Solucao impares Calculo 6ed CAP.3

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3 DIFFERENTIATION RULES

3.1 Derivatives of Polynomials and Exponential Functions

1. (a)h is the number such that lim k0 hk₁ k = 1. (b) { 2=7{_{ 1 0=001 0=9928 0=0001 0=9932 0=001 0=9937 0=0001 0=9933 { 2=8{_{ 1 0=001 1=0291 0=0001 1=0296 0=001 1=0301 0=0001 1=0297

From the tables (to two decimal places), lim k0 2=7k 1 k = 0=99 and limk0 2=8k 1 k = 1=03. Since 0=99 ? 1 ? 1=03, 2=7 ? h ? 2=8.

3. i({) = 186=5 is a constant function, so its derivative is 0, that is, i0({) = 0. 5. i(w) = 2 2₃w i0(w) = 0 2₃ =2₃ 7. i({) = {3 4{ + 6 i0({) = 3{2 4(1) + 0 = 3{2 4 9. i(w) =1₄(w4+ 8) i0(w) = 1₄(w4+ 8)0=1₄(4w41+ 0) =w3 11. | = {2@5 |0=2₅{(2@5)1=2₅{7@5= 2 5{7@5 13. Y (u) = 4₃u3 Y0(u) = 4₃3u2= 4u2 15. D(v) = 12 v5 =12v5 D0(v) = 12(5v6) = 60v6 or 60@v6 17. J({) ={ 2h{={1@2 2h{ J0({) = 1₂{1@2 2h{= 1 2{ 2h { 19. I ({) = (1₂{)5=1₂5{5= ₃₂1{5 I0({) = ₃₂1(5{4) =₃₂5{4 21. | = d{2+e{ + f |0= 2d{ + e 23. | = { 2_{+ 4}_{{ + 3} { ={ 3@2_{+ 4}_{1@2_{+ 3}_{1@2 |0₌ 3 2{1@2+ 4 ₁ 2 {1@2_{+ 3}1 2 {3@2₌3 2 { +2 { 3 2{{ k note that{3@2={2@2· {1@2={{l

The last expression can be written as 3{ 2 2{{+ 4{ 2{{ 3 2{{ = 3{2+ 4{ 3 2{{ . 25. | = 42 |0= 0since 42is a constant. 87

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27. We rst expand using the Binomial Theorem (see Reference Page 1). K({) = ({ + {1₎3 ₌_{3_{+ 3}_{2_{1_{+ 3}_{({1₎2_{+ (}_{1₎3₌_{3_{+ 3}_{{ + 3{}1₊_{3 K0₍_{{) = 3{}2_{+ 3 + 3(}_1{2_{) + (}_3{4_{) = 3}_{2_{+ 3}_3{2_3{4 29. x =5w + 4w5=w1@5+ 4w5@2 x0=1₅w4@5+ 4 5 2w3@2 = 1₅w4@5+ 10w3@2 or 1@ 55w4 + 10w3 31. } = D |10 +Eh|=D|10+Eh| }0=10D|11+Eh|= 10D |11 +Eh| 33. | =4{ = {1@4 |0=1₄{3@4= 1 44{3. At (1> 1), | 0₌ 1

4and an equation of the tangent line is | 1 =1

4({ 1) or | =14{ +34.

35. | = {4+ 2h{ |0= 4{3+ 2h{. At (0> 2), |0= 2and an equation of the tangent line is| 2 = 2({ 0) or| = 2{ + 2. The slope of the normal line is 1₂(the negative reciprocal of 2) and an equation of the normal line is | 2 = 1

2({ 0) or | = 12{ + 2. 37. | = 3{2 {3 |0= 6{ 3{2.

At (1> 2), |0= 6 3 = 3, so an equation of the tangent line is | 2 = 3({ 1) or | = 3{ 1.

39. i({) = h{ 5{ i0({) = h{ 5.

Notice thati0({) = 0 when i has a horizontal tangent, i0is positive wheni is increasing, and i0is negative wheni is decreasing.

41. i({) = 3{15 5{3+ 3 i0({) = 45{14 15{2.

Notice thati0({) = 0 when i has a horizontal tangent, i0is positive wheni is increasing, and i0is negative wheni is decreasing.

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SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ¤ 89 43. (a) (b) From the graph in part (a), it appears thati0is zero at{₁ 1=25, {₂ 0=5,

and{3 3. The slopes are negative (so i0is negative) on (> {1)and ({2> {3). The slopes are positive (soi0is positive) on ({1> {2)and ({3> ).

(c)i({) = {4 3{3 6{2+ 7{ + 30 i0₍_{{) = 4{}3_9{2_{12{ + 7}

45.i({) = {4 3{3+ 16{ i0({) = 4{3 9{2+ 16 i00({) = 12{2 18{ 47.i({) = 2{ 5{3@4 i0({) = 2 15₄{1@4 i00({) = 15₁₆{5@4

Note thati0is negative wheni is decreasing and positive when i is increasing.i00is always positive sincei0is always increasing.

49. (a)v = w3 3w y(w) = v0(w) = 3w2 3 d(w) = y0(w) = 6w (b)d(2) = 6(2) = 12 m@s2

(c)y(w) = 3w2 3 = 0 when w2= 1, that is,w = 1 and d(1) = 6 m@s2.

51.The curve| = 2{3+ 3{2 12{ + 1 has a horizontal tangent when |0= 6{2+ 6{ 12 = 0 6({2+{ 2) = 0 6({ + 2)({ 1) = 0 { = 2 or { = 1. The points on the curve are (2> 21) and (1> 6).

53.| = 6{3+ 5{ 3 p = |0= 18{2+ 5, but{2 0 for all {, so p 5 for all {.

55.The slope of the line 12{ | = 1 (or | = 12{ 1) is 12, so the slope of both lines tangent to the curve is 12.

| = 1 + {3 _|0_{= 3}_{2_{. Thus, 3}_{2_{= 12} _{2_{= 4} _{{ = ±2, which are the {-coordinates at which the tangent} lines have slope 12. The points on the curve are (2> 9) and (2> 7), so the tangent line equations are | 9 = 12({ 2) or| = 12{ 15 and | + 7 = 12({ + 2) or | = 12{ + 17.

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57. The slope of| = {2 5{ + 4 is given by p = |0= 2{ 5. The slope of { 3| = 5 | =1₃{ 5₃ is1₃,

so the desired normal line must have slope1₃, and hence, the tangent line to the parabola must have slope3. This occurs if 2{ 5 = 3 2{ = 2 { = 1. When { = 1, | = 12 5(1) + 4 = 0, and an equation of the normal line is | 0 =1

3({ 1) or | =13{ 13.

59. Letd> d2be a point on the parabola at which the tangent line passes through the point (0> 4). The tangent line has slope 2d and equation | (4) = 2d({ 0) | = 2d{ 4. Sinced> d2_{also lies on the line,}_d2_{= 2}_{d(d) 4, or d}2_{= 4. So} d = ±2 and the points are (2> 4) and (2> 4).

61. i0({) = lim k0 i({ + k) i({) k = limk0 1 { + k 1 { k = limk0 { ({ + k) k{({ + k) = limk0 k k{({ + k) = limk0 1 {({ + k) = 1 {2 63. LetS ({) = d{2+e{ + f. Then S0({) = 2d{ + e and S00({) = 2d. S00(2) = 2 2d = 2 d = 1.

S0_{(2) = 3} _{2(1)(2) + e = 3 4 + e = 3 e = 1.}

S (2) = 5 1(2)2_{+ (}_{1)(2) + f = 5 2 + f = 5 f = 3. So S ({) = {}2_{{ + 3.}

65. | = i({) = d{3+e{2+f{ + g i0({) = 3d{2+ 2e{ + f. The point (2> 6) is on i, so i(2) = 6 8d + 4e 2f + g = 6 (1). The point (2> 0) is on i, so i(2) = 0 8d + 4e + 2f + g = 0 (2). Since there are horizontal tangents at (2> 6) and (2> 0), i0(±2) = 0. i0(2) = 0 12d 4e + f = 0 (3) and i0(2) = 0 12d + 4e + f = 0 (4). Subtracting equation (3) from (4) gives 8e = 0 e = 0. Adding (1) and (2) gives 8e + 2g = 6, sog = 3 since e = 0. From (3) we have f = 12d, so (2) becomes 8d + 4(0) + 2(12d) + 3 = 0 3 = 16d d = 3

16. Nowf = 12d = 12 ₃

16

=9₄ and the desired cubic function is| =₁₆3{39₄{ + 3.

67. i({) = 2 { if { 1 and i({) = {2 2{ + 2 if { A 1. Now we compute the right- and left-hand derivatives dened in Exercise 2.8.54: i0 (1) = lim k0 i(1 + k) i(1) k = limk0 2 (1 + k) 1 k = limk0 k k = limk01 = 1 and i0 +(1) = lim k0+ i(1 + k) i(1) k = limk0+ (1 +k)2 2(1 + k) + 2 1 k = limk0+ k2 k = limk0+k = 0.

Thus,i0(1)does not exist sincei0(1)6= i₊0(1), soi is not differentiable at 1. Buti0({) = 1 for { ? 1 andi0({) = 2{ 2 if { A 1.

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SECTION 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ¤ 91 69. (a) Note that{2 9 ? 0 for {2? 9 |{| ? 3 3 ? { ? 3. So

i({) = ; A A ? A A = {2₉ _if _{{ 3} {2_{+ 9} _if _{3 ? { ? 3} {2₉ _if _{{ 3} i0₍_{{) =} ; A ? A = 2{ if { ? 3 2{ if 3 ? { ? 3 2{ if { A 3 = + 2{ if |{| A 3 2{ if |{| ? 3 To show thati0(3)does not exist we investigate lim

k0

i(3 + k) i(3)

k by computing the left- and right-hand derivatives dened in Exercise 2.8.54. i0 (3) = lim k0 i(3 + k) i(3) k = limk0 [(3 + k)2+ 9] 0 k = limk0(6 k) = 6 and i0 +(3) = lim k0+ i(3 + k) i(3) k = limk0+ (3 +k)2 9 0 k = limk0+ 6k + k2 k = limk0+(6 +k) = 6.

Since the left and right limits are different, lim

k0

i(3 + k) i(3)

k does not exist, that is,i0(3) does not exist. Similarly,i0(3) does not exist. Therefore,i is not differentiable at 3 or at 3.

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71.Substituting{ = 1 and | = 1 into | = d{2+e{ gives us d + e = 1 (1). The slope of the tangent line | = 3{ 2 is 3 and the slope of the tangent to the parabola at ({> |) is |0= 2d{ + e. At { = 1, |0= 3 3 = 2d + e (2). Subtracting (1) from

(2) gives us 2 = d and it follows that e = 1. The parabola has equation | = 2{2_{.

