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5 - 178 46 47 j CHAPTER3 DIFFERENTIATION RULES Let |PR| ac...

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Unformatted text preview: 178 46. 47. j CHAPTER3 DIFFERENTIATION RULES Let |PR| : ac. Then we get the following formulas for r and h in terms of 6 and an: sing—Z => r—ccsingandcosg—E => h— 6 A 2 T ac _ 2 2 _ 9: _ “OS 2' P I Q Now A(6) 2 %7rr2 and 3(9) : 5mm : m. So - A(9) . l7”; . r . xsin(6/2) x 1 —— : 1 2 = A _ 2 ; ___ 9361+ B 6) 69—13(1)le Th ”6133+ h 2W6E151+ xcos(6/2) a : %7r 6133+ tan(6/2) : 0. R By the definition of radian measure. 3 : r6, where r is the radius of the circle. 6 d 6 By drawing the bisector of the angle 6. we can see that sin ~2- = 7/72 2 d = 2r sin 5. So lim f = lim ——7fi—— : lim 2 ' (0/2) : lim —6—/2—— = 1. [This is just the reciprocal of the limit 6—.0+ d 9—.0+ 2r sin(6/2) 960+ 28in(6/2) 6—»0 sin(6/2) sinz lim z—>0 : 1 combined with the fact that as 6 —+ 0. g —> 0 also] 3.5 The Chain Rule 1. 10. 11. Letu : g(:I:) = 4m and y : flu) : sinu. Then gal—c : %§% : (cosu)(4) : 4cos4x. d d du 3 3 — — z : :1/2 _y:l_:l’1/23:__:._———. Letu7g(:c)—4+3mandy f(u) fl u .Then dw dudm 2U () 2W 2m d .Letu : 9(13) : 1 ~ 902 andy = flu) = um. Then % : gig—1;: : (10u9)(—2:c) : —20:c(1 - x2)9. Letu — (a3) — since and : f(u) : tanu Then 13—, : fld—u : (see2 u) (coszc) : sec2(sinac) ~ cossc or —g — y ’ dsc du dm ’ equivalently, [sec(sin 16)]2 cos x. .Letu 2 9(30) 2 fiandy : f(u) : e“. fi dy-d£@_ u 1 71/2 _ fi._1___ e The“ dm * dudm '(6 )(295 )’6 zfi ’ 2fi Letu = (x) = 6”” andy : f(u) = sinu. Then (—13% = d—yd—u = (cosu)(em) 2 ex 0036’”. 9 ‘ da: du dw . F(:c) : (x3 + 4:07 :> F/(x) : 7(;.:3 + 4m)6(3w2 + 4) [or 735%»? + 4)6(3x2 + 4)] pm : (x2 _ a: +1)3 => F'(;I:) : 3(z2 — a: + 1)2 (2x # 1) F(:1:) : \4/1 +2zc+sc3 : (1 +2$+$3)1/4 => I ,1 34/511 2 3: 1 -2+3x2 F(m)_4(1+2x+w) dw< "I" 554—9”) 4(1+2m+m3)3/4 ( I 2+3m2 _ 2+3x2 4(1 + 206 + 933W“ 4 {/(1 + 2x + m3)3 rm , (1 H4)“ 2 f’(m) = 2(1 +m“)‘”3<4m3) = —§””:— ‘ 1 3 3m g(t) : —-i— : (t4 + 1)‘3 => g'(t) : —3(t4 + 1)‘4(4t3) = #12t3(t4 + 1Y4 : _%12_t3_4 (t4+1)3 (t +1) 12. 13 14 15. 16 17. 18. 19. 20. 21. 23. 24. 