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Unformatted text preview: 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 cc 2 a is y'(a) 2 —a—2. Its equation [S w a c c 0 2c , ‘ . . 20 . _ . _ y — a 2 —§(m 2 a) org 2 ill—2m + 3, so its y-mtercept is 3. Setting y — 0 gives a: — 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 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 m 2 \$1000 _ 1 Power Rule) that f'(:c) : 1000x999, so f’(1) = 1000(1)999 = 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 3:21 (5—1 m—bl 11—1 x—A =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 (12 2 1 Z <=> a 2 So the lines intersect at (0, -41). 3.2 The Product and Quotient Rules 1. Product Rule: y 2 (x2 +1)(a:3 + 1) => 1 y 2 (x2 +1)(3\$2) + (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). 160 CHAPTER 3 DIFFERENTIATION RULES 3: 7 3:3 ﬂ at i 3303/2 2. Quotient Rule: = T = 71/2— => 361/2 (1 _%x1/2) 7 (ac _ 3w3/2>(%\$—1/2) F'W) * 2 (ml/2) _ 561/2 _ gm —%\$1/2 + gas i 511/2 _ 3x I lm_1/2 _ 3 x :c 2 Simplifying ﬁrst: : w = ﬂ — 3m : 11/2 — 3110 => F'(:c) = inc—U2 — 3 (equivalent). at For this problem. simplifying ﬁrst seems to be the better method. 3. By the Product Rule. f(ac) : mazem => f'(a:) : x2 6% (ex) + e‘” % (1:2) = 9:261 + 63129:) = meza + 2). 2 4. B theProductRule.gm : xexzzclﬂem => 9' so 2931/2 e1 +63: 1534/2 :lx_1/2em(2ac+1). y 1} 5. By the Quotient Rulet y : €— 2} 332 d d 2 m x 2 , x 505 >—6 g (95) : ﬂew—exam) _ mew—2) : ewe—2) y _ (m2)2 \$4 \$4 \$3 , # ex , (1+x)ewvez(1) _e””—l—:I:ea”—e”c _ acem 6. By the Quotient Rule. y — 1+ 30 —> y — (1 + my (36 + D2 — (x + U2. The notations 2% and Q2]; indicate the use of the Product and Quotient Rules, respectively. 3.271 R , (23c+1)(3)—(3ac~1)(2)_6m+3'6x+2# 5 7'9(\$):2w+1 Q: 9Im)‘ (n+1)2 (2x+1)2 _(2w+1)2 2t QR , _ (4+t2)(2)—(2t)(2t) _ 8+2t2—4t2 _ 8—2252 s'fIt):4+t2 :5 fItI— (4+t2)2 # (4+2?)2 7(4+t2)2 9. V(z) = (2x3 + 3W4 # 2x) 2% V’(x) = (25c3 + 3)(4:c3 — 2) + (:34 - 2\$)(6m2) : (8:56 + 8:163 —— 6) + (6:166 i 12363) = 143:6 — 4x3 7 6 10. Y(u) : (qr-2 + ur3)(u5 — 2u2) 2i Y'(u) : (If? + u-3)(5u4 — 4n) + (11.5 ! 2u2)(—2u—3 — 3u'4) : (5112 — 41F1 + 51L — 111L’2)—l—(—2u2 - 3n + 41f1 + 6u_2) : 3n2 + 211, + 2u‘2 ' 2 r 7>(y + 5313) : (th #3y_4)(y+5y3) 2‘; F'(y) : (y—‘2 — 3y’4)(1 + 153/2) + (y + 5313) (—2y*3 + 12yr5) : (3f2 + 15 — 3y_4 — 4574—2) + (Jar2 +12y’4 — 10 + Gog—2) : 5 +14y‘2 + 9y’4 or 5 + 14/342 + 9/y4 ‘E PR 12. R(t) : (t+e‘)(3 7 ﬂ) :> R’(t) : (t +e‘)(—§t‘1/2) + (3 — x/i) (1 +e) : (_%t1/2#%t—1/Zet) +(3+3et_\/i_\/Eet) :3+36t; %ﬂ_\/Eet_et/(2\/t‘) 13. 