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Unformatted text preview: 3 Cl DIFFERENTIATION RULES 3.1 Derivatives of Polynomials and Exponential Functions eh 1 1 1. (a) e is the number such that gin?) = 1. (b) x a: a: (2.7 —1)/:c :1: (2.8 —1)/:c —0.001 0.9928 —0.001 1.0291 —0.0001 0.9932 —0.0001 1.0296 0.001 0.9937 0.001 1.0301 0.0001 0.9933 0.0001 1.0297 . . 2.7h — 1 . 2.8" — 1 . From the tables (to two dec1mal places). 111111) h : 0.99 and Ami) : 1.03. Since 0.99 <1<1.03.2.7 < e < 2.8. 2. (a) (b) flax) : e” is an exponential function and g(:c) : we is a - d a: fl 2: d e _ e—l power function. E (e ) — e and day (m ) 7 ex . (c) f(x) : e1 grows more rapidly than g($) = we when x is large. The function value at as : 0 is 1 and the slope at :r : 0 is 1. 3. f(:c) : 186.5 is a constant function. so its derivative is 0. that is. f'(m) : 0. 4. f(a:) : m is a constant function. so its derivative is 0. that is. f'(:L‘) = 0. 5. f(x):5m~1 => f'(:z)=5—0:5 6. Fm : —4x10 => F'(z) : —4(10;v10_ 1) : ~40m9 7. f(x) =$2+3m—4 => f’(x)=2x2_1+3—0=2x+3 s. g(;c) : 52:8 — 2x5 + 6 :5 g’(2:) : 5(8zn8_ 1) ~ 2(55" 1) + o : 40:257 —10:1:4 9. f(t) = 4—104 + 8) :> f’(t) : flt“ + 8)’ : §(4t4‘1 + 0) = t3 10. flat) 2 gts — 314 +t => f’(t) = %(6t5) — 3(413) +1 : 3t5 1 122:3 +1 2/5 :> y’ : —3x(_2/5)—1 : ¥g$—7/s : 2 11. y 2 cc- 5 5 _5$7/5 12. 74:561—1—3 => y’ :5(61)+0=561 13. V(r) : gm"3 2) V’(r) = §7r(3r2) 2 47M"2 14. R(t) : 5t‘3/5 :> R’(t) : 5[—§t(’3/5) ’ 1] = —3t‘8/5 151 152 15. 16. 17. 18. 19. 20. 21. 23. 24. 25. 26. 27. 28. 29. 30. 31. 3 CHAPTER3 DIFFERENTIATION RULES Y(t) = 62:49 :> Y’(t) : 6(—9)t-1° : —54t—‘° @ —\/m47 :> R’( ):_7./*om—8__ 7m R<w>— $8 1 2* 12 w -1 712 z, I G($):\/_*2e —a:/ —2e :> G’($)#§m /_2€ #2fi_28 1 _ f 1/3 223 y#€’/E x :> 31' 3:1: / _3x2/3 Fm): éw>5—(%)5m5—3—12w5 : mm—asm—gx4 f(t)= it—i=t1/2EI_1/2 :> f'(t) 1171/2 <‘lr3/2) —1 +——1 ‘5 2 2 2x/i 2t\/Z V 2 1 _ 2 #2 I _2 2 3_2 2 9(w)Aw+F—x +:c => g(x)_;1;+(—)x _ $_; : fiw _ 1) : w3/2 _ $1/2 : y, : gx1/2 _ lmrl/2_ 1x-1/2(3:I: , 1) [factor GUI—2‘ 1 9321/2] 2 2.7! , 350—1 or y , Zfl' 2 y=gC +4m+3— 3/2+4m1/2+3:c*1/2 => fl 2 3 y’—%w”2+4(%)w‘“2+3( @fs/LWWE’ 2W5 r_l :3 O v-v ('D H :7 m n N 3/2 : 33/2431” : z m] y = 4772 => 3," 2 0 since 47r2 is a constant. g<u> : fiu+¢3‘u= fiwfifi i g'<u> = fi<1>+¢§(;u-W) : fiwé/afi) yzax2+bx+c 23> y':2am+b b —2 I 1) ‘2 #3_ v_£___E yzae” +— +—=ae“ +bv‘1 +012 —> y ‘ae bv 20v ae” v2 v3 1 , _ 2 _ 2 73/4 I_ _ _§ 7/4: : v—t — 4t3—t —t => 11 —2t (4)t 2t+4t7/4 2t+4t4t3 23% 10A 2 2 $6 + Bey— — Ay‘10 + Bey :> z' : —1OAgf11 +363! 2 —F + Bey 35. 37. 38. SECTION 3.1 DERIVATIVES 0F POLYNOMIALS AND EXPONENTIAL FUNCTIONS E 153 34. f(:c) = 32:5 — 20503 + 50$ => f’($) : 15m4 — 60102 + 50. Notice that INN”) : 0 when f has a horizontal Notice that J" (an) 2 0 when f has a horizontal tangent. f, 15 p0s1tive when f ‘5 increasmg. and f tangent and that f’ is an even function while f is an is negative when f is decreasing. odd function. f(x):3:c15—5w3+3 => 36. f(ac):ac+1/m=x+zc‘l => f’(m) = 459314 —151L'2. f'($) : 1 i .7;_2 =1—1/m2. 8 —l.2' 1.2 —9 Notice that f’(;c) : 0 when f has a horizontal tangent. f’ is positive when f is increasing. and f' - / ¥ . is negative when f is decreasing. Notice that f (x) — 0 when f has a horizontal tangent. f' is positive when f is increasing, and f' is negative when f is decreasing. To graphically estimate the value of f’(1) for f(a:) : 3m2 — m3. we‘ll graph f in the viewing rectangle [1 — 0.1,1 + 0.1] by [f(0.9), f(1.1)]. as shown in the figure. [When assigning values to the window variables, it is convenient to use Y1(0.9) for Ymm and Y1(1.1) for Ymax] If we have sufficiently zoomed in on the graph of f, we should obtain a graph that looks like a diagonal line; if not. graph again with 1 — 0.01 and 1 + 001. etc. Estimated value: 2.299 2.299 ~ 1.701 0.589 / 1 W x : — : 2 , f( ) 1.1 — 0.9 0.2 99 Exact value: f(:c) 2 3a:2 7 m3 2 f’(:c) = 6x — 3x2. sof’(1):6~3:3. 0.9 1.1 1.701 See the previous exercise. Since f is a decreasing function. assign Y1(3.9) to Ymax and Y1 (4.1) to Ymin. N 049386 w 0.50637 —0.01251 Estimated value: f’(4) ~ 4 1 3 9 : T : —0.06255. Exact value: f(ac) : 26—1/2 2 f’(;c) 2 —§$‘3/2, so f’(4) : —%(4‘3/2) : ~%(§) : —fi — —0.0625. 154 I! CHAPTER 3 DIFFERENTIATION RULES 39. y : m4 + 265”” :> y’ : 4m3 + 261. At (0, 2), y’ : 2 and an equation of the tangent line is y — 2 2 2(30 — O) or y : 2:1: + 2. 40. y : (1 + 2m)2 : 1+ 4:6 + 4:02 => 1/ = 4 + 850. At (1.9), y’ 2 12 and an equation ofthe tangent line is y—9=12(m—1)0ry=12m—3. 41.y:3w2~$3 : y'z6m—3m2.At(1,2),y':6—323 so an equation of the tangent line is y i 2 : 3(30 ~ 1). or y : 3:0 7 1. 42, yam/5:323” => 3/ : 3951/2. At (4,8). 12 y' = %(2) = 3, so an equation of the tangent line is y-8:3(z—4),ory:3x—4. 43. (a) 50 (b) - 10 From the graph in part (a). it appears that f’ is zero at an m r125. 2:2 m 0.5. and 5123 m 3. The slopes are negative (so f’ is negative) on (—00, :01) and (2:2, 3:3). The slopes are positive (so f’ is positive) on (1mm) and ($37 00). (C) f(:c) = $4 — 3x3 — 6922 + 73c + 30 :> 100 f'(:c) 2 4w3 — 9:02 — 122: + 7 _3 A. 5 —4O SECTION 3.1 DERIVATIVES 0F POLYNOMIALS AND EXPONENTIAL FUNCTIONS 155 44. (a) 8 (b) From the graph in part (a)‘ it appears that f’ is zero at m1 2 0.2 and 2:2 2 218. The slopes are positive (so f’ is positive) on (~00, $1) and (mg. 00). The slopes are negative (so fI is negative) on ($112). I 4 45. The curve y 2 2933 + 3x2 — 12310 + 1 has a horizontal tangent when y' 2 6x2 —I— 63: ~ 12 2 0 42 6(9:2 + at ~ 2) 2 0 <2 6(fE + 2)(m — 1) 2 0 42) x 2 —2 orm 2 1. The points on the curve are (‘2, 21) and (1‘ —6). (C) 9C”) 2 ex — 3:32 => g'(x) 2 eI 2 6m 46. f(:c) 2 x3 + 3x2 + as + 3 has a horizontal tangent when f’(a:) 2 32:2 + 6x + 1 2 0 <2 : T6iv636h122—1iéx/6. m 47.y—6m3l5w 3 m—y'218m2—l—5‘butx220f0rallx.som25forallxi 481Theslopeofy:14—26“—3misgivenbym2y’22eI—3. y=I+2€"23x 6 Theslopeof333~y25 <2 y23w—5is3. 