25. SECTION 3.5 THE CHAIN RULE C 2 _ sec t f(t) = \3/1 +tant : (1 +ta.nt)1/3 => f’(t) = %(1+tant) 2”380%: —3—2 3 (/ (1 + tan t) . y : cos(a3 + $3) :> y’ = — sin(a3 + 5133) ~3x2 [a3 is just a constant] : —3x2 sin(a3 + $3) . y : a3 + C083 :5 :> y' = 3(cos ac)2(7 sin cc) [a3 is just a constant] 2 *3 sinmcos2 ac —mx I —mcl: d *m1 —m1 34:6 :> y:e E(imx):e (—m):—me . y : 486C511} => 3/ 2 4sec 5mtan 5112(5) = 208% 5m tan 5:0 got) = (1 + 4m)5(3 + cc — W3 2 g’(w) : (1+ 4:15)5 - 8(3 + a: — m2)7(1— 25) + (3 + a: — $2)8~5(1+ 42:)4 - 4 : 4(1 + 4$)4(3 + a: — x2)7 [2(1+ 41')(1— 2x) + 5(3 + m — 52)] = 4(1+ 4z)4(3 + :1: — $2)7 [(2 + 4x — 1652) + (15 + 5:5 — 552)] = 4(1+ 4x)4(3 + z — m2)7 (17 + 9x — 215:2) h(t) 2 (t4 ¥ 1)3(t3 +1)4 :> h’(t) = (t4 — 1)3 4(53 +1)3(3t2) + (:3 +1)4~3(t4 —1)2(4t3) :12252034L —1)2(t3 +1)3 [(254 *1) +is(t3 + 1)] : 122:2(14 —1)2(t3 +1)3 (22:4 + t — 1) y = (2x - 5)4(8$2 — 5)_3 => 3/ : 4(233 — 5)3(2)(8m2 1 5) ‘3 + (2x — 5)4(—3)(85~2 ~ 5)‘4 (1630) : 8(2x — 5)3(8:02 ~ 5)‘3 — 4833(22: — 5)4(8m2 — 5) ‘4 [This simplifies to 8(250 ~ 5)3(82:2 — 5) ”(—452 + 305 — 5).] y: (m2+1)($2+2)1/3 : 9’ 2 25(52 + 2W3 + (962 +1)(§)(2:2 + 2)‘2/3 (2x) : 2;;(52 + 2)1/3[1+ 75232)] y : ave—$2 ¢ y' = xe‘m2(—2x) + 6‘12 -1: e‘zg(—2x2 +1) 2 6—762 (1 — 2:52) y : 6’5”” cos 336 => 3/ : 6’5”” (—BSin 3x) + (cos 3x)(—5e’51) = —e‘51(3sin 3x + 5cos3zc) _ mcosz / zcosn: d Ecosz ~ y—e :> y :6 -d—m(xcosm)=e [w(—sm$)+(cosx)-1]=e“°”(cosm—xsinm) Using Formula 5 and the Chain Rule, y : 10H2 :> I _ 1—z2 d ~12 y _ 10 (mm) - E (1 —:52) = —2:c(1n10)101 . 2~1 2—1 1/2 F : : (Z) 2+1 (2+1) :> _1/2 1 2 F,(Z) :1(221 .1 2—1 _1z+1 /'(2+1)(1)—(2—1)(1) 2 2+1 d2 2+1 2 2~1 (2+1)2 _1(z+1)1/2 z+1—z+1_1(z+1)1/2 2 1 war/2‘ (2+1)? ‘2<z—1)1/2<z+1>2‘<z~i)v2<z+i>3/2 179 180 C CHAPTER3 DIFFERENTIATIONRULES 26. C(y) : -(y(—g;2—1):)5 :> 2(312 + 2y)4(y e1)3[2(y2 + 2y) , 5(y -1)<y +1)l (92+2y)1° _ 2(y e 1)3i(2y2 +4y) + (—5212 +5)] 2 2(2/ —1)3(—3y2 MM?) (212 + 2206 (212 + 2206 27. =—’"— => y x/r2+1 2 2 2 2 _ PF _ r mm” y, \/7’2+1(1)—7‘~%(7’2+1)1/2(27‘) T + «(7‘2+1 _ r2+1 <v~2+1>2 (r241)? <r2+1>2 r2+1 —r2 _ : _——( ) : —————-1 or (r2 + 1) 3/2 ( r2 +1)3 (r2 + 1)3/2 Another solution: Write y as a product and make use of the Product Rule. y : r(r2 + 1) ‘1/2 fi 11’ = r - —§(r2 + 1)’3/2 (27") + (T2 +1)’”2 . 