14. 15. 16. 17. 18. 19. 20. 21. 24. SECTION 3.2 THE PRODUCT AND OUOTIENT RULES : L g; y 32:2 — 21: +1 , (32:2 — 2t + 1)(2t) — t2(6t — 2) 21132:2 — 2t + 1 ~ t(3t — 1)] y (37:2 — 2t + 1)2 7 (32:2 — 215 +1)2 2t(3t2 — 2t + 1 — 3152 + t) 2t(1 — t) _ (3t? — 2t+1)2 _ (3222 — 21t+1)2 t3 +t QR , (t4 — 2) (312 + 1) i (t3 + t) (49) (31:6 + 11“ ~ 6262 ~ 2) 7 (4t6 + 4114) : é : _ y t4 ~ 2 y (t4 — 2)2 (t4 i 2)2 _ —t6~3t4~6t2—2 _ t6+3t4+6t2+2 _ (It4 — 2V (t4 — 2)2 y 2 (T2 * 2r)er :1; y' : (r2 — 2r)(er) + 6(27" — 2) : er(r2 — 27" + 27‘ — 2) = er(r2 ~ 2) 1 QR , (3+kes)(0) — (1)(1+ke5) 1+kes y = n y n 2 — —*2 s+kes (3+kes) (3+kes) 3— yzw :v2~2\/27=v2—2v1/2 => yl:2v—2(%)v-1/222117171/2. v _ 3/2 _ We can change the form of the answer as follows: 21) — 11—1/2 = 21) — i : M = M ﬁ ﬁ \/27 z : HIS/2(a) + ceu’) = 1115/2 + cw3/2ew => 2/ : gw3/2 +C(w3/2.ew +621) . gun/2) : Ems/2 + écwl/26w(2w+3) 1 , (x4+z2+1)(0)—1(4x3+2;c) 2\$(2w2+1) YFW —> 31— 2 “%2 at +3: +1 (w4+m2+1) (m4+z2+1) _ «5—1 y‘ «5+1 1 1 1 1 1 1 y,_(\/E+1)(2ﬁ) («5 1>(2ﬂ)_2+2WE +2ﬁ 1 («inf («inf VEwa IL‘ fa) : 93+c/m :> I i (x+c/x)(1)~z(1—C/x2) a:+c/z~m+c/ac 2c/x 9:2 20x “95) C 2 — 2 2 —ﬁ“2=‘2 (Hg) (1" +0) (w +6) 9” (962“) {13 1:2 161 ﬁx): am-f-b :> f,(x) : (cm+d)(a)—(am+b)(c) acm—i—ad—acx—bc _ ad—bc ca: + d (car: + d)2 W (cm + d)2 _ (cm + d)2 220 / (w + 1)(2) — (230(1) I —> —_ _ ‘ I 2 l . . y m +1 3/ (x +1)2 (a: + Dz. At (1,1), y 2. and an equatlon of the tangent lme isy~1=%(x—1),ory=%m+%. (acumﬁ) 45(1) y: x/5 y/_ _(\$+1)—(2x)_ l—m AWN”) 93“ (m+1)2 2¢5(x+1)2 ‘2ﬁ(z+1)2' " * y, = 1%)?) 2 —0.03, and an equation ofthe tangent line is y — 0.4 : 7003(25 — 4), or y : —0.03a: + 052, 162 CHAPTER 3 DIFFERENTIATlON RULES 25. y : 2:456m => y' : 2(w-ez +6I - 1) :2ex(:c+1). At(0,0).y' : 260(0—l— 1) : 2-1-1: 2.andan equation of the tangent line is y i 0 = 2(m - 0). or y = 29:. ez chem—em-l em —1 , . 26. y 7 Z 9 y' g 962 = (:2 At (1,6). 3; : 0, and an equation of the tangent line is y—e:0(a:71).ory=e. 1 27. = : (a)?! ﬁx) 1+\$2 (b) 1+ac2 0 —1 2:1: ,2 f’(a:) —( ) 2( )2 x 2.Sotheslopeofthe <1+z2> (was?) i . . 