771—3 2636 3—3 —> 655—3 => $21n3.This occurs at the point (ln37 7 — 3ln 3) z (1.1., 3.7). Let (a. a2) be a point on the parabola at which the tangent line passes through the point (0. —4). The tangent line has slope 2a and equation y ~ (—4) 2 2a(m — 0) 42} y 2 2cm: 2 4. Since (a, a2) also lies on the line. a2 2 2a(a) — 4. or a2 2 4. So a 2 i2 and the points are (2, 4) and (—2, 4). 156 50. 51. 52. 53. CHAPTER 3 DIFFERENTIATION RULES Ify = x2 + cc. then y' = 2.7: + 1. If the point at which a tangent meets the parabola is ((1412 + a), then the slope of 2 the tangent is 2a + 1. But since it passes through (2. *3). the slope must also be % : (Lia—SL3. a: a — 2 3 . . . Therefore. 2a + 1 : %. Solvmg this equation for a we get a2 + a + 3 = 2a2 — 3a — 2 4:) a2 # 4a a 5 : (a 7 5)(a + 1) = 0 4:) a : 5 or *1. Ifa = 71. the point is (~1.()) and the slope is *1. so the equation is y i O : (—1)(:c + 1) ory : —:c ~ 1. lfa : 5. the point is (5,30) and the slope is 11. so the equation isyg30:11(m—5)ory=11:c~25. y = f(:c) = 1 a m2 => f’(:c) : —293. so the tangent line at (2, —3) has 1 slope f’(2) : —4. The normal line has slope —_—4 : i and equation y+3:%(m—2)ory:fi:c#%. y = f(x) = x — m2 => f’(:c) : 1 # 2x. So f’(1) = —1, and the slope of the normal line is the negative reciprocal of that of the tangent line. that is. —1/(—1) : 1. So the equation of the normal line at (1, 0) is y — 0 : 1(56 — 1) 4:) y : :c — 1. Substituting this into the equation of the parabola, we obtain a: — 1 = x — x2 4:; ac : :El. The solution as = —1 is the one we require. Substituting z : —1 into the equation of the parabola to find the y—coordinate. we have y : —2. So the point of intersection is (—1. —2). as shown in the sketch. 1 1 , ,. f(m+h)—f(m)1. m+hT;_- w—(w+h) W-m h w. 1. 21.13:. WM) #11 -1 1 =Ifimflflm= x2 54. Substituting 2: : 1 and y : 1 into y = (1:02 + bx gives us a + b = 1 (l). The slope of the tangent line 3; = 3a: — 2 is 3 and the slope of the tangent to the parabola at (93, y) is y' — 2am I b. At an 7 1, y’ — 3 —> 3 7 2a + b (2). Subtracting (1) from (2) gives us 2 = a and it follows that b : ~1. The parabola has equation 3; : 2m2 - ac. SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS C 55. f(cc) : 2 — ac ifm g 1 and f(:c) : 9:2 , 2x + 2 ifz > 1. Now we compute the right— and left—hand derivatives defined in Exercise 2.9.46: , f(1+h)—f(1) 2~(1+h)—1 7h . : _ — 1 ~ 2 1 ,1: ~1 