73.| = i({) = d{2 i0({) = 2d{. So the slope of the tangent to the parabola at { = 2 is p = 2d(2) = 4d. The slope of the given line, 2{ + | = e | = 2{ + e, is seen to be 2, so we must have 4d = 2 d = 1₂. So when { = 2, the point in question has |-coordinate 1

2· 22=2. Now we simply require that the given line, whose equation is 2{ + | = e, pass through the point (2> 2): 2(2) + (2) = e e = 2. So we must have d = 1₂ ande = 2. 75.i is clearly differentiable for { ? 2 and for { A 2. For { ? 2, i0({) = 2{, so i0(2) = 4. For{ A 2, i0({) = p, so

i0

+(2) =p. For i to be differentiable at { = 2, we need 4 = i0(2) =i+0(2) =p. So i({) = 4{ + e. We must also have continuity at{ = 2, so 4 = i(2) = lim

{2+i({) = lim_{2+(4{ + e) = 8 + e. Hence, e = 4.

77.Solution 1: Let i({) = {1000. Then, by the denition of a derivative,i0(1) = lim {1

i({) i(1) { 1 = lim{1

{1000₁ { 1 . But this is just the limit we want to nd, and we know (from the Power Rule) thati0({) = 1000{999, so

i0_{(1) = 1000(1)}999_{= 1000. So lim} {1

{1000₁ { 1 = 1000.

Solution 2: Note that ({1000_{1) = ({ 1)({}999₊_{998₊_{997₊_{· · · + {}2₊_{{ + 1). So} lim {1 {1000₁ { 1 = lim{1 ({ 1)({999+{998+{997+· · · + {2+{ + 1) { 1 = lim{1({ 999₊_{998₊_{997₊_{· · · + {}2₊_{{ + 1)} = 1 + 1 + 1 +· · · + 1 + 1 + 1 ~} = 1000, as above. 1000 ones

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79. | = {2 |0= 2{, so the slope of a tangent line at the point (d> d2)is|0= 2d and the slope of a normal line is 1@(2d), ford 6= 0. The slope of the normal line through the points (d> d2)and (0> f) is d

2_f d 0, sod 2_f d = 1 2d d2_{f =}1

2 d2=f 12. The last equation has two solutions iff A12, one solution iff =12, and no solution if f ?1

2. Since the|-axis is normal to | = {2regardless of the value off (this is the case for d = 0), we have three normal lines iff A 1₂and one normal line iff 1₂.

3.2 The Product and Quotient Rules

1. Product Rule: | = ({2+ 1)({3+ 1)

|0_{= (}_{2_{+ 1)(3}_{2_{) + (}_{3_{+ 1)(2}_{{) = 3{}4_{+ 3}_{2_{+ 2}_{4_{+ 2}_{{ = 5{}4_{+ 3}_{2_{+ 2}_{. Multiplying rst:| = ({2+ 1)({3+ 1) ={5+{3+{2+ 1 |0= 5{4+ 3{2+ 2{ (equivalent). 3. By the Product Rule,i({) = ({3+ 2{)h{

i0₍_{{) = ({}3_{+ 2}_{)(h{₎0₊_h{₍_{3_{+ 2}_{)0_{= (}_{3_{+ 2}_{)h{₊_h{₍₃_{2_{+ 2)} =h{[({3+ 2{) + (3{2+ 2)] =h{({3+ 3{2+ 2{ + 2)

5. By the Quotient Rule,| = h { {2 |0= {2 g g{(h{) h{g{g ({2) ({2)2 = {2₍_h{₎_h{₍₂_{) {4 ={h {₍_{{ 2)} {4 =h {₍_{{ 2)} {3 . The notations PR and QR indicate the use of the Product and Quotient Rules, respectively.

7. j({) = 3{ 1 2{ + 1 QR j0₍_{{) =} (2{ + 1)(3) (3{ 1)(2) (2{ + 1)2 = 6{ + 3 6{ + 2 (2{ + 1)2 = 5 (2{ + 1)2 9. Y ({) = (2{3+ 3)({4 2{) PR Y0₍_{{) = (2{}3_{+ 3)(4}_{3_{2) + ({}4_2{)(6{2_{) = (8}_{6_{+ 8}_{3_{6) + (6{}6_12{3_{) = 14}_{6_4{3₆ 11. I (|) = 1 |2 3 |4 (| + 5|3) =|2 3|4| + 5|3 PR I0₍_{|) = (|}2_3|4_{)(1 + 15}_|2_{) + (}_{| + 5|}3₎₍_2|3_{+ 12}_|5₎ = (|2+ 15 3|4 45|2) + (2|2+ 12|4 10 + 60|2) = 5 + 14|2+ 9|4 or 5 + 14@|2+ 9@|4 13. | = { 3 1 {2 QR |0₌(1 {2) (3{2) {3(2{) (1 {2)2 = {2₍₃_3{2_{+ 2}_{2₎ (1 {2)2 = {2₍₃_{2₎ (1 {2)2

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SECTION 3.2 THE PRODUCT AND QUOTIENT RULES ¤ 93 15.| = w 2_{+ 2} w4_3w2_{+ 1} QR |0₌ (w4 3w2+ 1)(2w) (w2+ 2)(4w3 6w) (w4 3w2+ 1)2 = 2w[(w4 3w2+ 1) (w2+ 2)(2w2 3)] (w4 3w2+ 1)2 = 2w(w 4_3w2_{+ 1}_2w4_4w2_{+ 3}_w2_{+ 6)} (w4 3w2+ 1)2 = 2w(w4 4w2+ 7) (w4 3w2+ 1)2 17.| = (u2 2u)hu |PR 0= (u2 2u)(hu) +hu(2u 2) = hu(u2 2u + 2u 2) = hu(u2 2) 19.| = y 3_2y_y y =y2 2y = y2 2y1@2 |0= 2y 2 ₁ 2 y1@2_{= 2}_{y y}1@2_. We can change the form of the answer as follows: 2y y1@2= 2y 1

y = 2yy 1 y = 2y3@2 1 y 21.i(w) = 2w 2 +w QR i0₍_{w) =} (2 +w 1@2₎₍₂₎_2w1 2w1@2 (2 +w )2 = 4 + 2w1@2 w1@2 (2 +w )2 = 4 +w1@2 (2 +w )2 or 4 +w (2 +w )2 23.i({) = D E + Fh{ QR i0₍_{{) =}(E + Fh{)· 0 D(Fh{) (E + Fh{)2 = DFh{ (E + Fh{)2 25.i({) = { { + f@{ i0({) = ({ + f@{)(1) {(1 f@{2) { + f_{2 = { + f@{ { + f@{ {2₊_f { 2 = 2f@{ ({2+f)2 {2 · {2 {2 = 2f{ ({2+f)2 27.i({) = {4h{ i0({) = {4h{+h{· 4{3={4+ 4{3h{ or {3h{({ + 4) i00₍_{{) = ({}4_{+ 4}_{3₎_h{₊_h{₍₄_{3_{+ 12}_{2_{) = (}_{4_{+ 4}_{3_{+ 4}_{3_{+ 12}_{2₎_h{ = ({4+ 8{3+ 12{2)h{ or {2h{({ + 2)({ + 6) 29.i({) = { 2 1 + 2{ i 0₍_{{) =} (1 + 2{)(2{) {2(2) (1 + 2{)2 = 2{ + 4{2 2{2 (1 + 2{)2 = 2{2+ 2{ (1 + 2{)2 i00₍_{{) =}(1 + 2{)2(4{ + 2) (2{2+ 2{)(1 + 4{ + 4{2)0 [(1 + 2{)2]2 = 2(1 + 2{)2(2{ + 1) 2{({ + 1)(4 + 8{) (1 + 2{)4 =2(1 + 2{)[(1 + 2{) 2_{4{({ + 1)]} (1 + 2{)4 = 2(1 + 4{ + 4{2 4{2 4{) (1 + 2{)3 = 2 (1 + 2{)3 31.| = 2{ { + 1 |0= ({ + 1)(2) (2{)(1) ({ + 1)2 = 2 ({ + 1)2.

At (1> 1), |0=1₂, and an equation of the tangent line is| 1 =₂1({ 1), or | = 1₂{ +1₂. 33.| = 2{h{ |0= 2({ · h{+h{· 1) = 2h{({ + 1).

At (0> 0), |0= 2h0(0 + 1) = 2· 1 · 1 = 2, and an equation of the tangent line is | 0 = 2({ 0), or | = 2{. The slope of the normal line is1₂, so an equation of the normal line is| 0 = 1₂({ 0), or | = 1₂{.

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35. (a)| = i({) = 1 1 +{2 i0₍_{{) =}(1 +{2)(0) 1(2{)

(1 +{2)2 = 2{

(1 +{2)2. So the slope of the tangent line at the point1>1₂isi0(1) = 2

22 = 1 2and its equation is| 1₂ =1₂({ + 1) or | =1₂{ + 1. (b) 37. (a)i({) = h { {3 i0({) = { 3₍_h{₎_h{₍₃_{2₎ ({3)2 = {2_h{₍_{{ 3)} {6 = h {₍_{{ 3)} {4

(b) i0= 0wheni has a horizontal tangent line, i0is negative when i is decreasing, and i0_{is positive when}_{i is increasing.}

39. (a)i({) = ({ 1)h{ i0({) = ({ 1)h{+h{(1) =h{({ 1 + 1) = {h{. i00₍_{{) = {(h}{_{) +}_h{_{(1) =}_h{₍_{{ + 1)}

(b) i0= 0wheni has a horizontal tangent and i00= 0wheni0has a horizontal tangent.i0is negative wheni is decreasing and positive when i is increasing.i00is negative wheni0is decreasing and positive wheni0is increasing.i00is negative wheni is concave down and positive when i is concave up. 41. i({) = { 2 1 +{ i 0₍_{{) =} (1 +{)(2{) {2(1) (1 +{)2 = 2{ + 2{2 {2 (1 +{)2 = {2_{+ 2}_{ {2_{+ 2}_{{ + 1} i00₍_{{) =}({2+ 2{ + 1)(2{ + 2) ({2+ 2{)(2{ + 2) ({2+ 2{ + 1)2 = (2{ + 2)({2+ 2{ + 1 {2 2{) [({ + 1)2]2 =2({ + 1)(1) ({ + 1)4 = 2 ({ + 1)3, soi00(1) = 2 (1 + 1)3 = 2 8 = 1 4.

43. We are given thati(5) = 1, i0(5) = 6,j(5) = 3, and j0(5) = 2.

(a) (ij)0(5) =i(5)j0(5) +j(5)i0(5) = (1)(2) + (3)(6) = 2 18 = 16 (b) i j 0 (5) = j(5)i 0₍₅₎_i(5)j0₍₅₎ [j(5)]2 = (3)(6) (1)(2) (3)2 = 20 9 (c) j i 0 (5) = i(5)j 0₍₅₎_j(5)i0₍₅₎ [i(5)]2 = (1)(2) (3)(6) (1)2 = 20 45. i({) = h{j({) i0({) = h{j0({) + j({)h{=h{[j0({) + j({)]. i0(0) =h0[j0(0) +j(0)] = 1(5 + 2) = 7

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SECTION 3.2 THE PRODUCT AND QUOTIENT RULES ¤ 95 47. (a) From the graphs ofi and j, we obtain the following values: i(1) = 2 since the point (1> 2) is on the graph of i;

j(1) = 1 since the point (1> 1) is on the graph of j; i0_{(1) = 2}_{since the slope of the line segment between (0}_{> 0) and (2> 4)} is4 0

2 0= 2;j

0_{(1) =}_{1 since the slope of the line segment between (2> 4) and (2> 0) is} 0 4

2 (2) =1. Nowx({) = i({)j({), so x0(1) =i(1)j0(1) +j(1) i0(1) = 2· (1) + 1 · 2 = 0.