2 . tangent line atthe pomt (—1, IS f’(—1) = 2—2 = i and its equationisy~ % — %(w+1)ory : %:c+ 1. x = : :> (a) y ftm) 1 +332 (b) 1+ 9:2 1— :0 2:5 1 — 2 f’(;v) # ( ) 2( ) = w 2.Sothe slopeofthe (ll-\$2) (1+ac2) tangent line at the point (3, 0.3) is f’(3) : 1:0?) and its equation is y — 0.3 : —0.08(ac — 3) or y = ~0.08m + 054. 3 m x 2 2 e“: a? (e )—6 3m x 62(fti3) er(x—3) 29. (a) f(a:):F => f’(;c);———(\$—3)2(—l:__—mG——:T (b) f’ = 0 when f has a horizontal tangent line, f’ is negative when f is decreasing, and f’ is positive when f is increasing # g; (mg—1)1~m(2m) _ -x2—1 (a) _ I _ (m2 7 1)2 _ (\$2 _ 1)2 (b) Notice that the slopes of all tangents to f are negative and f’(a:) < 0 always. 31. We are given that f(5) : 1. f'(5) : 6, 9(5) : —3. and g'(5) = 2. (a) (fg)'(5) : f(5)g'(5) + 9(5)f'(5) = UK?) + (—3)(6) = 2 e 18 : —16 f ’ _g(5)f’(5)—f(5)g’(5) _<a3><6>-<1><2> by (b) H (5* [9(5)]? 3 (-3)? 9 g I _ f(5)9'(5) '9(5)f/(5) : (1X2) — (—3)(6) I (C) (f) (5)” “6(5)? (1)2 20 SECTION 3.2 THE PRODUCT AND QUOTIENT RULES 163 32. We are given that f(3) = 4, 9(3) : 2. f’(3) : 46‘ and g’(3) : 5_ (a) (f +9)’ (3) = f’(3 +9’(3) : —6 + 5 = 71 (b) (f9)’ (3) = f(3)9'(3) + 9(3)f'(3) : (4)(5) + (2)(—6) : 20 i 12 : 8 i ’ _ 9(3) '(3) — f(3)g/(3) 7 mm ~ <4>(s> _ ~32 g (C) (9) (3) ‘ ig<s>r (2)2 4 L ’ _ [f(3) — 9mm) — f(3) [f’(3) —g'<3>1 “” (f-g) (3) [f(3)—9(3)l2 (4 — 2)(—6) — 4(—6 4 5) —12 + 44 Z 8 ~8 (4 a 2)2 _ 22 33- f(w) : ez9(4) => f’(m) = ems/(m) + we)? = ex lg'(w) + gm]- f’(0) : 6” lg'(0) + 9(0)] :1(5 + 2) Z 7 d h(:c) mh'(m)—h(m)~1 d hm) 2h’(2)—h(2) 2(—3)~(4) —10 34'E[m]: x2 9 dx[xL:2_ i _ —2.5 22 4 4 35. (a) From the graphs off and 9. we obtain the following values: f(1) : 2 since the point (1, 2) is on the graph of f; g(1) = 1 since the point (1. 1) is on the graph ofg; f’(1) : 2 since the slope of the line segment between (0, 0) 4 _ and (2,4) is 0 = 2; g’(1) : —1 since the slope of the line segment between (—2, 4) and (2, 0) is 2 (i (:12) : ~1. Now we) = f(:c)g(x). so am = f(1)g’(1)+g(1)f’(1) = 2- <—1>+1-2 : o. _ ,1 a 9(5)f’(5) — f(5)g’(5) 2(—§) — 3.; _§ 2 (b) Mir) * soc (5) _ [9(5)]2 ﬁ 22 T 3 36. (a) Pm = F(.”L‘)G(w). so P’(2) : F(2)G’(2) + G(2)F’(2) : 3 . g + 2 . 