1 : (r2 +1)’3/2[_r2 + (r2 + 1)1] : (r2 +1),3/2(1) : (T2 +1)_3/2 The step that students usually have trouble with is factoring out (r2 + 1) 73/2. But this is no different than factoring out :02 from $2 + 205; that is. we are just factoring out a factor with the smallest exponent that appears on it. In this case‘ —% is smaller than 7% €211 28. = —— => / (eu +e~u) (62” 2) #e2u(eu _e—u) 82u(2€u +2e—u _eu +67”) €2u(eu+3e—u) y — _ . (en + 67102 (6“ + 6—“)Z (6“ + e‘“)2 Another solution: Eliminate negative exponents by first changing the form of y. 271. u 3u e e e eu +€~u eu 6211 + 1 I (6214 + 1) (363”) _ e3u (262“) e3u (36211. + 3 _ 2621A) eBu (621; + 3) (e2u + 1)2 (82“ + 1)2 (62“ +1)2 29. y : tan(cos m) :> y' : secg(cos :8) - (i sin :12) : # sinazsec2(cos :3) 30. y : sin2 :3 cos x , cosx (2 sin cc cos :c) ; sin2 17(i sin :c) _ sinac (2 cos2 w + sin2 :16) _ sinx (1 + cos2 :3) y — cos2 w cos2 90 cos2 ac : sing: (1 + sec2 :10) . . 2 . . Another method: 3; : tan cc SlIl m :> y' : sec2 :13 $111 as + tan ac cos a: : sec at Sin m + smzc 31. Using Formula 5 and the Chain Rule, 3; : 2“in 7” => y' : 2>in "’3 (1n 2) - ;— (sinnx) : 233“” (In 2) ‘ cos 7m ‘ 7r 2 25"" 7”” (7r 1n 2) cos 7m :6 32. y : tan2(36) : (tan 36)2 :> y' 2 2(tan 36) - % (tan 30) : 2tan 30 ~ sec2 30 - 3 : 6 tan 30 sec2 39 33. 34. 35. y 2 sec2$ + tan2w 2 (see (B)? + (tan 3:) 36. 37 38. 39. 40. 41. 42. 45. 47. .y2x2e‘ SECTION3.5 THE CHAIN RULE 3 181 y 2 (1 + cos2 z)6 2> y' 2 6(1 —t—cos2 x)5 ~260sm(—sin$) 2 —12 C051: sinx(1 14105293)5 '1 '2'1 1 ~i —sinl~lcosl y2ms1n; 2> y—31n;+$cos$ $2 _ at m x 2 => 3/ 2 2(sec w)(sec :0 tan :5) + 2(tan ar:)(sec2 an) 2 2 sec2 9: tan m + 2 sec2 :5 tan .2 2 4 sec2 :15 tan :6 2 an a: an x d In an a: 2 1 —1 2 __ kSGC fl ktanfi 31:51” W :> inekt ‘r-E(ktan\/E)2e t ‘/_(ksec\/E-§w /)—2—\/_$—€ . y 2 cot2(sin t9) 2 [cot(sin 0)]2 2 y' 2 2 [cot(sin 6)] - 3% [cot(sin (9)] 2 2cot(sin 0) ~ [~ csc2(sin 0) - cos 0] 2 —2<3036 cot(sin 0) csc2(sin 0) y 2 sin(sin(sin 13)) 2> y' 2 cos(sin(sin :13) ad; (sin(sin an» 2 cos(sin(sin m)) cos(sin 11:) 005.7: _ 1:; —1/2 1 _1/2 2—1— 1 1 y—\/$+\/EE 2> y 2(m+\/E) (1+2w ) 2 sc+\/E( +2“; y=m 2 y'2%(x+ m+\/E)_1/2[1+%(m+fi)‘l/2(1+%x’l/2)] y2sin(tan\/si—HE) y’2cos(tan\/siTw 2cos(tan\/siTa: ( §|¢~ U d . tan x/sinx) 2 cos<tan Vsinx) sec:2 x/sinx- d— (smzv)1/2 a: sec Vsinx- %(sinx)’1/2.cosx (sec2x/SiTx) (Ngfiycosz) d —23"‘ => '~23”2(1 2)—(3w2)—23121 239021 3 2 y— y— n dx — (n) (n)($) tan V sin a: I0 . y 2 (1+ 2mm => 3/ = 10(1+ 2:12)9 ~ 2 2 20(1 + 2x)? At (0, 1), y' = 20(1+ 0)9 = 20, and an equation of the tangent line is y — 1 2 20(x — 0), or y 2 20m +1. . y 2 since + sin2 30 2> y' 2 005:1: + 2 sin :ccos m. At (0, 0). y’ 2 1, and an equation of the tangent line is y—021(x—0),ory2z. y 2 sin(sinm) 2> y’ 2 cos(sinx) ‘COSIL‘. At (7130), y' 2 c0s(sin 7r) - cos1r 2 cos(0) - (—1) 2 1(—1) 2 —1. and an equation of the tangent line is y — 0 2 —1(x — 71'). or y 2 —:c + 7r. 3 2> y’ 2 m2(~e‘w) + e”(2w) 2 2am” — x2e”. At (1, i) y’ 2 26‘1 — (2‘1 2 %. So an equation of the tangent line is y — i 2 am ~ 1) org 2 1m. e _ 2 , * (1 + €_I)(0) 2 2(—e’z) 26—73 (a) y 1 +6” fl 3/ (1+e‘1)2 — m- (b) 0 At (071): :1], — 26 _ 2(1) 2 (1+e°)2 ‘ (1+1? *¥* . . . ‘3 So an equation of the tangent line 18 y — 1 2 am — 0) or y 2 %m + 1. - 21.5 1 D. II! 182 CHAPTER 3 DIFFERENTIATION RULES 48. (a) Form > 0, Irc| : :13, andy 2 flan) : «2+7 y/2 — x2 (1) — x(§)(2 — x2)_1/2(—2a:) (2 ~ 22)”? 11,013): < H2) (2—m2)+$2 2 (2 — 1:2)3/2 _ (2 — x2)3/2 So at (1, 1). the slope of the tangent line is f’(1) : 2 and its equation isy—1:2(x71)0ry:2x71. Ila/(1:): m %(1—:B2) 1/2 (—277) 4 V1 ”$2(1).\/1—m2 x2 m - 2 - (1 -x2) —1 50. (a) (b) f(x) = sin(:c + Sin 2512) => f’(:c) : cos(ac + sin 2x) ~ dim (x + sin 2x) : cos(m + sin 2:6)(1 + 2 cos 293) (b) 4 .1- Notice that all tangents to the (b) 8 — l graph of f have negative slopes and f’(a:) < 0 always. From the graph of f, we see that there are 5 horizontal tangents, so there must be 5 zeros on the graph of f’. From the symmetry of the graph of f . we must have the graph of f’ as high at :c = O as it is low at x = 7r. The intervals of increase and decrease as well as the signs of f’ are indicated in the figure. SECTl0N3.5 THE CHAIN RULE :1 183 51. For the tangent line to be horizontal. f'(m) : 0. f(w) 2 28mm + sin2 ac => f’(m):2cosm+2sin:ccosm:0 41> 2cosa:(1+sinm):0 4:) coszc=00rsinx=—1,so x = g + 27m or 37" + 2n7r, where n is any integer. Now f(-’25) = 3 and f(37”) : —1, so the points on the curve with a horizontal tangent are (g + 27m, 3) and (37" + 2n7r, —1). where n is any integer. 52. f($)=sin2m—2sinx => f’(:1:):2c052m—2cosm:4cos2w—2cosx—2,and 4c0s2w—2cosw—2:0 c) (cosm—1)(4cosm+2)=0 4:) cosx:lorcosm=——;—.Som=2n7ror (2n + 1)7r j: 7—; 71 any integer. 53- FOE) = f(9($)) => F'($) = f’(9($)) 'g’(ac), so F'(3) = f'(g(3)) -g'(3) = f’(6) - g’(3) = 7- 4 : 28. Notice that we did not use f’(3) : 2. 54. w : u 0 v :> w($) = u(v(:c)) => 11/ (at) = u'(v(:c)) ~ v'(x), so w’(0) = u'(v(0)) -v'(0) = u'(2) -v'(0) = 4-5 = 20. The other pieces ofinformation, u(0) : 1, u'(0) = 3, and v'(2) = 6, were not needed. 55. (a)h(a:)=f(g(m)) => h’(w)=f’(g(w))-g’(m)~soh’(1)=f’(g(1))-g’(1)=f’(2)-6=5-6=30. (b) Ho) 2 gum) :» H’(m) = 9’(f(:r)) ~ fix), so H'(1> = 9’(f(1))-f’(1)= g'<3) -4 = 9 ~ 4 = 36. 55- (a) PM = f(f(-’I?)) => F'(~”6) = f'(f(x)) - f’($)9 SO F'(2) = f'(f(2)) - f'(2) = NH - 5 = 4 - 5 = 20- (b) C(96) = g(9(x)) => 0’06) = 9’(9(93)) ~9'($)~ 50 9(3) = 9’(9(3)) -9’(3) = g'(2) - 9 = 7 ' 9 = 63. 57- (3)1455) : f(9($)) => “'03) = f'(g($))9'($)~ SO 11(1) = f'(9(1))9'(1) = f'(3)9'(1). To find f/(3), note that f is linear from (2,4) to (6, 3), so its slope is 2:: = —i. To find g'(1), note that g is linear from (0,6) to (2,0), so its slope is 2 0 = —3. Thus, f'(3)g'(1) = (—i)(—3) = Aim (b) WC) = g(f($)) => Wit) = 9'(f($))f’(~”v). SO 1(0) = 9'(f(1))f'(1) = 9’(2)f'(1)~Which does not exist since g’(2) does not exist. (C) 10(16) = 9(9($)) => w'CE) : 9'(9($))9'($)~ SO 11/0) = that g is linear from (2,0) to (5, 2). so its slope is g _ 0 : g g'(9(1))9'(1) = 9’(3)9'(1).T0 find g’(3)~ note . Thus, g'(3)g'(1) : (§)(—3) = —2. 53. (a) ht”) = f(f(x)) => h’($) = f’(f($))f’(w)- 30 W2) : f'(f(2))f’(2) = f’(1)f’(2) % (~1)(-1) = 1- 0» gm : f(:v2) 2 g’tw) : f’(m“‘) ~ i (302) : f’(m2)(2w)4 So g’(2) : f’ (22)(2 . 2) : 4f’(4) z 4(1.5) : 6. 184 CHAPTER 3 DIFFERENTIATION RULES 59- WC) : f(g(w)) => h’(56) = f’(g(w))g’(w)- SO h'(0~5) = f’(9(0-5))g’(0~5) = f’(0.1)9’(0-5)~ We can estimate the derivatives by taking the average of two secant slopes. 14.8 — 12.6 18 4 i 14 8 m1 + m2 22 + 36 For ' 0.1 :m — A _ ' ' i ’ ' ~ _ f ( ) 1 0_1_0 22.m2 0.2_0.1 —36.Sof(0.l)~ 2 _ 2 :29. 0.10 — 0.17 0.05 — 0.10 m m For I 0.5 I : —— : — : __— _ _ I ~ 1 + 2 i g( ) ml 0.5 _ 0.4 0.7, mg 0.6 _ 0'5 — 0.5. $09 (0.5) ~ 2 — —0.6. Hence. h'(0.5) : f’(0.1)g'(0.5) z (29)(—0.6) = —17.4. 60«(313): f(f($)) => 9’06) = f’(f($))f'($)- 30 9'0) : f'(f(1))f'(1) = f'(2)f'(1)~ 3.1 e 2.4 4.4 — 3.1 m1 + m2 For ,2: Z :1 ~ : —_ I - _ f( ) 7711 2.0‘ 1.5 4 m2 2572.0 2.6. Sof (2) 2 72. 