(b)y({) = i({)@j({), so y0(5) = j(5)i

0₍₅₎_i(5)j0₍₅₎ [j(5)]2 = 21₃ 3 ·2₃ 22 = 8 3 4 = 2 3 49. (a)| = {j({) |0={j0({) + j({) · 1 = {j0({) + j({) (b)| = { j({) |0= j({) · 1 {j 0₍_{) [j({)]2 = j({) {j0₍_{) [j({)]2 (c)| = j({) { |0= {j 0₍_{{) j({) · 1} ({)2 = {j0₍_{{) j({)} {2 51.If| = i({) = { { + 1, theni0({) = ({ + 1)(1) {(1) ({ + 1)2 = 1

({ + 1)2. When{ = d, the equation of the tangent line is | d_{d + 1} = 1

(d + 1)2({ d). This line passes through (1> 2) when 2 dd + 1= 1

(d + 1)2(1 d) 2(d + 1)2 d(d + 1) = 1 d 2d2+ 4d + 2 d2 d 1 + d = 0 d2+ 4d + 1 = 0. The quadratic formula gives the roots of this equation asd =4 ±

s 42 4(1)(1) 2(1) = 4 ±12 2 =2 ± 3, so there are two such tangent lines. Since

i2 ±3= 2 ± 3 2 ±3 + 1= 2 ±3 1 ±3· 13 1 3 =2± 2 33 3 1 3 = 1 ±3 2 = 13 2 ,

the lines touch the curve atD

2 +3>1 ₂3

(0=27> 0=37) andE2 3>1 +₂3 (3=73> 1=37).

53.IfS (w) denotes the population at time w and D(w) the average annual income, then W (w) = S (w)D(w) is the total personal income. The rate at whichW (w) is rising is given by W0(w) = S (w)D0(w) + D(w)S0(w)

W0_{(1999) =}_{S (1999)D}0_{(1999) +}_D(1999)S0_{(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 termS (w)D0(w) $1.346 billion represents the portion of the rate of change of total income due to the existing population’s increasing income. The termD(w)S0(w) $281 million represents the portion of the rate of change of total income due to increasing population.

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We will sometimes use the formi0j + ij0rather than the formij0+ji0for the Product Rule.

55. (a) (ijk)0= [(ij)k]0= (ij)0k + (ij)k0= (i0j + ij0)k + (ij)k0=i0jk + ij0k + ijk0 (b) Puttingi = j = k in part (a), we have g

g{[i({)]3= (iii)0=i0ii + ii0i + iii0= 3iii0= 3[i({)]2i0({). (c) g

g{(h3{) = g{g (h{)3= 3(h{)2h{= 3h2{h{= 3h3{

57. Fori({) = {2h{,i0({) = {2h{+h{(2{) = h{({2+ 2{). Similarly, we have i00₍_{{) = h}{₍_{2_{+ 4}_{{ + 2)}

i000₍_{{) = h}{₍_{2_{+ 6}_{{ + 6)} i(4)₍_{{) = h}{₍_{2_{+ 8}_{{ + 12)} i(5)₍_{{) = h}{₍_{2_{+ 10}_{{ + 20)}

It appears that the coefcient of{ 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 · 2, 12 = 4 · 3, 20 = 5 · 4. So a reasonable guess is that i(q)₍_{{) = h}{_[_{2_{+ 2}_{q{ + q(q 1)].}

Proof: Let Vqbe the statement thati(q)({) = h{[{2+ 2q{ + q(q 1)]. 1.V1is true becausei0({) = h{({2+ 2{).

2. Assume thatV_nis true; that is,i(n)({) = h{[{2+ 2n{ + n(n 1)]. Then i(n+1)₍_{{) = g}

g{ k

i(n)₍_{)l₌_h{₍₂_{{ + 2n) + [{}2_{+ 2}_{n{ + n(n 1)]h}{ =h{[{2+ (2n + 2){ + (n2+n)] = h{[{2+ 2(n + 1){ + (n + 1)n] This shows thatV_n+1is true.

3. Therefore, by mathematical induction,V_qis true for allq; that is, i(q)({) = h{[{2+ 2q{ + q(q 1)] for every positive integerq.

3.3 Derivatives of Trigonometric Functions

1. i({) = 3{2 2 cos { i0({) = 6{ 2( sin {) = 6{ + 2 sin { 3. i({) = sin { +1₂cot{ i0({) = cos { 1₂csc2{

5. j(w) = w3cosw j0(w) = w3( sin w) + (cos w) · 3w2= 3w2cosw w3sinw or w2(3 cosw w sin w)

7. k() = csc + hcot k0() = csc cot + h( csc2) + (cot )h= csc cot + h(cot csc2)

9. | = { 2 tan { | 0₌(2 tan {)(1) {( sec2{) (2 tan {)2 = 2 tan { + { sec2{ (2 tan {)2 11. i() = sec 1 + sec

i0₍_{) =} (1 + sec)(sec tan ) (sec )(sec tan )

(1 + sec)2 =

(sec tan ) [(1 + sec ) sec ]

(1 + sec)2 =

sec tan (1 + sec)2

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SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ¤ 97 13.| = sin{ {2 |0= {2_cos_{{ (sin {)(2{)} ({2)2 = {({ cos { 2 sin {) {4 = { cos { 2 sin { {3 15.Using Exercise 3.2.55(a),i({) = {h{csc{

i0₍_{{) = ({)}0_h{_csc_{{ + {(h}{₎0_csc_{{ + {h}{_(csc_{)0_{= 1}_h{_csc_{{ + {h}{_csc_{{ + {h}{₍_{cot { csc {)} =h{csc{ (1 + { { cot {) 17. g g{(csc{) = gg{ 1 sin{ = (sin{)(0) 1(cos {) sin2{ = cos { sin2{ = 1 sin{· cos{ sin{ = csc { cot { 19. g g{(cot{) = gg{ _cos_{ sin{

= (sin{)( sin {) (cos {)(cos {)

sin2{ = sin2{ + cos2{ sin2{ = 1 sin2{= csc 2_{

21.| = sec { |0= sec{ tan {, so |0(₃) = sec₃ tan₃ = 23. An equation of the tangent line to the curve| = sec { at the point₃> 2is| 2 = 23{ ₃or| = 23{ + 2 2₃3.

23.| = { + cos { |0= 1 sin {. At (0> 1), |0= 1, and an equation of the tangent line is| 1 = 1({ 0), or | = { + 1. 25. (a)| = 2{ sin { |0= 2({ cos { + sin { · 1). At₂> ,

|0_{= 2}

2cos2 + sin2

= 2(0 + 1) = 2, and an equation of the tangent line is| = 2{ ₂, or| = 2{.

(b)

27. (a)i({) = sec { { i0({) = sec { tan { 1

(b) Note thati0= 0wherei has a minimum. Also note that i0is negative wheni is decreasing and i0is positive wheni is increasing.

29.K() = sin K0() = (cos ) + (sin ) · 1 = cos + sin K00₍_{) = ( sin ) + (cos ) · 1 + cos = sin + 2 cos}

31. (a)i({) = tan{ 1

sec{

i0₍_{{) =} sec{(sec2{) (tan { 1)(sec { tan {)

(sec{)2 =

sec{(sec2{ tan2{ + tan {)

sec2{ = 1 + tan{ sec{ (b)i({) = tan{ 1 sec{ = sin{ cos{ 1 1 cos{ = sin{ cos { cos{ 1 cos{

= sin{ cos { i0({) = cos { ( sin {) = cos { + sin {

(c) From part (a),i0({) =1 + tan{ sec{ =

1 sec{+

tan{

sec{ = cos{ + sin {, which is the expression for i

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33. i({) = { + 2 sin { has a horizontal tangent when i0({) = 0 1 + 2 cos { = 0 cos { = 1₂ { =2

3 + 2q or43 + 2q, where q is an integer. Note that43 and23 are±3 units from. This allows us to write the solutions in the more compact equivalent form (2q + 1) ±₃,q an integer.

35. (a){(w) = 8 sin w y(w) = {0(w) = 8 cos w d(w) = {00(w) = 8 sin w (b) The mass at timew =2₃ has position{2₃ = 8 sin2₃ = 8

3 2

= 43, velocityy2₃ = 8 cos2₃ = 81₂=4, and accelerationd2₃ =8 sin2₃ =8

3 2

=43. Sincey2₃? 0, the particle is moving to the left.

37. From the diagram we can see that sin = {@10 { = 10 sin . We want to nd the rate of change of{ with respect to , that is, g{@g. Taking the derivative of { = 10 sin , we get g{@g = 10(cos ). So when = 3, g{g = 10 cos3 = 10 ₁ 2 = 5ft@rad. 39. lim {0 sin 3{ { = lim{0 3 sin 3{

3{ [multiply numerator and denominator by 3] = 3 lim 3{0 sin 3{ 3{ [as{ 0, 3{ 0] = 3 lim 0 sin [let = 3{] = 3(1) [Equation 2] = 3 41. lim w0 tan 6w sin 2w = limw0 sin 6w w · 1 cos 6w· w sin 2w = lim w0 6 sin 6w 6w · limw0 1 cos 6w· limw0 2w 2 sin 2w = 6 lim w0 sin 6w 6w · limw0 1 cos 6w· 1 2limw0 2w sin 2w= 6(1)· 1 1· 1 2(1) = 3 43. lim 0 sin(cos) sec = sin lim 0cos lim 0sec = sin 1 1 = sin 1

45. Divide numerator and denominator by. (sin also works.)

lim 0 sin + tan = lim0 sin 1 +sin · 1 cos = lim 0 sin 1 + lim 0 sin 0lim 1 cos = 1 1 + 1· 1= 1 2 47. lim {@4 1 tan {

sin{ cos { = lim{@4

1sin{ cos{

· cos {

(sin{ cos {) · cos { = lim{@4

cos{ sin {

(sin{ cos {) cos { = lim{@4 1 cos{= 1 1@2 = 2 49. (a) g g{tan{ = gg{ sin{ cos{ sec

2_{{ =} cos{ cos { sin { ( sin {)

cos2{ = cos2{ + sin2{ cos2{ . So sec 2_{{ =} 1 cos2{. (b) g g{sec{ = gg{ 1

cos{ sec { tan { =

(cos{)(0) 1( sin {)

cos2{ . So sec{ tan { = sin{ cos2{.

