (m/s)/s or m/sz. (C) The particle is speeding up when 1) and a have the same sign. This occurs when 2 < t < 4 and when t > 6. It is slowing down when 1; and a have opposite signs; that is, when 0 S t < 2 and when 4 < t < 6. g t A _ , _ (1+t2)(1)—t(2t) 1-132 50. (a) :c(t) — 1+ t2 v(t) x (t) (1+t2)2 — ——(1+t2)2. a(t) 2 v’(t) — 2:10: . a(t) — 0 —> 21f(t2 — 3) 2 0 2 t 2 0 or x/g (c) v and a have the same sign and the particle is speeding up when 1 < t < The particle is slowing down and v and a have opposite signs when O<t<1andwhent>\/§. 51. (a) y(t) 2 Asinwt 2> v(t) 2 y/(t) 2 Aw coswt 2> a(t) 2 v'(t) A4112 sin wt (b) a(t) 2 ~Aw2 sinwt 2 —w2(A sinwt) 2 —w2y(t), so a(t) is proportional to (c) speed 2 |v(t)| 2 Au; Icoswtl is a maximum when cos wt 2 il But when coswt 2 i1. we have sinwt 2 O, and a(t) 2 —Aw2 sinwt 2 —Aw2(0) 2 0. 203 204 52. CHAPTER 3 DIFFERENTIATION RULES do do d3 d1) E I E E : E 1’ velocity with respect to time (in other words, the acceleration) whereas the derivative d1) / ds is the rate of change of do By the Chain Rule, a(t) = (t) : v(t) The derivative dv/dt is the rate of change of the the velocity with respect to the displacement. 53. Let P($) : (1:32 + bx + 0. Then P'(:c) : 2am + b and P”(;c) : 2a. P"(2) : 2 => 2a : 2 => a : 1. P’(2):3 => 2(1)(2)+b—3 —> 4 I bi?) —> b: 1. 13(2) :5 :> 1(2)2+(71)(2)+c=5 :> 2+c:5 => c:3.SoP(w) 2x2 —a:+3. 54. Let Q(:c) : amg —I— but2 + 0:3 + d. Then Q'(;v) : 3a562 + 2ba: + c, Q”(:c) : firm + 2b and Q"’(:n) = 6a. Thus. Q(1):a+b+c+d=1.Q'(1)= 3a+2b+c=3.Q"(1) =6a+2b:6andQ"'(1) 26a 2 12. Solving these four equations in four unknowns a, b, c and d we get a = 2. b : —3, c : 3 and d = —1. so Q(m) = 23:3 — 3x2 + 3x # 1. 55. y = Asinrc + Bcosm :> y' : Acosw — Bsinx :> y" : AAsinm — Bcosm. Substituting into y” -I— y' — 2y : sing: gives us (73A — B) sinac + (A — 3B) 00525 = lsinx, so we must have —3A — B 2 1 and A — 3B : 0. Solving for A and B. we add the first equation to three times the second to get B = —1—10 and A = ;%. 56. y 2 A562 + Bsc + C => 3/ : 2A2: + B => y” : 2A. We substitute these expressions into the equation y” + yI A 2y : 9:2 to get (2A) + (2Arc + B) — 2(Ax2 + Bx + C) : x2 2A+2Aa2+Be2Am2—2Bzc—202x2 (—2A)m2 + (2A # 2B)z + (2A + B — 20) z (m2 + (on + (0) The coefficients of :02 on each side must be equal. so 72A 2 1 => A = 7%. Similarly, 2A ; 2B 2 0 => A:B:—§and2A+B—2C_0 4 1 § 20—0 4; C——§. 57. y 7 e” _> y' — re” —> y” # r26”, so y" + 5y' * 6y = r26” + 5mTm — 6eTI = 6”” (r2 + 57" — 6) : e”(r + 6)(r — 1): 0 : (r+6)(r—1) :0 :> rzlor—G. y # ekm _> y! _ AeAz y/I _ A26Am‘ Thug, y + y/ : yll <2> CA2: + Aekx : A26A1} <:> e’\I()\2 —/\— 1) :0 <=> A: 1—i.f@,sincee’\”” 9&0. 59- flail) = 1179(w2) :> HIE) : $91062) ~2x +9($2)-1= 9(932) + 206240132) => f"(w) = 9'($2) ~ 2w + 2962 ‘9"(902) ' 2m + 9'60?) '455 : 6$9'($2) + 4w39”($2) w ’ a: — :L‘ n w2I9'(€C) + 1’9"(36) - 9'(€6)I ' 213 [avg/(SC) - 906)] $2g"(w) - 2m9'66) + 29(93) f ' $4 — $3 / , 7 9 (fl) 51- HI) =9(x/E) => f (56) =g’(\/5_6)-%$ 1/2 = 77; => 2fi‘gu(\/E) . émrl/2 _g/(\/E).2.