d f_(1) hl—llg— h hl—l>o~ h hi? h h—llif- an , . f(1+h)—f(1) . (1+h)2—2(1+h)+2—1 . h2 . = _ _ _ : 1 h: f+(1) Ill—Iii); h 111—133le h Ill—133+ h hill): Thus. f’(1) does not exist since f'_(1) yé fill). so f is not differentiable at 1. But f’(m) 2 —1 form<1andf’(:c):2wi2if:c>1. —172$ ifx<71 56.9(95): m2 if—lgargl :L' if $ > 1 lim w — lim [—1 _ 2(—1+h)] ‘1 A lim _—2h : 11m (—2) = ~2 and h—»0- h h—+O’ h h—>0’ h h-—;O_ , g(—1+h)~g(~1) _ (—1+h)2—1 —2h+h2 l : l E = ‘ = _ I _ .35: I. .2231 h .1135: h .133: < 2 + h) 2’ so 9 is differentiable at —1 and g’(—1) : —2. . g(1) —g(1) (1)2 7 1 , 2h+h2 . 11—13(1): h h—»0- h hi}: it hard: (2 + h) 2 and . 9(1) — 9(1) . (1) -1 ~ h - - 1 : l : — : 2 I 1133+ h 1133+ h ill—1.155r h hl—lfg‘l' 1 1. so 9 (1) does not exist. Thus, 9 is differentiable except when a: = 1, and —2 if$<—1 g'(m)= 22: if —1S$<1 1 ifat>1 57.(a)N0tethatx2—9<Oform2<9 41> Iml<3 41> ~3<x<3.So 352—9 ifz5—3 f(x): —:c2+9 if—3<$<3 :> 362—9 ”9523 2:1: ifx<—3 f’(x): —2:c if ~3<x<3 : {2x if |w|>3 2m ifw>3 —2ac if [2:] < 3 To show that f’(3) does not exist we investigate fling) “3 + h]: _ “3) by computing the left- and right—hand derivatives defined in Exercise 2.9.46. \ 157 158 58. 59. 60. 61. :1 CHAPTER3 DIFFERENTIATION RULES hed)‘ h hao- h . 3 h — 3 h 2 a 7 11(3): hm f( + > f(3) 2 m [< + > 9] 0: hHO‘I' h h~—+0+ h Since the left and right limits are different. (b) . 3 h 7 11m M does not exist. that is. hat) h f'(3) does not exist. Similarly. f’(i3) does not exist. Therefore. f is not differentiable at 3 or at —3. —3 Ifo1.thenh(:c):Ix—lI—I—Im+2|=wil+$+2:2m+1. If—2<:c<1,thenh(9:) -(a:e1)+ac+2=3. Ifm S i2.then h(:c) f(ac—1)—(:c+2)= 72x—1.Therefore. —2ac—1 ifx£;2 —2 ifm<i2 ha): 3 if—2<:c<1 :> h’(:c): 0 if—2<a:<1 2m+1 ifol 2 ifsc>1 To see that h'(1) : lim1 M does not exist. 22—» QC — observe that lim M : lim 3—:2 :Obut za1_ m~1 z—»1' 371 lim M = lirn 2:” _ 2 : 2. Similarly. z—>1+ it — 1 $411+ 2: e 1 h'(—2) does not exist. y : f(w) 2 am2 :> f’(:t:) : 2am. So the slope of the tangent to the parabola at m = 2 is m : 2a(2) : 4a. The slope of the given line. 293 + y : b (i) y = #2m + b. is seen to be #2, so we must have 4a : —2 (i) a = 7%. So when a: = 2. the point in question has y-coordinate —% - 22 = *2. Now we simply require that the given line. whose equation is 2:0 + y = b, pass through the point (2. —2): 2(2) + (—2) : b 4: b : 2. So we must have a : —% and b : 2. f is clearly differentiable for ac < 2 and for ac > 2. For ac < 2, f'(:c) = 236. so f’_ (2) : 4. For as > 2. f'(ac) : m. so 11(2) : m. For f to be differentiable at :c : 2. we need 4 : f’_ (2) : ffir(2) 2 m. So f(at) : 4x + b. We must also have continuity at x = 2. so 4 : f(2) = lim+ (w) : lim+(4w + b) : 8 + b. Hence, b : ~4. z—>2 x—>2 y : f(m) : ax3 + bx2 + ca: + d => f’(:c) : 3am2 + 2bac + c. The point (72.6) is on f. so f(—2) : 6 => —8a + 4b # 20 + d : 6(1). The point (2. 0) is on f. so f(2) : 0 => 8a + 4b + 2c + d : O (2). Since there are horizontal tangents at (—2. 6) and (2, 0). f’(:E2) : 0. f/(—2) = 0 => 12a — 4b + c = O (3) and f'(2) : 0 => 12a + 4b + c : 0 (4). Subtracting equation (3) from (4) gives 8b : 0 i b : 0. Adding (1) and (2) gives 8b + 2d : 6. so d : 3 since b = O. From (3) we have c : —12a. so (2) becomes 8a + 4(0) + 2(—12a) + 3 : 0 :> 3 : 16a => (1 : %. Now c : -12a : —12(f—6) : —3 and the desired cubic function is y : 1%."? — 235 + 3. SECTION 3.2 THE PRODUCT AND OUOTIENT RULES 159 . c . . 62. (a) any 2 c 2> y 2 3. Let P 2 ((1.2). The slope of the tangent line at ca 2 a is y'(a) 2 —a—2. Its equation [S w a c c 0 2c , ‘ . . 20 . _ . _ y — d 2 —§(m 2 a) org 2 ill—2m + 3, so its y-mtercept is 3. Setting y — 0 gives ac — 2a. so the . 2 . c ac—intercept is 2a. The midpoint of the line segmentjoimng (0. EC) and (2a, 0) IS (a, a) 2 P. (b) We know the ac— and y—intercepts of the tangent line from part (a). so the area of the triangle bounded by the axes and the tangent is %(base)(height) 2 émy 2 %(2a)(2c/a) 2 2c. a constant. 63. Solution 1: Let f (cc) 2 21000. Then, by the definition of a derivative, . flat) 2 f(1) . 21°00 21 . . . l . f’(1) — 11m 1 — hm 1 . But this IS JUSI the limit we want to find, and we know (from the 121 {E — 2221 .T 2 $1000 _ 1 Power Rule) that f'(:c) : 1000x999, so f’(1) = 1000(1)9‘99 = 1000. So lim — :1000 1—»1 :E—l Solution 2: Note that (101000 — 1) 2 (:0 — 1)(:I:999 + 9:998 + x997 + - - o + m2 + a: + 1). So _ 9000-1 ‘ ($—1)(m999+w998+x997+~-+x2+z+1) 11m 11m ma] (5—1 m—bl 11—1 2—»1 =lim($999+x998+x997+---+x2+x+1):1+1+1+w+1+1+1 “ 1000 ones 2 1000, as above. 64. In order for the two tangents to intersect on the y—axis, the points of tangency must be at equal distances from the y-axis. since the parabola y 2 x2 is symmetric about the y-axis. Say the points of tangency are (a, a2) and (—0,,(12). for some a > 0. Then since the derivative ofy 2 :62 is dy/dx 2 2x. the left—hand tangent has slope —2a and equation 3; — a2 2 ~2a(m + a), or y 2 —2aw — a2. and similarly the right—hand tangent line has equation y — a2 2 2a(x ~ a), or y 2 2am — (12‘ So the two lines intersect at (0, —a2). Now if the lines are perpendicular. then the product of their slopes is 71, so (—2a)(2a) 2 —1 (2 a2 2 1 2 <2 (1 2 5 So the lines intersect at (0, -41). 3.2 The Product and Quotient Rules 1. Product Rule: y 2 (x2 + 1) (3:3 + 1) 2> I y 2 (x2 + 1) (3x2) + (2:3 + 1)(2a:) 2 3m4 + 3202 + 2x4 + 2x 2 5x4 + 3:102 + 2x. Multiplying first: y 2 (m2 + 1) (m3 + 1) 2 m5 + .223 + 2:2 + 1 2> y' 2 59:4 + 3:62 + 2x (equivalent). ...
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