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SECTION 3.4 THE CHAIN RULE ¤ 99 (c) g g{(sin{ + cos {) = gg{ 1 + cot{ csc{ cos{ sin { = csc{ ( csc 2_{{) (1 + cot {)( csc { cot {)} csc2{ = csc{ [ csc2{ + (1 + cot {) cot {] csc2{ = csc 2_{{ + cot}2_{{ + cot {} csc{ = 1 + cot { csc{ So cos{ sin { = cot{ 1

csc{ .

51.By the denition of radian measure,v = u, where u is the radius of the circle. By drawing the bisector of the angle , we can see that sin

2= g@2 u g = 2u sin 2. So lim0+ v g = lim0+ u 2u sin(@2) = lim0+ 2· (@2) 2 sin(@2) = lim0 @2 sin(@2) = 1. [This is just the reciprocal of the limit lim

{0 sin {

{ = 1combined with the fact that as 0,2 0 also=]

3.4 The Chain Rule

1.Letx = j({) = 4{ and | = i(x) = sin x. Then g|

g{= g|gxgxg{ = (cosx)(4) = 4 cos 4{. 3.Letx = j({) = 1 {2and| = i(x) = x10= Then g|

g{= g|gxgxg{= (10x9)(2{) = 20{(1 {2)9. 5.Letx = j({) ={ and | = i(x) = hx. Theng|

g{= g|gxgxg{= (hx) 1 2{1@2 =h{· 1 2{ = h{ 2{. 7.I ({) = ({4+ 3{2 2)5 I0({) = 5({4+ 3{2 2)4· g g{ {4_{+ 3}_{2₂_{= 5(}_{4_{+ 3}_{2₂₎4₍₄_{3_{+ 6}_{) or 10{({4+ 3{2 2)4(2{2+ 3) 9.I ({) =41 + 2{ + {3= (1 + 2{ + {3)1@4 I0₍_{{) =}1 4(1 + 2{ + {3)3@4· g_g{(1 + 2{ + {3) = 1 4(1 + 2{ + {3)3@4· (2 + 3{ 2_{) =} 2 + 3{2 4(1 + 2{ + {3)3@4 = 2 + 3{ 2 4s4 (1 + 2{ + {3)3 11.j(w) = 1 (w4+ 1)3 = (w 4_{+ 1)}3 _j0₍_{w) = 3(w}4_{+ 1)}4₍₄_w3_{) =}_12w3₍_w4_{+ 1)}4₌ 12w3 (w4+ 1)4 13.| = cos(d3+{3) |0= sin(d3+{3)· 3{2 [d3is just a constant] =3{2sin(d3+{3) 15.| = {hn{ |0={hn{(n)+hn{· 1 = hn{(n{ + 1) or (1 n{)hn{ 17.j({) = (1 + 4{)5(3 +{ {2)8 j0₍_{{) = (1 + 4{)}5_{· 8(3 + { {}2₎7₍₁_{2{) + (3 + { {}2₎8_{· 5(1 + 4{)}4_{· 4} = 4(1 + 4{)4(3 +{ {2)72(1 + 4{)(1 2{) + 5(3 + { {2) = 4(1 + 4{)4(3 +{ {2)7(2 + 4{ 16{2) + (15 + 5{ 5{2)= 4(1 + 4{)4(3 +{ {2)7(17 + 9{ 21{2)

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19. | = (2{ 5)4(8{2 5)3 |0 _{= 4(2}_{{ 5)}3₍₂₎₍₈_{2₅₎3_{+ (2}_{{ 5)}4₍_3)(8{2₅₎4₍₁₆_{) = 8(2{ 5)3(8{2 5)3 48{(2{ 5)4(8{2 5)4 [This simplies to 8(2{ 5)3(8{2 5)4(4{2+ 30{ 5).] 21. | = {2_{+ 1} {2₁ 3 |0_{= 3}{2+ 1 {2₁ 2 · g_g{ {2_{+ 1} {2₁ = 3 {2_{+ 1} {2₁ 2 ·({2 1)(2{) ({2+ 1)(2{) ({2 1)2 = 3 {2_{+ 1} {2₁ 2 ·2{[{2 1 ({2+ 1)] ({2 1)2 = 3 {2_{+ 1} {2₁ 2 · 2{(2) ({2 1)2 = 12{({2_{+ 1)}2 ({2 1)4 23. | = h{ cos { |0=h{ cos {· g

g{({ cos {) = h{ cos {[{( sin {) + (cos {) · 1] = h{ cos {(cos{ { sin {) 25. I (}) = u } 1 } + 1 = } 1 } + 1 1@2 I0₍_{}) =} 1 2 } 1 } + 1 1@2 · g g} } 1 } + 1 =1 2 } + 1 } 1 1@2 ·(} + 1)(1) (} 1)(1) (} + 1)2 = 1 2 (} + 1)1@2 (} 1)1@2· } + 1 } + 1 (} + 1)2 = 1 2 (} + 1)1@2 (} 1)1@2 · 2 (} + 1)2 = 1 (} 1)1@2(} + 1)3@2 27. | = u u2_{+ 1} |0 ₌ u2_{+ 1 (1)}_{u ·}1 2(u2+ 1)1@2(2u) u2+ 12 = u2_{+ 1} u2 u2_{+ 1} u2+ 12 = u2_{+ 1}_u2_{+ 1}_u2 u2_{+ 1} u2+ 12 = u2_{+ 1}_u2 u2_{+ 1}3 = 1 (u2+ 1)3@2 or (u 2_{+ 1)}3@2

Another solution: Write | as a product and make use of the Product Rule. | = u(u2_{+ 1)}1@2 |0₌_{u ·}1

2(u2+ 1)3@2(2u) + (u2+ 1)1@2· 1 = (u2+ 1)3@2[u2+ (u2+ 1)1] = (u2+ 1)3@2(1) = (u2+ 1)3@2. The step that students usually have trouble with is factoring out (u2+ 1)3@2. But this is no different than factoring out{2 from{2+{5; that is, we are just factoring out a factor with thesmallest exponent that appears on it. In this case, 3₂ is smaller than1₂.

29. | = sin(tan 2{) |0= cos(tan 2{) · g

g{(tan 2{) = cos(tan 2{) · sec2(2{) · gg{(2{) = 2 cos(tan 2{) sec2(2{) 31. Using Formula 5 and the Chain Rule,| = 2sin {

|0_{= 2}sin {_{(ln 2)}_{· g}

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SECTION 3.4 THE CHAIN RULE ¤ 101 33.| = sec2{ + tan2{ = (sec {)2+ (tan{)2

|0_{= 2(sec}_{{)(sec { tan {) + 2(tan {)(sec}2_{{) = 2 sec}2_{{ tan { + 2 sec}2_{{ tan { = 4 sec}2_{{ tan {} 35.| = cos 1 h2{ 1 +h2{ |0₌_sin1 h2{ 1 +h2{ · g_g{ 1 h2{ 1 +h2{ = sin 1 h2{ 1 +h2{ ·(1 +h2{)(2h2{) (1 h2{)(2h2{) (1 +h2{)2 = sin 1 h2{ 1 +h2{ ·2h2{ (1 +h2{) + (1 h2{) (1 +h2{)2 = sin 1 h2{ 1 +h2{ · _{(1 +}2h2{_h_2{(2)₎₂ = 4h 2{ (1 +h2{)2 · sin 1 h2{ 1 +h2{

37.| = cot2(sin) = [cot(sin )]2 |0_{= 2[cot(sin}_{)] · g}

g[cot(sin)] = 2 cot(sin ) · [ csc2(sin) · cos ] = 2 cos cot(sin ) csc2(sin) 39.i(w) = tan(hw) +htan w i0(w) = sec2(hw)· g

gw(hw) +htan w· ggw(tanw) = sec2(hw)· hw+htan w· sec2w 41.i(w) = sin2hsin2w=ksinhsin2wl2

i0₍_{w) = 2}k_sin_hsin2_wl · g_gwsin hsin2_w = 2 sin hsin2_w · coshsin2_w · g_gwhsin2_w = 2 sin hsin2_w cos hsin2_w · hsin2_w · g gw sin2w = 2 sin hsin2_w cos hsin2_w hsin2_w · 2 sin w cos w = 4 sinhsin2wcoshsin2whsin2wsinw cos w

43.j({) = (2udu{+q)s j0₍_{{) = s(2ud}u{₊_q)s1_{· g}

g{(2udu{+q) = s(2udu{+q)s1· 2udu{(lnd) · u = 2u2s(ln d)(2udu{+q)s1du{ 45.| = cosssin(tan{) = cos(sin(tan {))1@2

|0₌_{sin(sin(tan {))}1@2_{· g}

g{(sin(tan{))1@2= sin(sin(tan {))1@2·12(sin(tan{))1@2· g_g{(sin(tan{)) = sin

s

sin(tan{)

2ssin(tan{) · cos(tan {) · gg{tan{ =

sinssin(tan{)

2ssin(tan{) · cos(tan {) · sec

2₍_{{) ·} = cos(tan {) sec 2₍_{{) sin}s_sin(tan_{) 2ssin(tan{) 47.k({) ={2+ 1 k0({) = 1 2({ 2_{+ 1)}1@2₍₂_{{) =} { {2_{+ 1} k00₍_{{) =} {2_{+ 1}_{· 1 {}k1 2({2+ 1)1@2(2{) l {2_{+ 1}2 = {2_{+ 1}1@2 ({2+ 1) {2 ({2+ 1)1 = 1 ({2+ 1)3@2

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49. | = h{sin{ |0=h{· cos { + sin { · h{ =h{( cos { + sin {) |00₌_h{₍2_sin_{{ + cos {) + ( cos { + sin {) · h}{

=h{(2sin{ + cos { + cos { + 2sin{) = h{(2sin{ 2sin{ + 2 cos {) =h{(2 2) sin{ + 2 cos {

51. | = (1 + 2{)10 |0= 10(1 + 2{)9· 2 = 20(1 + 2{)9.

At (0> 1), |0= 20(1 + 0)9= 20, and an equation of the tangent line is| 1 = 20({ 0), or | = 20{ + 1.

53. | = sin(sin {) |0= cos(sin{) · cos {. At (> 0), |0= cos(sin) · cos = cos(0) · (1) = 1(1) = 1, and an equation of the tangent line is| 0 = 1({ ), or | = { + .

55. (a)| = 2 1 +h{ | 0₌(1 +h{)(0) 2(h{) (1 +h{)2 = 2h{ (1 +h{)2. At (0> 1), |0= 2h 0 (1 +h0)2 = 2(1) (1 + 1)2 = 2 22 = 1 2. So an equation of the tangent line is| 1 =1₂({ 0) or | =1₂{ + 1. (b) 57. (a)i({) = {2 {2 ={(2 {2)1@2 i0₍_{{) = { ·}1 2(2 {2)1@2(2{) + (2 {2)1@2· 1 = (2 {2)1@2 {2_{+ (2}_{2₎₌2 2{2 2 {2 (b) i0= 0wheni has a horizontal tangent line, i0is negative wheni is

decreasing, andi0is positive wheni is increasing.