0 : g. (b) CM) :F(w)/G(x).so m7) 7 G(7)F’(7) —F(7)G’(7) : 1-Z ~5- 7) 1 10 43 [G(7)]2 1 ‘ 4 + 3 * 12 37- (a) y = 239(96) 3 y’ = \$9496) + 9(06) ~ = 179/(93) + 9(90) _ i A I _ 9(92) - 1 ~ wg’(m) 9(56) — mg’(x) “0 y gm y [ng ‘ mm)? (c) y — % —> y, _ \$9'(\$)(;)g(zr) ~1 _ \$949312— 9(4) 38. (a) y = afﬁx) 2 y’ 2 Wm + f(m)(2x) (b) y _ fig?) y, _ m2f’(m)( —2;‘2(\$)(2\$) _ WNW); 2f(\$) I 962 , _ f(\$)(2\$) exzf’W) (C) y ftx) ;‘ y ‘ mm? (d) 9: 1+5?” => I «a? were) + M] ~ [1+ mm] 2 1ﬁ y ﬂ 2 (ﬁt) - x3/2f1(\$)+1,1/2f(\$) _ ém~1/2 _ %x1/2f(x) 2:61/2 + gaff/(it) _1 w ' 2w”? h 2103/2 164 C CHAPTER3 DIFFERENTIATIONRULES 39. If P(t) denotes the population at time t and A(t) the average annual income, then T(t) = P(t)A(t) is the total personal income. The rate at which T(t) is rising is given by T'(t) : P(t)A'(t) + A(t)P’(t) : T’(1999) : P(1999)A’(1999) + A(1999)P’(1999) : (961 .400)(\$1400/yr) + (\$30.593)(9200/yr) = \$1,345,960.000/yr + \$281.455.600/yr : \$1.627.415.600/yr So the total personal income was rising by about \$1.627 billion per year in 1999. The term P(t)A'(t) z \$1.346 billion represents the portion of the rate of change of total income due to the existing population‘s increasing income The term A(t)P'(t) 2: \$281 million represents the portion of the rate of change of total income due to increasing population. 40. (a) f(20) : 10.000 means that when the price of the fabric is \$20/yard, 10.000 yards will be sold. f '(20) : —350 means that as the price of the fabric increases past \$20/ yard. the amount of fabric which will be sold is decreasing at a rate of 350 yards per (dollar per yard). (b) R(p) = pm?) => R’(P) : pf’(p) + ﬂ?) ‘ 1 => R'(20) : 20f’(20) + f(20) - 1 = 20(7350) + 10000 = 3000. This means that as the price of the fabric increases past \$20/ yard. the total revenue is increasing at \$3000/ (\$ / yard). Note that the Product Rule indicates that we will lose \$7000/ (\$/yard) due to selling less fabric. but that that loss is more than made up for by the additional revenue due to the increase in price. w . then f'(:c) 7 (m + 1x1) ‘ 36(1) 2 1 . When an : a, the equation of the tangent 41.1fy=f(:v): 5H1 (x+1)2 (ac+1)2 a 1 . . a 1 a + 1 : W01: - a). This line passes through (1,2) when 2 — a + 1 z m 2(a+1)2—a(a+1):1—a <4 2a214a+2 a2 a 1+a—0 \ a2+4a+1=0. a4 42 —4(1)<1> e4 _ git/3, The quadratic formula gives the roots of this equation as a _ 2(1) : 2 (l—a) <=> line is y — so there are two such tangent lines. Since 6 —2i\/§ ~2ix/3. #wﬁ f(T2:\/§):-2i\/§+1:—1i\/§.—1¥\/§ -6 . 6 2::2\/'3‘;\/§—3 # —1i\/§_1\$\/§ ' 1—3 _ —2 2 *6 the lines touch the curve at A(—2 + z (#027. —0.37) and B<g2 — J3 kgé) z (#373137). 