2.0—1.8 2.4—2.0 m +m For ,1: : = .4 : _ 1 2 7 f()m1 10—05 0 .m2 1.5—1.0 0—.8.'1~Sof() 2 70.6. Hence. g'(1) = f'(2)f’(1)%(2)(0.6) = 1.2. 61. (a) F<m> : new) :» mm) = f’(e”) f; (em) = New (b) G(a:)=efm => G’(at)=ef(m)%f(m)=ef(x)f'(ac) 62. (a) PM = We) => mm) = my”) % (we) : f’(a:“)a$a’1 (b) C(50) = If($)la 3 G'(93)— — 010013010"1 f'(~”0) 63. (a) f(;t) L(a:4) => f (:13)— L (x4)- 4:83 — (1/m4)- 42:3 = 4/50 form > 0. (b) g(a:) : L(4a:) => g'($)— — L (49:) -4— — (1/(4x)) 4— — 1/2: forav > 0. (C) F($) = [L0014 if F“) = 4IL(VB)l3 ' UGO) = 4IL(~’C)I3 ~ (1/1”) = 4IL(€€)I3/$ ((1)000): 13(1/95) => 0'66) = ”(1/36) ' (21/132) = (1/(1/1?» ' (—1/22) = 3014/13) = *1/30 for a: > 0. 54- T06) = f(9(h(93))) => “(00 = f'(9(h($))) ~ Q'UIW) ' ”(96% SO 7"(1) = f'(9(h(1))) '9'(h(1)) 'h'(1) = f'(9(2)) ~9'(2) '4 = f'(3) ' 5 - 4 : 6 * 5 ~ 4 : 120 65. 3(t) : 10 +4 1 sin(107rt) :> the v:locity after t seconds is v(t)— — s'(t)- — icos(107rt)(107r)= — 52"cos(107rt) cm/s. 66. (a) s : Acos(wt + 6) => velocity : s : —wA sin(wt + 6). (b)IfA7éOandw7$0.thens'=0 4:) sin(wt+6)=0 4:) wt+6:mr (i) 15:71::6. n an integer. 27Tt dB 27rt 27r 0.77r 27rt 771' 27ft 61(3) B(t)=40+0.35sin5—4 => E—(0.35COS—fi)(a>—ECOSE—4_ ~5—4COS—5—4‘ dB 77r 271' b A — _._:._ ._,r§5 () (t 1 dt 54cos54 0.16. 68. L(t)— # 12 + 2. 8sm(365 (t — 80)) => L'(t) = 2.8cos(§g——5(t 80)) (32—35). On March 21.13 : 80. and L'(80) m 0.0482 hours per day. On May 21, t— — 141. and L'(141) m 0.02398. which is approximately one—half of L' (80). SECTION 3.5 THE CHAIN RULE 185 69. 8(1‘) : 2671'5l811127t't é v(t) : s'(t) : 2[e’l’5'(cos 27rt)(27r) + (511127T[)63—1 5t(71.5)] : 26’1“”(27T cos 27ft — 1.5811127rt) 2 Graph ot~ Graph of position - velocity ““2 *l 70 (’) liIn , (2‘) — lirn —1§ : ; = 1. since 1' > 0 :> ikrt fl 700 => 8’“ H 0. 'd t—bocpi _t—Hxl+a€’kt 1+a>0 11065;“ 1 7 (in ,x, 7‘2 ,. (b) 71(1): (1+01z’“) 1 => Ff:—(l+ae M) (~kae “)=x(1+a€_kt)2 (c) l From the graph ot'p(t) : (1 + 1063—0“) *1. it seems that A t : 0.8 (indicating that 80% of the population has heard the p l - rumor) when 1‘ z 7.4 hours. 0 lo 71. (a) Using a calculator or CAS. we obtain the model Q : (117’ with a : 1000124369 and b : 0000045115933. We can change this model to one with base 5’ and exponent hi 1) lb” 2 e’ l” b from precalculus mathematics or from Section 7.3|: Q :(1t1H“l' : 1000124376“W” * '3“. (b) Use Q'(t.) : 0])” 111 b or the calculator command nDeriv(Y1, X .04) with Y1=ab" to get / (2/0104) % 4570.63 [1A. The result of Example 2 in Section 2 I was 4570 11A. 72. (a) P : (11/ with a, : 1.502711 >< 1040:1110 b : 1029953851. 31000 (P in thousands) where P is measured in thousands of people. The fit appears to he very good. I785 1865 0 . 5308 1 3.929 7210 1 3308 (b For 1800: : 7 : 13*,9. : \ :: I, _ l "1‘ 1800 — 1790 ' m" 1810 1 1800 19; 2 So P’(180()) x (rm + 77l2)/2 : 105.?5 thousand people/year. 23.192 — 17063 31413 1 23.102 For 1850: , : , : 612.9. . : \% : "’1 1850 — 1840 "" 1800 ~ 1850 82‘) 1‘ So P/(1850) x (ml + 7123)/2 : 7'19 thousand people/)ettr. (c) Use the calculator command nDeriv (Y1 , X, year) Vt ith Y1:a.o“ to get P'(1800) % 156.85 and P/(1850) x ($86.07. These estimates are somew (d) P(1870) % 41.946. hat less than the ones in part (b). 56. The difference of 3.4 million people is most likely due to the Civil War (18604865). 186 D CHAPTER3 DIFFERENTIATIONRULES 45(122)8 . , .. . . ——_(2t + 1)” Without simplifying. With either Maple 0r Mathematiea. we first get 1‘ , . 8 7 i o W) = 0g)— 18 (t 2) . (2t + ”9 ——(2t + 1)“). and the simplification command results in the above expression. 73. (a) Derive gives g'(t) : . . _ l . r; ., , , (b) Derive gives 3/ : 1(1‘2 7 17 + l) (21'. + 1)4(17.1‘” + (3.1” ’ 91‘ + 3) without simplifying. With either Maple or Mathematiea. we first get /_ i _ 4 ,3 ‘ 4 i 3 . 3 . . y — 10(21 +1) (1. .1. t 1) l 4(2;i- : l) (.1‘5 .z' i 1) (31:2 ~ 1). [t we use Mathematiea’s Factoror 81mplify. or Maple's factor. we get the above expression. but Maple‘s simpl ify gives the polynomial expansion instead. For locating horizontal tangents. the factored form is the most helpful. 4 1/2 :' ~ .' | 74. (a) f.(r) # (g) . Derive gives f’(,r) : .1" + J. + 1 (1" + .r + ”(£4 _ :7: 7 ll whereas either Maple or M’ h‘ , - r/ 3.1.7 ,1 ~ . _‘ _ at emittiea give j (17) : ——4—’1~7 altersimplification, :r i .1: + > N + 1~+1 (11"‘+v1‘+1)' (h) f’(.;:) 7 t) s 314 i — 0 ,i- ~ + [“13 z +o7598. (e) f’(.1r) : 0 where I has horizontal tangents, f' has two maxima and 1 one minimum where f has inflection points. f [ | 4 4 72 75. (a) If If is exen. then f(:l'> : ff’l) Using the Chain Rule to differentiate this equation. we get I f/(J') : f'(!.1') (— (-3) : *f/(il‘). Thus. flt 7!) : af’(1‘)i so f’ is odd. (1.1' (b) If f is odd. then f(1) : ‘f(;;i:), Differentiating this equation. we get f,(.l‘) : af’(~.r)(#1) : ’(<17)_ so f’ is even, 76' liml : {fm taint" t’ : .f’<.r>i.1<:r>i"‘ . <7 1 i [girl] g’trifm f/(II') f(-l‘).