59. For the tangent line to be horizontal,i0({) = 0. i({) = 2 sin { + sin2{ i0({) = 2 cos { + 2 sin { cos { = 0 2 cos{(1 + sin {) = 0 cos { = 0 or sin { = 1, so { =₂ + 2q or3₂ + 2q, where q is any integer. Now i

2

= 3andi3₂ =1, so the points on the curve with a horizontal tangent are₂ + 2q> 3and3₂ + 2q> 1, whereq is any integer.

61. I ({) = i(j({)) I0({) = i0(j({)) · j0({), so I0(5) =i0(j(5)) · j0(5) =i0(2) · 6 = 4 · 6 = 24 63. (a)k({) = i(j({)) k0({) = i0(j({)) · j0({), so k0(1) =i0(j(1)) · j0(1) =i0(2)· 6 = 5 · 6 = 30.

(b)K({) = j(i({)) K0({) = j0(i({)) · i0({), so K0(1) =j0(i(1)) · i0(1) =j0(3)· 4 = 9 · 4 = 36.

65. (a)x({) = i(j({)) x0({) = i0(j({))j0({). So x0(1) =i0(j(1))j0(1) =i0(3)j0(1). To ndi0(3), note thati is linear from (2> 4) to (6> 3), so its slope is3 4

6 2 = 1

4. To ndj

0_{(1), note that}_{j is linear from (0> 6) to (2> 0), so its slope}

is0 6 2 0=3. Thus, i 0₍₃₎_j0_{(1) =}1 4 (3) =3₄.

(b)y({) = j(i({)) y0({) = j0(i({))i0({). So y0(1) =j0(i(1))i0(1) =j0(2)i0(1), which does not exist since j0₍₂₎_{does not exist.}

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SECTION 3.4 THE CHAIN RULE ¤ 103 (c)z({) = j(j({)) z0({) = j0(j({))j0({). So z0(1) =j0(j(1))j0(1) =j0(3)j0(1). To ndj0(3), note thatj is

linear from (2> 0) to (5> 2), so its slope is 2 0 5 2= 2 3. Thus,j 0₍₃₎_j0_{(1) =}2 3 (3) = 2. 67. (a)I ({) = i(h{) I0({) = i0(h{) g g{(h{) =i0(h{)h{ (b)J({) = hi({) J0({) = hi({) g g{i({) = hi({)i0({) 69.u({) = i(j(k({))) u0({) = i0(j(k({))) · j0(k({)) · k0({), so u0_{(1) =}_i0₍_{j(k(1))) · j}0₍_{k(1)) · k}0_{(1) =}_i0₍_{j(2)) · j}0₍₂₎_{· 4 = i}0₍₃₎_{· 5 · 4 = 6 · 5 · 4 = 120} 71.I ({) = i(3i(4i({))) I0₍_{{) = i}0₍₃_{i(4i({))) · g}

g{(3i(4i({))) = i0(3i(4i({))) · 3i0(4i({)) · gg{(4i({)) =i0(3i(4i({))) · 3i0(4i({)) · 4i0({), so

I0_{(0) =}_i0₍₃_{i(4i(0))) · 3i}0₍₄_{i(0)) · 4i}0_{(0) =}_i0₍₃_{i(4 · 0)) · 3i}0₍₄_{· 0) · 4 · 2 = i}0₍₃_{· 0) · 3 · 2 · 4 · 2 = 2 · 3 · 2 · 4 · 2 = 96.} 73.| = Dh{+E{h{

|0₌_D(h{_{) +}_E[{(h{_{) +}_h{_{· 1] = Dh}{₊_Eh{_E{h{_{= (}_{E D)h}{_E{h{

|00_{= (}_{E D)(h}{₎_E[{(h{_{) +}_h{_{· 1] = (D E)h}{_Eh{₊_E{h{_{= (}_{D 2E)h}{₊_E{h{_, so |00+ 2|0+| = (D 2E)h{+E{h{+ 2[(E D)h{ E{h{] +Dh{+E{h{

= [(D 2E) + 2(E D) + D]h{+ [E 2E + E]{h{= 0.

75.The use ofG, G2,= = =, Gqis just a derivative notation (see text page 157). In general,Gi(2{) = 2i0(2{), G2_{i(2{) = 4i}00₍₂_{{), = = =, G}q_{i(2{) = 2}q_i(q)₍₂_{{). Since i({) = cos { and 50 = 4(12) + 2, we have} i(50)₍_{{) = i}(2)₍_{{) = cos {, so G}50_{cos 2}_{{ = 2}50_{cos 2}_{.

77.v(w) = 10 + 1₄sin(10w) the velocity after w seconds is y(w) = v0(w) =1₄cos(10w)(10) = 5₂ cos(10w) cm@s. 79. (a)E(w) = 4=0 + 0=35 sin2w

5=4 gEgw = 0=35 cos2w 5=4 2 5=4 =0=7 5=4 cos 2w 5=4 = 7 54cos 2w 5=4 (b) Atw = 1, gE gw = 7 54cos 2 5=4 0=16. 81.v(w) = 2h1=5wsin 2w

y(w) = v0₍_{w) = 2[h}1=5w_{(cos 2}_{w)(2) + (sin 2w)h}1=5w₍_{1=5)] = 2h}1=5w₍₂_{cos 2w 1=5 sin 2w)}

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83. By the Chain Rule,d(w) = gy

gw =gygvgvgw = gygvy(w) = y(w) gygv. The derivativegy@gw is the rate of change of the velocity with respect to time (in other words, the acceleration) whereas the derivativegy@gv is the rate of change of the velocity with respect to the displacement.

85. (a) Using a calculator or CAS, we obtain the modelT = dewwithd 100=0124369 and e 0=000045145933. (b) UseT0(w) = dewlne (from Formula 5) with the values of d and e from part (a) to get T0(0=04) 670=63 A.

The result of Example 2 in Section 2.1 was670 A. 87. (a) Derive givesj0(w) = 45(w 2)

8

(2w + 1)10 without simplifying. With either Maple or Mathematica, we rst get j0₍_{w) = 9}(w 2)8

(2w + 1)9 18

(w 2)9

(2w + 1)10, and the simplication command results in the expression given by Derive. (b) Derive gives|0= 2({3 { + 1)3(2{ + 1)4(17{3+ 6{2 9{ + 3) without simplifying. With either Maple or

Mathematica, we rst get|0= 10(2{ + 1)4({3 { + 1)4+ 4(2{ + 1)5({3 { + 1)3(3{2 1). If we use

Mathematica’s Factor or Simplify, or Maple’s factor, we get the above expression, but Maple’s simplify gives the polynomial expansion instead. For locating horizontal tangents, the factored form is the most helpful.

89. (a) Ifi is even, then i({) = i({). Using the Chain Rule to differentiate this equation, we get i0₍_{{) = i}0₍_{{) g}

g{({) = i0({). Thus, i0({) = i0({), so i0is odd.

(b) Ifi is odd, then i({) = i({). Differentiating this equation, we get i0({) = i0({)(1) = i0({), so i0is even.

91. (a) g

g{(sinq{ cos q{) = q sinq1{ cos { cos q{ + sinq{ (q sin q{) [Product Rule] =q sinq1{ (cos q{ cos { sin q{ sin {) [factor outq sinq1{] =q sinq1{ cos(q{ + {) [Addition Formula for cosine]

=q sinq1{ cos[(q + 1){] [factor out{]

(b) g

g{(cosq{ cos q{) = q cosq1{ ( sin {) cos q{ + cosq{ (q sin q{) [Product Rule]

=q cosq1{ (cos q{ sin { + sin q{ cos {) [factor outq cosq1{] =q cosq1{ sin(q{ + {) [Addition Formula for sine]

=q cosq1{ sin[(q + 1){] [factor out{]

93. Since=₁₈₀ rad, we have g

g(sin) =gg

sin₁₈₀ = ₁₈₀ cos₁₈₀ =₁₈₀ cos.

95. The Chain Rule says thatg|

g{ =g|gxgxg{, so g2_| g{2 =_g{g g| g{ = g g{ g| gxgxg{ = g g{ g| gx gx g{+g|gxg{g gx g{ [Product Rule] = g gx g| gx gx g{ gx g{+g|gxg 2_x g{2 = g2_| gx2 gx g{ 2 +g| gxg 2_x g{2

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SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 105

3.5 Implicit Differentiation

1. (a) g g{({| + 2{ + 3{2) = g{g (4) ({ · |0+| · 1) + 2 + 6{ = 0 {|0=| 2 6{ |0₌| 2 6{ { or |0=6 | + 2 { . (b){| + 2{ + 3{2= 4 {| = 4 2{ 3{2 | = 4 2{ 3{ 2 { = 4 { 2 3{, so |0= 4 {2 3. (c) From part (a),|0= | 2 6{

{ =(4@{ 2 3{) 2 6{{ =4@{ 3{{ = 4 {2 3. 3. (a) g g{ 1 {+ 1 | = g g{(1) 1 {2 1 |2 |0= 0 1 |2|0= 1 {2 |0= | 2 {2 (b) 1 {+ 1 | = 1 1 | = 1 1 {= { 1{ | = {{ 1, so|0= ({ 1)(1) ({)(1) ({ 1)2 = 1 ({ 1)2. (c)|0= | 2 {2 = [{@({ 1)]2 {2 = { 2 {2₍_{{ 1)}2 = 1 ({ 1)2 5. g g{ {3₊_|3₌ g g{(1) 3{2+ 3|2· |0= 0 3|2|0=3{2 |0={ 2 |2 7. g g{({2+{| |2) = g{g (4) 2{ + { · |0+| · 1 2| |0= 0 {|0_{2| |}0₌_{2{ | ({ 2|) |}0₌_{2{ | |}0₌2{ | { 2| = 2{ + | 2| { 9. g g{ {4₍_{{ + |)}₌ g g{ |2₍₃_{{ |)} _{4_{(1 +}_|0_{) + (}_{{ + |) · 4{}3₌_|2₍₃_|0_{) + (3}_{{ |) · 2| |}0 {4₊_{4_|0_{+ 4}_{4_{+ 4}_{3_{| = 3|}2_|2_|0_{+ 6}_{{| |}0_2|2_|0 _{4_|0_{+ 3}_|2_|0_{6{| |}0_{= 3}_|2_5{4_4{3_| ({4+ 3|2 6{|) |0= 3|2 5{4 4{3| |0=3| 2_5{4_4{3_| {4_{+ 3}_|2_6{| 11. g

g{({2|2+{ sin |) = gg{(4) {2· 2| |0+|2· 2{ + { cos | · |0+ sin| · 1 = 0 2{2| |0+{ cos | · |0=2{|2 sin | (2{2| + { cos |)|0=2{|2 sin | |0=2{|

2_{sin |} 2{2| + { cos | 13. g

g{(4 cos{ sin |) = gg{(1) 4 [cos { · cos | · |0+ sin| · ( sin {)] = 0 |0_{(4 cos}_{{ cos |) = 4 sin { sin |} _|0₌ 4 sin{ sin |