42. y : x 7 1 4 y’ — (x + 1)(1)_ (x J “(1) — 1fthe tangent intersects the curve when x : a, z+1 (:c+1)2 (x—1—1) then its slope is 2/(a + 1)2. But ifthe tangent is parallel to ac ~ 2y : 2. 1 that is. y : ix — 1. then its slope is Thus. ——-— : :> (MI)2 (a+1)2:4 :> a+1::l:2 => a:1ori3.Whena—1.y*0 and the equation of the tangent is y — 0 = as; — 1) or y : éx - When a : —3. y : 2 and the equation of the tangent is y e 2 : + 3) 1 7 ory:§x+§. SECTION 3.3 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES I: 165 We will sometimes use the form f’g + fg' rather than the form fg’ + gf’ for the Product Rule. 43- (a) (fgh)' : I(f9)hl’ : (fg)’h + (f9)h' = (f’g + fg’)h + (fg)h’ = f’gh + fg'h + fgh’ (b) Putting f : g = h in part (a) we have i was)? : (fff)’ : f’ff + ff’f + m/ = 3m“ : 3lf(x)l2f'(w)~ (c) i (832:) : % (em)3 : 3(ez)2 ex = 362%2 : 36335 d d d 1 _9(\$)‘d—\$(1)*1'£Ig(\$)l _ 44. (3) dz 7 [amp [Quotient Rule] _ 9(56) ~0—1‘g’(x) ﬂ 0-9/(36) ‘ _M [9(96)I2 [903)]2 [9%)]2 b) _ 1 _> , _ 4x3 + 21: or —2:c(2:c2 + 1) ( y_x4+:1:2+1 y‘ (\$4+:c2+1)2 (:c4+ac2+1)2 d gn _ d 1 ¥ (mu), - _ inxn—l _ _ n—172n _ _ —n71 (c) a (x ) _ ﬂ _ —(mn)2 [by the Reelprocal Rule] — \$271 7 my — m: 3.3 Rates of Change in the Natural and Social Sciences —\ 1. (a) s = f(t) : t2 — 10t+ 12 => 22(t) : f’(t) : 2t — 10 (b) 12(3) 2 2(3) — 10 2 —4 ft/s (c) The particle is at rest when v(t) = 0 <:> 2t — 10 = 0 <:> t = 5 s. (d) The particle is moving in the positive direction when -v(t) > 0 <:> 2t ~ 10 > 0 <:> 225 > 10 ¢> t > 5, (e) Since the particle is moving in the positive direction (f) t= 8. and in the negative direction. we need to calculate the S = —4 distance traveled in the intervals [0, 5] and [5. 8] t= 5. separately. If(5) — f(0)| : [—13 —12|: 25 ft and S: _13 I: 0: ms) —f(5)[ = 1—44—13» 29ft. The total 3212 4. g 0 distance traveled during the ﬁrst 8 s is 25 + 9 : 34 ft. 2. (a) s : f(t) : t3 — 9i2 + 152: + 10 :> v(t) : f’(t) : 32:2 i 18t + 15 = 3(t — 1)(t — 5) (b) 11(3) : 3(2)(—2) : ~12ft/s (f) —8, =66 (c)v(t):0 <=> t=lsor55 ,:5' Hh—s (d)v(t)>0 ¢> ogt<1ort>5 5315 t=l, t=0, 5:17 (6) If(1)— f(0)I = I17*10I= 7~ j: 10 /F a 116(5); f(1)| : [~15 —17i= 32. and 0 If(8) e f(5)I = I66 e (E15)I : 81~ Total distance : 7 + 32 + 81 : 120 ft. 3. (a) s : f(t) = t3 —12t2 + 36t => 12(t) : f’(t) : 3t2 — 242: + 36 (b) v(3) : 27 — 72 + 36 = —9 ft/s ...
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