¢'1'(¢") f/(J'fl/(vll , flj'lsl/(J') 7 Kill") [girllg 7 lat-I‘lll 77. (it) ;.l (siii‘l .1‘ cos rm) : It siti'"1 .1' cos .1‘ (03» 111' Jr sin" 1' (kn sin/111’) [Product Rule] ( .i' : n siii"’1 J: (cos n.1' (on .r ~ siii 11.1'Sil) .1) [factor out nsiii” .IT[ : n siii”’1 .1‘ (‘()b(ll.1‘ + .l') [Addition Formulator eosine[ : ll siii"71 .i' cos[(n + l).ri [factor out .i'] (h) 11]— (insH ,rt-Os 12.1') : Ii cos'Hl .r ( 7 sin .l') cos n.1' + t-os' ,i' (7715“) 11.1') [Product Rule[ ( .r : in eos'HJ .t' (cos 11.1' sin .I' T siii 11.1' (’os .tt) [factor out ~11 cos“ l .1‘[ : 7r1(‘(')s”’1 .i'siiitni' + .rl [Addition Formula for siiie[ : #IICOSH ’1 .i'siii[(li ,i, llJ'l [factor out .‘l'[ SECTION 3.5 THE CHAIN RULE 3 187 78. “The rate of change of y5 with respect to at is eighty times the rate of change of y with respect to so” 4:) — y5 = 80 — 4:) 5y — = 80 — (i) 5y4 : 80 (Note that dy/da: 7’: 0 since the curve never has a due dac d1: dac horizontal tangent) (I) y4 = 16 <=> y : 2 (since y > 0 for all m) 79. Since 0" 2 (1—3))6rad,we have 0316 (sin0°) 2 die (sinfi )2 £5 cos 1:709 2 F’focosfl". 80. (a) f($) : [:13| : Var—2 = (9:2)1/2 => f'(:c)= %(m2)_1/2(2$)= w/x/F = x/lxl form 7E 0. f is not differentiable at a: : 0. (b) f(9:) = |sinxl : singa: : . a1 2 . sing: C081: if sins: > 0 f'(;v) 2 %(sm2 ac) / 2smxcosa: = cosm ={ lsinml —cosz if sinx < O 5(6) : W : filQWHlH , Where 56). Suppose the result is true for n = k, where k 2 1. Then A103) = Q($)P'(x) — P($)Q’( \ f(k+1)($) : ([QI?;)(]9;L1)I : [Q($)]k+1A;g($) — Ak($) - (k + 1)[Q($)]k ~ Q'(x) {l62(93)l’“+1l2 : [QWHHIAMfl — (19 +1)Ak($)lQ($)le’(w) [62(90)l2"+2 : [Q(x)lk{lQ(w)llA2(w) ~ (k + maiden Q(m)A’ (x) — (k +1 A w , [Q($)]'“[Q(w)l’°+2 = WM : Ak+1($)/[Q($)]k+2, where Ak+1($) : Q(T)A§c(x) __ (k + 1)Ak($)Q/(m). We have shown that the formula holds for n = 1, and that when it holds for n : . . kitalsoholdsforn=k+1. Thus, by mathematical induction, the formula holds for all positive integers n. ...
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