%x—1/2 mil/2 wire/E) 7 91m] (NE)? 490 f”($) — _ _ 43: fl SECTION3J HIGHER DERIVATNES :1 205 fa) : 3x5 _ 10.23 + 5 :> f'(q:) : 15.734 — 30x2 => 62. f”(a:) = 60353 — 60m = 60:3(902 — 1) : 60m(:z: + 1)(:c — 1) So f”(x) > 0 when —1 < x < 0 or m > 1, and on these intervals the graph off lies above its tangent lines; and f"(a:) < 0 when w < —1 or O < ac < 1. and on these intervals the graph of f lies below its tangent lines. _ 1 , V — (2x +1) 63 we) 7 x2” :» f (m) 4 (WW 2 ,, ($2 + x)2(—2) + (2x + 1)(2)(x2 + x)(2$ +1) 2(3z2 + 3x +1) f : 4 — 3 __> (m2 + an) (1:2 + x) 16mm _ (m2 + x)3(2)(6w + 3) — 2(3cm + 3x +1)(3)(gc2 + x)2(2w +1) (5132 + x)6 —6(4x3 + 6x2 + 4:1: +1) : fi 3 (m2 + $)4 fwm _ (x2 + x)4(—6)(12$2 +122: + 4) + 6(4ac3 + 6x2 + 4x +1)(4)(x2 + x)3(2a: +1) ($2 + $)8 _ 24(5x4 +109:3 +10$2 + 5:1: +1) (:62 + m)5 f<5><w> — 7 1 1 1 (b) “70) I m = E — x+1 :> f/(ml = —$fi2+($+1)_2 2’ f"($) =2mh3_2(93+1)_3 => f”’(x) : (—3><2)w*4 + (3><2><x +1)“ : --- :> f‘"’(w)=(—1)"n![x“”+1)—(x +1)“"+”l 756 +17 1 —6(56x4 + 544E — 2184 2 + 6184 — 6139 64. a F0 : ‘ "I 2 11? ac ) ( ) r f(m) 2352 _ 736 g 4, 3 CAS gives us f (2x2 ‘ 7x _ 4)4 . 718 + 17 ~3 5 (b) Usmg a CAS we get f (:13) = m = 2m + 1 + x _ 41 Now we differentiate three times to obtain 144 30 filly”) : (2x+1)4 _ (5(1- 4)4‘ 65. For f(a:) : $262, f’(m) 2 $263” + e$(2x) = 61(x2 + 2:10). Similarly, we have f”(m) : em($2 + 4x + 2) f”'($) : 61(1132 + 63: + 6) f<4>(;c) : emf + 8x +12) f(5)(x) = ez($2 +10x + 20) It appears that the coefficient of a: in the quadratic term increases by 2 with each differentiation. The pattern for the constant terms seems to be 0 = 1 - 0, 2 = 2 ~ 1. 6 = 3 e 2. 12 = 4 ~ 3, 20 2 5-4. So a reasonable guess is that f(")(m) : 6" [x2 + 2nd; + n(n ~ Proof: Let Sn be the statement that f(")(.7c) = em [$2 + 27135 + n (n — 1)]. l. 51 is true because f’($) = e“6 (932 + 2x). 206 CHAPTER 3 DIFFERENTIATION RULES 2. Assume that Sk is true; that is, f('“)(m) : ex [37:2 + 2km + k (k — 1)]. Then f<k+1)(w) = % [f(k)($)] = e$(2$ + 2k) + [x2 + 2km + k(k —1)]ew : e9” [:02 + (2k + 2)a: + (k2 + k)] = e“6 [m2 + 2(k +1):z: + (k + 1)k] This shows that Sk+1 is true. 3. Therefore, by mathematical induction, Sn is true for all n; that is, f (")(33) 2 cm [x2 + 27150 + n(n — 1)] for every positive integer n. 66. (a) Use the Product Rule repeatedly: F = fg :> F' : fly -I- fgI : F" = (f"g + f’g') + (f’g’ + f9") = f”g + 2f’g’ + f9”. (b) = f”’g + f”g’ + 2(f”g’ + M”) + f’g” + f9’” = f”’g + 3f”g’ + 3f’g” + f9’” 2 F“) = M; + f”’g’ + 3(f’”g’ + f”g”) + 3(f”g” + M”) + f’g’” + f9“) = f (4)9 + 4f”’g' + 6f”g” + 4f’g’" + f9“) (0) By analogy with the Binomial Theorem, we make the guess: n n7 n n— n 711* n~ n F‘")=f()g+nf( 1)9'+(2>f( 2)9"+"'+(k>f( k)g(k)+~-+nf'g( 1)+fg() where n! "("‘1)("'2)“'(n—k+1). k 2 Mn — k)! k! . asaa 67. The Chain Rule says that d1: — du dx, so 2 d—y- : 1 (1—3! 2 —d— — d dy du + dy d (in) [Product Rule] (1302 drc dsc dx du dm dcc du dw du dw dx :[d (E du] du+dyd2u : dzy (du)2+ dygiu EB du E a du dsc2 duz dam at: (1:132 . d2y dgy du 2 dy dzu 68. From Exerc1se 67, a; : ——u§ E + a a; APPLIED PROJECT WHERE SHOULD A PILOT START DESCENT? 207 69. We will show that for each positive integer n. the nth derivative f (n) exists and equals one of f, f’, f”, f’”, . . . , f(”_1). Since 1””) = f, the firstpderivatives off are f’. f”. f’”, . ,. ,f(p_1), and f. In particular. our statement is true for n = 1. Suppose that k: is an integer. k 2 1, for which f is k-times differentiable with f0“) in the set S : {f, f’, f", . .. , f(p’l)}. Since f is p—times differentiable, every member ofS' [including f(k)] is differentiable, so f (“1) exists and equals the derivative of some member of S . Thus, f (19“) is in the set {f’, f". f’”, . . , , fm}, which equals 8' since f”) = f. We have shown that the statement is true for n = 1 and that its truth for n : k implies its truth for n 2 k + 1. By mathematical induction. the statement is true for all positive integers n. APPLIED PROJECT Where Should a Pilot Start Descent? 1. Condition (i) will hold if and only if all of the following four conditions hold: (a) P(0) = 0 (,6) P’(0) = 0 (for a smooth landing) (7) P'(E) : 0 (since the plane is cruising horizontally when it begins its descent) (6) 13(2) = h. First of all, condition a implies that P(0) = d = 0, so P(m) : of + bx2 + cm :> P’(z) : 3am2 + be + c. But P’(0) : c = 0 by condition ,8. So P'(€) 2 3a€2 + 21% = 8 (3a€ + 2b). Now by condition 'y, 3a! + 2b = 0 :> a : — Therefore, P(m) 2 ——§st + bacz. Setting PM) = h for condition 6. we get 21) 2 1 3h 2h PZ=——€3 M2: —— 2 2— —> 2 — () 3! + h => 31)! +b€ h 31)! h > b 32 a £3 . So 2h 3h y 2 = —e—3$3 + 6—2232. . . ., d 2 2. By condition (11), d—f = —v for all t. so as (t) = f — vt. Condition (iii) states that S k. By the Chain Rule. dt2 dy dy d2: 2h 2 day 3h dac 6hx2v 6hscv h _ _ __ _ __ _ we ave dt dm dt [3 (3a: ) dt + (2 (2m) dt £3 £2 (form 3 Z) :> dzy 6hr; dx 6hv dz 12m;2 6hv2 . W — ~ h — [3 35+ [2 .Inpamcular,whent=0.:c=€andso (123, 12m;2 6hv2 6hv2 dzy 6hv2 . . . F ‘20 ~ — £3 2 £2 = * £2 . Thus ? L0 : e2 3 k. (This condition also follows from taking .1 : O.) 1 mi 3. We substitute k = 860 mi/h2. h = 35000 ft x 52 ft, and v = 300 mi/h into the result of part (b): 6(35,000- F180)(300)2 35 000 Ram => (>3001/6.’§g , ~ , £2 _ _ 5280.860 645miles ...
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This note was uploaded on 12/08/2009 for the course MATH 101 taught by Professor Dr.tahir during the Fall '08 term at King Fahd University of Petroleum & Minerals.

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7 - 198 C CHAPTERS DIFFERENTIATION RULES 67. (a) If y : f...

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