4 cos{ cos | = tan{ tan |

15. g g{(h{@|) = g{g ({ |) h{@|· gg{ { | = 1 |0 h{@|_{· | ·}1 { · |0 |2 = 1 |0 h{@|· 1 | {h {@| |2 · |0= 1 |0 |0 {h {@| |2 · |0= 1 h {@| | |0₁_{h{@| |2 =| h {@| | |0= | h{@| | |2_{h{@| |2 =|(| h {@|₎ |2_{h{@|

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17. s{| = 1 + {2| 1₂({|)1@2({|0+| · 1) = 0 + {2|0+| · 2{ { 2s{|| 0₊ | 2s{| ={ 2_|0_{+ 2}_{| |0 # { 2s{| { 2 $ = 2{| | 2s{| | 0 # { 2{2s_{| 2s{| $ =4{| s {| | 2s{| | 0₌ 4{| s {| | { 2{2s_{| 19. g

g{(h|cos{) = gg{[1 + sin({|)] h|( sin {) + cos { · h|· |0= cos({|) · ({|0+| · 1)

h|_sin_{{ + h}|_cos_{{ · |}0₌_{{ cos({|) · |}0₊_{| cos({|) h}|_cos_{{ · |}0_{{ cos({|) · |}0₌_h|_sin_{{ + | cos({|)} [h|cos{ { cos({|)] |0=h|sin{ + | cos({|) |0= h

|_sin_{{ + | cos({|)} h|_cos_{{ { cos({|)} 21. g g{ i({) + {2_[_i({)]3₌ g

g{(10) i0({) + {2· 3[i({)]2· i0({) + [i({)]3· 2{ = 0. If { = 1, we have i0_{(1) + 1}2_{· 3[i(1)]}2_{· i}0_{(1) + [}_i(1)]3_{· 2(1) = 0 i}0_{(1) + 1}_{· 3 · 2}2_{· i}0_{(1) + 2}3_{· 2 = 0} i0_{(1) + 12}_i0_{(1) =}_{16 13i}0_{(1) =}_{16 i}0_{(1) =}16 13. 23. g g|({4|2 {3| + 2{|3) = g|g(0) {4· 2| + |2· 4{3{0 ({3· 1 + | · 3{2{0) + 2({ · 3|2+|3· {0) = 0 4{3|2{0 3{2| {0+ 2|3{0=2{4| + {3 6{|2 (4{3|2 3{2| + 2|3){0=2{4| + {3 6{|2 {0₌ g{ g| = 2{ 4_{| + {}3_6{|2 4{3|2 3{2| + 2|3 25. {2+{| + |2= 3 2{ + { |0+| · 1 + 2||0= 0 { |0+ 2| |0=2{ | |0({ + 2|) = 2{ | |0₌ 2{ |

{ + 2| . When{ = 1 and | = 1, we have |0=1 + 22 1· 1 = 3

3 =1, so an equation of the tangent line is | 1 = 1({ 1) or | = { + 2.

27. {2+|2= (2{2+ 2|2 {)2 2{ + 2| |0= 2(2{2+ 2|2 {)(4{ + 4| |0 1). When { = 0 and | =1₂, we have 0 +|0= 2(1₂)(2|0 1) |0= 2|0 1 |0= 1, so an equation of the tangent line is| 1₂ = 1({ 0) or | = { +1₂.

29. 2({2+|2)2 = 25({2 |2) 4({2+|2)(2{ + 2| |0) = 25(2{ 2| |0)

4({ + | |0)({2+|2) = 25({ | |0) 4| |0({2+|2) + 25||0= 25{ 4{({2+|2) |0₌25{ 4{({2+|2)

25| + 4|({2+|2). When{ = 3 and | = 1, we have |

0₌75 120

25 + 40 =4565 =139 , so an equation of the tangent line is| 1 = ₁₃9({ 3) or | = ₁₃9{ +40₁₃.

31. (a)|2= 5{4 {2 2| |0= 5(4{3) 2{ |0=10{ 3_{

| .

So at the point (1> 2) we have |0=10(1) 3₁

2 =

9

2, and an equation of the tangent line is| 2 =9₂({ 1) or | =9₂{ 5₂.

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SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 107 33.9{2+|2 = 9 18{ + 2| |0= 0 2| |0=18{ |0 =9{@| |00₌₉| · 1 { · |0 |2 =9 | {(9{@|) |2 =9 · | 2_{+ 9}_{2 |3 =9 · 9

|3 [sincex and y must satisfy the original equation, 9{2+|2= 9]. Thus,|00=81@|3. 35.{3+|3 = 1 3{2+ 3|2|0 = 0 |0 ={ 2 |2 |00₌_|2(2{) {2· 2| |0 (|2)2 = 2{|2 2{2|({2@|2) |4 = 2{|4+ 2{4| |6 = 2{|(|3+{3) |6 = 2{ |5, since{ and | must satisfy the original equation, {3+|3= 1.

37. (a) There are eight points with horizontal tangents: four at{ 1=57735 and four at{ 0=42265. (b)|0= 3{ 2_{6{ + 2} 2(2|3 3|2 | + 1) | 0₌_{1 at (0> 1) and |}0₌1 3 at (0> 2). Equations of the tangent lines are| = { + 1 and | =1₃{ + 2.

(c)|0= 0 3{2 6{ + 2 = 0 { = 1 ±1₃3

(d) By multiplying the right side of the equation by{ 3, we obtain the rst graph. By modifying the equation in other ways, we can generate the other graphs. |(|2_{1)(| 2)} ={({ 1)({ 2)({ 3) |(|2_{4)(| 2)} ={({ 1)({ 2) |(| + 1)(|2_{1)(| 2)} ={({ 1)({ 2) (| + 1)(|2 1)(| 2) = ({ 1)({ 2) {(| + 1)(|2_{1)(| 2)} =|({ 1)({ 2) |(|2_{+ 1)(}_{| 2)} ={({2 1)({ 2) |(| + 1)(|2₂₎ ={({ 1)({2 2)

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39. From Exercise 29, a tangent to the lemniscate will be horizontal if|0 = 0 25{ 4{({2+|2) = 0 {[25 4({2₊_|2_{)] = 0} _{2₊_|2₌25

4 (1). (Note that when { is 0, | is also 0, and there is no horizontal tangent at the origin.) Substituting25₄ for{2+|2in the equation of the lemniscate, 2({2+|2)2= 25({2 |2), we get {2_|2₌25

8 (2). Solving (1) and (2), we have {2=7516and|2= 2516, so the four points are ±53 4 > ±54 . 41. { 2 d2 | 2 e2 = 1 2{ d2 2||0 e2 = 0 |0= e 2_{

d2_| an equation of the tangent line at ({0> |0)is | |0=e

2_{ 0

d2_|₀({ {0). Multiplying both sides by | 0 e2 gives| 0| e2 | 2 0 e2 ={ 0{ d2 { 2 0

d2. Since ({0> |0)lies on the hyperbola, we have{0{ d2 |_e02| = { 2 0 d2 | 2 0 e2 = 1.

43. If the circle has radiusu, its equation is {2+|2=u2 2{ + 2||0= 0 |0={

|, so the slope of the tangent line atS ({₀> |₀)is{0

|0. The negative reciprocal of that slope is 1 {0@|0 =

|0

{0, which is the slope ofRS , so the tangent line at S is perpendicular to the radius RS .

45. | = tan1{ |0= 1 1 +{2 · g_g{{= 1 1 +{ 1 2{1@2 = 1 2{ (1 + {) 47. | = sin1(2{ + 1) |0₌_s 1 1 (2{ + 1)2 · gg{(2{ + 1) = 1 s 1 (4{2+ 4{ + 1)· 2 = 2 4{2_4{= 1 {2_{ 49. J({) =1 {2arccos{ J0({) =1 {2·1 1 {2 + arccos{ · 1 2(1 { 2₎1@2₍_{2{) = 1 {}arccos{ 1 {2 51. k(w) = cot1(w) + cot1(1@w) k0₍_{w) =} 1 1 +w2 1 1 + (1@w)2 · ggw 1 w = 1 1 +w2 w 2 w2_{+ 1}· 1 w2 = 1 1 +w2 + 1 w2_{+ 1} = 0. Note that this makes sense becausek(w) =

2 forw A 0 and k(w) = 3 2 forw ? 0. 53. | = cos1(h2{) |0=s 1 1 (h2{)2 · gg{(h 2{_{) =}2h2{ 1 h4{ 55. i({) =1 {2arcsin{ i0({) =1 {2· 1 1 {2 + arcsin{ · 1 2 1 {21@2(2{) = 1 {arcsin{ 1 {2 Note thati0= 0where the graph ofi has a horizontal tangent. Also note thati0is negative wheni is decreasing and i0is positive wheni is increasing.

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SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 109 57.Let| = cos1{. Then cos | = { and 0 | sin | g|

g{ = 1 g| g{ = 1 sin| = 1 s 1 cos2| = 1

1 {2. [Note that sin| 0 for 0 | .] 59.{2+|2=u2is a circle with centerR and d{ + e| = 0 is a line through R [assume d

ande are not both zero]. {2+|2=u2 2{ + 2||0= 0 |0={@|, so the slope of the tangent line atS₀({₀> |₀)is{₀@|₀. The slope of the lineRS₀is|₀@{₀, which is the negative reciprocal of{₀@|₀. Hence, the curves are orthogonal, and the families of curves are orthogonal trajectories of each other.

61.| = f{2 |0= 2f{ and {2+ 2|2=n [assume n A 0] 2{ + 4||0= 0 2||0={ |0= {

2(|)= {2(f{2) = 1

2f{, so the curves are orthogonal if f 6= 0. If f = 0, then the horizontal line | = f{2_{= 0}_intersects_{2_{+ 2}_|2₌_{n orthogonally} at±n> 0, since the ellipse{2+ 2|2=n has vertical tangents at those two points.

63.To nd the points at which the ellipse{2 {| + |2= 3crosses the{-axis, let | = 0 and solve for {.

| = 0 {2_{{(0) + 0}2_{= 3} _{{ = ±}_{3. So the graph of the ellipse crosses the}_{{-axis at the points}_±₃_{> 0}_. Using implicit differentiation to nd|0, we get 2{ {|0 | + 2||0= 0 |0(2| {) = | 2{ |0= | 2{

2| {. So|0at3> 0is 0 2 3 2(0)3= 2and| 0_at₃_{> 0}_is 0 + 2 3

2(0) +3= 2. Thus, the tangent lines at these points are parallel. 65.{2|2+{| = 2 {2· 2||0+|2· 2{ + { · |0+| · 1 = 0 |0(2{2| + {) = 2{|2 | |0₌2{|2+| 2{2| + {. So 2{|2+| 2{2| + { =1 2{| 2₊_{| = 2{}2_{| + { |(2{| + 1) = {(2{| + 1)} |(2{| + 1) {(2{| + 1) = 0 (2{| + 1)(| {) = 0 {| = 1 2 or| = {. But {| = 12 {2_|2₊_{{| =}1

4 12 6= 2, so we must have { = |. Then {2|2+{| = 2 {4+{2= 2 {4+{2 2 = 0 ({2+ 2)({2 1) = 0. So {2=2, which is impossible, or {2= 1 { = ±1. Since { = |, the points on the curve where the tangent line has a slope of1 are (1> 1) and (1> 1).

67. (a) If| = i1({), then i(|) = {. Differentiating implicitly with respect to { and remembering that | is a function of {,

we geti0(|) g| g{= 1, sog|g{= 1 i0₍_|) i10 ({) = 1 i0₍_i1₍_{)). (b)i(4) = 5 i1(5) = 4. By part (a),i10(5) = 1

i0₍_i1₍₅₎₎ = 1 i0₍₄₎= 1 ₂ 3 =3₂.

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69. {2+ 4|2= 5 2{ + 4(2||0) = 0 |0= {

4|. Now letk be the height of the lamp, and let (d> e) be the point of tangency of the line passing through the points (3> k) and (5> 0). This line has slope (k 0)@[3 (5)] = 1₈k. But the slope of the tangent line through the point (d> e) can be expressed as |0= d

4e, or as e 0

d (5) =d + 5e [since the line passes through (5> 0) and (d> e)], so d

4e= e

d + 5 4e2=d2 5d d2+ 4e2=5d. But d2+ 4e2= 5 [since (d> e) is on the ellipse], so 5 = 5d d = 1. Then 4e2=d2 5d = 1 5(1) = 4 e = 1, since the point is on the top half of the ellipse. Sok

8 = e d + 5 = 1 1 + 5= 1

4 k = 2. So the lamp is located 2 units above the {-axis.

3.6 Derivatives of Logarithmic Functions

1. The differentiation formula for logarithmic functions, g

g{(logd{) = 1

{ ln d, is simplest whend = h because ln h = 1. 3. i({) = sin(ln {) i0({) = cos(ln {) · g

g{ln{ = cos(ln {) · 1 {= cos(ln{) { 5. i({) = log₂(1 3{) i0({) = 1 (1 3{) ln 2 g g{(1 3{) =(1 3{) ln 23 or 3 (3{ 1) ln 2 7. i({) =5ln{ = (ln {)1@5 i0({) =1₅(ln{)4@5 g g{(ln{) = 1 5(ln{)4@5· 1 {= 1 5{s5 _(ln_{)4

9. i({) = sin { ln(5{) i0({) = sin { · 1

5{· gg{(5{) + ln(5{) · cos { = sin{ · 5 5{ + cos{ ln(5{) = sin{ { + cos{ ln(5{) 11. I (w) = ln(2w + 1) 3 (3w 1)4 = ln(2w + 1) 3_{ln(3w 1)}4 _{= 3 ln(2}_{w + 1) 4 ln(3w 1)} I0₍_{w) = 3 ·} 1 2w + 1· 2 4 · 1 3w 1· 3 = 6 2w + 1 12 3w 1, or combined, 6(w + 3) (2w + 1)(3w 1). 13. j({) = ln{{2 1= ln{ + ln({2 1)1@2= ln{ +1₂ln({2 1) j0₍_{{) =} 1 {+ 1 2· 1 {2₁· 2{ = 1 {+{2{₁ ={ 2_{1 + { · {} {({2₁₎ = 2{2 1 {({2₁₎ 15. i(x) = lnx 1 + ln(2x) i0₍_{x) =} [1 + ln(2x)] ·x1 ln x ·2x1 · 2 [1 + ln(2x)]2 = 1 x[1 + ln(2x) ln x] [1 + ln(2x)]2 = 1 + (ln 2 + lnx) ln x x[1 + ln(2x)]2 = 1 + ln 2 x[1 + ln(2x)]2 17. | = ln2 { 5{2 |0= 1 2 { 5{2 · (1 10{) = 10{ 1 2 { 5{2 or 10{ + 1 5{2+{ 2 19. | = ln(h{+{h{) = ln(h{(1 +{)) = ln(h{) + ln(1 +{) = { + ln(1 + {) |0₌_{1 +} 1 1 +{= 1 { + 1 1 +{ = {1 +{

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SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 111 21.| = 2{ log₁₀{ = 2{ log₁₀{1@2= 2{ ·1₂log₁₀{ = { log₁₀{ |0={ · 1

{ ln 10+ log10{ · 1 = 1

ln 10+ log10{

Note: _{ln 10}1 = lnh

ln 10= log10h, so the answer could be written as 1

ln 10+ log10{ = log10h + log10{ = log10h{. 23.| = {2ln(2{) |0 ={2· 1 2{· 2 + ln(2{) · (2{) = { + 2{ ln(2{) |00_{= 1 + 2}_{{ ·} 1 2{· 2 + ln(2{) · 2 = 1 + 2 + 2 ln(2{) = 3 + 2 ln(2{) 25.| = ln{ +1 +{2 |0₌ 1 { +1 +{2 g g{ { +1 +{2= 1 { +1 +{2 k 1 +1₂(1 +{2)1@2(2{) l = 1 { +1 +{2 1 + { 1 +{2 = 1 { +1 +{2 · 1 +{2+{ 1 +{2 = 1 1 +{2 |00₌1 2(1 +{2)3@2(2{) = _{(1 +}{_{2₎3@2 27.i({) = { 1 ln({ 1) i0₍_{{) =}[1 ln({ 1)] · 1 { · 1 { 1 [1 ln({ 1)]2 = ({ 1)[1 ln({ 1)] + { { 1 [1 ln({ 1)]2 = { 1 ({ 1) ln({ 1) + { ({ 1)[1 ln({ 1)]2 =2{ 1 ({ 1) ln({ 1) ({ 1)[1 ln({ 1)]2

Dom(i) = {{ | { 1 A 0 and 1 ln({ 1) 6= 0} = {{ | { A 1 and ln({ 1) 6= 1}

={ | { A 1 and { 1 6= h1={{ | { A 1 and { 6= 1 + h} = (1> 1 + h) (1 + h> ) 29.i({) = ln({2 2{) i0({) = 1 {2_2{(2{ 2) = 2({ 1) {({ 2). Dom(i) = {{ | {({ 2) A 0} = (> 0) (2> ). 31.i({) = ln{ {2 i0({) = { 2₍₁_{@{) (ln {)(2{)} ({2)2 = { 2{ ln { {4 = {(1 2 ln {)_{4 = 1 2 ln { {3 , soi0(1) = 1 2 ln 1 13 = 1 2 · 0 1 = 1. 33.| = ln {h{2_{= ln}_{{ + ln h}{2 _{= ln}_{{ + {}2 _|0₌ 1

{+ 2{. At (1> 1), the slope of the tangent line is |0_{(1) = 1 + 2 = 3, and an equation of the tangent line is}_{| 1 = 3({ 1), or | = 3{ 2.}

35.i({) = sin { + ln { i0({) = cos { + 1@{.

This is reasonable, because the graph shows thati increases when i0is positive, andi0({) = 0 when i has a horizontal tangent.

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37. | = (2{ + 1)5({4 3)6 ln | = ln(2{ + 1)5({4 3)6 ln | = 5 ln(2{ + 1) + 6 ln({4 3) 1 ||0= 5· 1 2{ + 1· 2 + 6 · 1 {4₃· 4{3 |0₌_| 10 2{ + 1+ 24{3 {4₃ = (2{ + 1)5({4 3)6 10 2{ + 1+ 24{3 {4₃ .

[The answer could be simplied to|0= 2(2{ + 1)4({4 3)5(29{4+ 12{3 15), but this is unnecessary.]

39. | = sin

2_{{ tan}4_{

({2+ 1)2 ln | = ln(sin

2_{{ tan}4_{{) ln({}2_{+ 1)}2

ln| = ln(sin {)2+ ln(tan{)4 ln({2+ 1)2 ln | = 2 ln |sin {| + 4 ln |tan {| 2 ln({2+ 1) 1 ||0= 2· 1 sin{· cos { + 4 · 1 tan{· sec 2_{{ 2 ·} 1 {2_{+ 1}· 2{ |0= sin2{ tan4{ ({2+ 1)2 2 cot{ +4 sec 2_{ tan{ 4{ {2_{+ 1} 41. | = {{ ln | = ln {{ ln | = { ln { |0@| = {(1@{) + (ln {) · 1 |0=|(1 + ln {) |0₌_{{_{(1 + ln}_{)

43. | = {sin { ln | = ln {sin { ln | = sin { ln { | 0 | = (sin{) · 1 {+ (ln{)(cos {) |0₌_|sin{ { + ln{ cos { |0₌_{sin {sin{ { + ln{ cos {

45. | = (cos {){ ln | = ln(cos {){ ln | = { ln cos { 1

| |0={ · 1

cos{· ( sin {) + ln cos { · 1 |0₌_|_{ln cos}_{{ {}sin{

cos{

|0_{= (cos}_{){_{(ln cos}_{{ { tan {)} 47. | = (tan {)1@{ ln | = ln(tan {)1@{ ln | = 1 {ln tan{ 1 ||0 = 1 {· 1 tan{· sec 2_{{ + ln tan { ·}1 {2 |0 ₌_|sec2{ { tan { ln tan{ {2 |0_{= (tan}_{)1@{sec2{ { tan { ln tan{ {2 or |0= (tan{)1@{· 1 { csc{ sec { ln tan{ { 49. | = ln({2+|2) |0= 1 {2₊_|2 _g{g ({2+|2) |0= 2{ + 2||0 {2₊_|2 {2|0+|2|0 = 2{ + 2||0 {2_|0₊_|2_|0_2||0_{= 2}_{{ ({}2₊_|2_2|)|0_{= 2}_{{ |}0₌ 2{ {2₊_|2_2| 51. i({) = ln({ 1) i0({) = 1 ({ 1) = ({ 1) 1 _i00₍_{{) = ({ 1)}2 _i000₍_{{) = 2({ 1)}3 i(4)₍_{{) = 2 · 3({ 1)}4 _{· · · i}(q)₍_{{) = (1)}q1_{· 2 · 3 · 4 · · · (q 1)({ 1)}q_{= (}₁₎q1(q 1)! ({ 1)q 53. Ifi({) = ln (1 + {), then i0({) = 1 1 +{, soi 0_{(0) = 1.} Thus, lim {0 ln(1 +{) { = lim{0 i({) { = lim{0 i({) i(0) { 0 =i0(0) = 1.

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SECTION 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ¤ 113

3.7 Rates of Change in the Natural and Social Sciences

1. (a)v = i(w) = w3 12w2+ 36w y(w) = i0(w) = 3w2 24w + 36 (b)y(3) = 27 72 + 36 = 9 ft@s

(c) The particle is at rest wheny(w) = 0. 3w2 24w + 36 = 0 3(w 2)(w 6) = 0 w = 2 s or 6 s. (d) The particle is moving in the positive direction wheny(w) A 0. 3(w 2)(w 6) A 0 0 w ? 2 or w A 6. (e) Since the particle is moving in the positive direction and in the

negative direction, we need to calculate the distance traveled in the intervals [0> 2], [2> 6], and [6> 8] separately.

|i(2) i(0)| = |32 0| = 32. |i(6) i(2)| = |0 32| = 32. |i(8) i(6)| = |32 0| = 32. The total distance is 32 + 32 + 32 = 96 ft.

(f )

(g)y(w) = 3w2 24w + 36 d(w) = y0₍_{w) = 6w 24.}

d(3) = 6(3) 24 = 6 (ft@s)@s or ft@s2_.

(h )

(i) The particle is speeding up wheny and d have the same sign. This occurs when 2 ? w ? 4 [y and d are both negative] and whenw A 6 [y and d are both positive]. It is slowing down when y and d have opposite signs; that is, when 0 w ? 2 and when 4? w ? 6.

3. (a)v = i(w) = cos(w@4) y(w) = i0(w) = sin(w@4) · (@4) (b)y(3) = ₄sin3₄ =₄ ·₂2 =₈2 ft@s [ 0=56]

(c) The particle is at rest wheny(w) = 0. ₄sinw₄ = 0 sinw₄ = 0 w₄ =q w = 0, 4, 8 s. (d) The particle is moving in the positive direction wheny(w) A 0. ₄sinw₄ A 0 sinw₄ ? 0 4 ? w ? 8. (e) From part (c),y(w) = 0 for w = 0> 4> 8. As in Exercise 1, we’ll

nd the distance traveled in the intervals [0> 4] and [4> 8]. |i(4) i(0)| = |1 1| = 2

|i(8) i(4)| = |1 (1)| = 2= The total distance is 2 + 2 = 4 ft.

(f ) (g)y(w) = 4sin w 4 d(w) = y0₍_{w) =} 4cos w 4 · 4 = 2 16cos w 4. d(3) = 2 16cos 3 4 = 2 16 2 2 = 2₂ 32 (ft@s)@s or ft@s 2_. (h )

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TION 3.7 TION 3.7

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(i) The particle is speeding up wheny and d have the same sign. This occurs when 0 ? w ? 2 or 8 ? w ? 10 [y and d are both negative] and when 4? w ? 6 [y and d are both positive]. It is slowing down when y and d have opposite signs; that is, when 2? w ? 4 and when 6 ? w ? 8.

5. (a) From the gure, the velocityy is positive on the interval (0> 2) and negative on the interval (2> 3). The acceleration d is positive (negative) when the slope of the tangent line is positive (negative), so the acceleration is positive on the interval (0> 1), and negative on the interval (1> 3). The particle is speeding up when y and d have the same sign, that is, on the interval (0> 1) when y A 0 and d A 0, and on the interval (2> 3) when y ? 0 and d ? 0. The particle is slowing down wheny and d have opposite signs, that is, on the interval (1> 2) when y A 0 and d ? 0.

(b)y A 0 on (0> 3) and y ? 0 on (3> 4). d A 0 on (1> 2) and d ? 0 on (0> 1) and (2> 4). The particle is speeding up on (1> 2) [y A 0, d A 0] and on (3> 4) [y ? 0, d ? 0]. The particle is slowing down on (0> 1) and (2> 3) [y A 0, d ? 0].

7. (a)v(w) = w3 4=5w2 7w y(w) = v0(w) = 3w2 9w 7 = 5 3w2 9w 12 = 0

3(w 4)(w + 1) = 0 w = 4 or 1. Since w 0, the particle reaches a velocity of 5 m@s at w = 4 s.

(b)d(w) = y0(w) = 6w 9 = 0 w = 1=5. The acceleration changes from negative to positive, so the velocity changes from decreasing to increasing. Thus, atw = 1=5 s, the velocity has its minimum value.

9. (a)k = 10w 0=83w2 y(w) = gk

gw = 10 1=66w, so y(3) = 10 1=66(3) = 5=02 m@s.

(b)k = 25 10w 0=83w2= 25 0=83w2 10w + 25 = 0 w = 10 ±₁₌₆₆17 3=54 or 8=51.

The valuew₁ =10 ₁₌₆₆17corresponds to the time it takes for the stone to rise 25 m andw₂=10 +₁₌₆₆17 corresponds to the time when the stone is 25 m high on the way down. Thus,y(w1) = 10 1=66

10 17

1=66

=17 4=12 m@s. 11. (a)D({) = {2 D0({) = 2{. D0(15) = 30mm2@mm is the rate at which

the area is increasing with respect to the side length as{ reaches 15 mm. (b) The perimeter isS ({) = 4{, so D0({) = 2{ =1₂(4{) =1₂S ({). The

gure suggests that if { is small, then the change in the area of the square is approximately half of its perimeter (2 of the 4 sides) times {. From the gure, D = 2{ ({) + ({)2. If { is small, then D 2{ ({) and so D@{ 2{.

13. (a) UsingD(u) = u2, we nd that the average rate of change is: (i) D(3) D(2) 3 2 = 9 4 1 = 5 (ii) D(2=5) D(2) 2=5 2 = 6=25 4 0=5 = 4=5 (iii) D(2=1) D(2) 2=1 2 = 4=41 4 0=1 = 4=1 (b)D(u) = u2 D0(u) = 2u, so D0(2) = 4.

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SECTION 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ¤ 115 (c) The circumference isF(u) = 2u = D0(u). The gure suggests that if u is small,

then the change in the area of the circle (a ring around the outside) is approximately equal to its circumference times u. Straightening out this ring gives us a shape that is approx-imately rectangular with length 2u and width u, so D 2u(u). Algebraically, D = D(u + u) D(u) = (u + u)2 u2= 2u(u) + (u)2.

So we see that if u is small, then D 2u(u) and therefore, D@u 2u. 15.V(u) = 4u2 V0(u) = 8u

(a)V0(1) = 8 ft2@ft (b)V0(2) = 16 ft2@ft (c)V0(3) = 24 ft2@ft

As the radius increases, the surface area grows at an increasing rate. In fact, the rate of change is linear with respect to the radius.

17.The mass isi({) = 3{2, so the linear density at{ is ({) = i0({) = 6{.

(a)(1) = 6 kg@m (b)(2) = 12 kg@m (c)(3) = 18 kg@m

Since is an increasing function, the density will be the highest at the right end of the rod and lowest at the left end. 19.The quantity of charge isT(w) = w3 2w2+ 6w + 2, so the current is T0(w) = 3w2 4w + 6.

(a)T0(0=5) = 3(0=5)2 4(0=5) + 6 = 4=75 A (b)T0(1) = 3(1)2 4(1) + 6 = 5 A

The current is lowest whenT0has a minimum.T00(w) = 6w 4 ? 0 when w ?2₃. So the current decreases whenw ?2₃ and increases whenw A2₃. Thus, the current is lowest atw =2₃ s.

21. (a) To nd the rate of change of volume with respect to pressure, we rst solve forY in terms of S . S Y = F Y = F_S gY_gS = F

S2.

(b) From the formula forgY@gS in part (a), we see that as S increases, the absolute value of gY@gS decreases. Thus, the volume is decreasing more rapidly at the beginning.

(c) = 1 Y gYgS = 1 Y F S2 = F (S Y )S = F FS = 1 S

23.In Example 6, the population function wasq = 2wq0. Since we are tripling instead of doubling and the initial population is 400, the population function isq(w) = 400 · 3w. The rate of growth isq0(w) = 400 · 3w· ln 3, so the rate of growth after 2=5 hours is q0(2=5) = 400 · 32=5· ln 3 6850 bacteria@hour= 25. (a)1920: p1= 1860 1750 1920 1910= 110 10 = 11,p2= 2070 1860 1930 1920= 210 10 = 21, (p₁+p₂)/ 2 = (11 + 21)@2 = 16 million@year 1980: p1= 4450₁₉₈₀ 3710₁₉₇₀= 740₁₀ = 74,p2= 5280₁₉₉₀ 4450₁₉₈₀= 830₁₀ = 83, (p1+p2)/ 2 = (74 + 83)@2 = 78=5 million@year

(b)S (w) = dw3+ew2+fw + g (in millions of people), where d 0=0012937063, e 7=061421911, f 12,822=97902, andg 7,743,770=396.

(c)S (w) = dw3+ew2+fw + g S0(w) = 3dw2+ 2ew + f (in millions of people per year)

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TION 3.7 TION 3.7

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(d)S0(1920) = 3(0=0012937063)(1920)2+ 2(7=061421911)(1920) + 12,822=97902 14.48 million@year [smaller than the answer in part (a), but close to it] S0₍₁₉₈₀₎_{75.29 million@year (smaller, but close)}

(e)S0(1985) 81.62 million@year, so the rate of growth in 1985 was about 81=62 million@year. 27. (a) Usingy = S

4o(U

2_u2₎_with_{U = 0=01, o = 3, S = 3000, and = 0=027, we have y as a function of u:}

y(u) = 3000 4(0=027)3(0=01

2_u2_). _{y(0) = 0=925 cm@s, y(0=005) = 0=694 cm@s, y(0=01) = 0.}

(b)y(u) = S 4o(U

2_u2₎ _y0₍_{u) = S}

4o(2u) = S u2o. Wheno = 3, S = 3000, and = 0=027, we have y0₍_{u) =} 3000u

2(0=027)3. y

0_{(0) = 0,}_y0₍₀_{=005) = 92=592 (cm@s)@cm, and y}0₍₀_{=01) = 185=185 (cm@s)@cm.}

(c) The velocity is greatest whereu = 0 (at the center) and the velocity is changing most where u = U = 0=01 cm (at the edge).

29. (a)F({) = 1200 + 12{ 0=1{2+ 0=0005{3 F0({) = 12 0=2{ + 0=0015{2 $@yard, which is the marginal cost function.

(b)F0(200) = 12 0=2(200) + 0=0015(200)2= $32@yard, and this is the rate at which costs are increasing with respect to the production level when{ = 200. F0(200)predicts the cost of producing the 201st yard=

(c) The cost of manufacturing the 201st yard of fabric isF(201) F(200) = 3632=2005 3600 $32=20, which is approximatelyF0(200). 31. (a)D({) = s({) { D0({) = {s 0₍_{{) s({) · 1} {2 ={s 0₍_{{) s({)} {2 .

D0₍_{{) A 0 D({) is increasing; that is, the average productivity increases as the size of the workforce increases.} (b)s0({) is greater than the average productivity s0({) A D({) s0({) A s({)

{ {s0({) A s({) {s0₍_{{) s({) A 0 {s}0({) s({) {2 A 0 D0({) A 0. 33. S Y = qUW W = S Y qU = (10)(0S Y=0821) = 1

0=821(S Y ). Using the Product Rule, we have gW gw = 1 0=821[S (w)Y 0₍_{w) + Y (w)S}0₍_{w)] =} 1 0=821[(8)(0=15) + (10)(0=10)] 0=2436 K@min. 35. (a) If the populations are stable, then the growth rates are neither positive nor negative; that is,gF

gw = 0andgZgw = 0. (b) “The caribou go extinct” means that the population is zero, or mathematically,F = 0.

(c) We have the equationsgF

gw =dF eFZ and gZgw =fZ + gFZ. Let gF@gw = gZ@gw = 0, d = 0=05, e = 0=001, f = 0=05, and g = 0=0001 to obtain 0=05F 0=001FZ = 0 (1) and 0=05Z + 0=0001FZ = 0 (2). Adding 10 times