{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 1 - 3 Cl DIFFERENTIATION RULES 3.1 Derivatives of...

This preview shows pages 1–9. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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). ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern