calculating the derivative - grosse pointe … 4.1 techniques for finding derivatives 243 40....

Chapter 4

CALCULATING THE DERIVATIVE

4.1 Techniques for FindingDerivatives

1. y = 12x3 ¡ 8x2 + 7x+ 5dy

dx= 12(3x3¡1)¡ 8(2x2¡1) + 7x1¡1 + 0= 36x2 ¡ 16x+ 7

2. y = 8x3 ¡ 5x2 ¡ x

12

dy

dx= 8(3x3¡1)¡ 5(2x2¡1)¡ 1

12x1¡1

= 24x2 ¡ 10x¡ 1

12

3. y = 3x4 ¡ 6x3 + x2

8+ 5

dy

dx= 3(4x4¡1)¡ 6(3x3¡1) + 1

8(2x2¡1) + 0

= 12x3 ¡ 18x2 + 14x

4. y = 5x4 + 9x3 + 12x2 ¡ 7xdy

dx= 5(4x4¡1) + 9(3x3¡1)

+ 12(2x2¡1)¡ 7(x1¡1)= 20x3 + 27x2 + 24x¡ 7

5. y = 6x3:5 ¡ 10x0:5dy

dx= 6(3:5x3:5¡1)¡ 10(0:5x0:5¡1)= 21x2:5 ¡ 5x¡0:5 or 21x2:5 ¡ 5

x0:5

6. f(x) = ¡2x1:5 + 12x0:5f 0(x) = ¡2(1:5x1:5¡1) + 12(0:5x0:5¡1)

= ¡3x0:5 + 6x¡0:5 or ¡ 3x0:5 + 6

x0:5

7. y = 8px+ 6x3=4

= 8x1=2 + 6x3=4

dy

dx= 8

μ1

2x1=2¡1

¶+ 6

μ3

4x3=4¡1

¶

= 4x¡1=2 +9

2x¡1=4

or4

x1=2+

9

2x1=4

8. y = ¡100px¡ 11x2=3= ¡100x1=2 ¡ 11x2=3

dy

dx= ¡100

μ1

2x1=2¡1

¶¡ 11

μ2

3x2=3¡1

¶

= ¡50x¡1=2 ¡ 22x¡1=3

3or¡50x1=2

¡ 22

3x1=3

9. g(x) = 6x¡5 ¡ x¡1g0(x) = 6(¡5)x¡5¡1 ¡ (¡1)x¡1¡1

= ¡30x¡6 + x¡2

or¡30x6

+1

x2

10. y = 10x¡3 + 5x¡4 ¡ 8xdy

dx= 10(¡3x¡3¡1) + 5(¡4x¡4¡1)¡ 8x1¡1

= ¡30x¡4 ¡ 20x¡5 ¡ 8 or ¡30x4

¡ 20x5¡ 8

11. y = 5x¡5 ¡ 6x¡2 + 13x¡1dy

dx= 5(¡5x¡5¡1)¡ 6(¡2x¡2¡1) + 13(¡1x¡1¡1)= ¡25x¡6 + 12x¡3 ¡ 13x¡2

or¡25x6

+12

x3¡ 13x2

12. f(t) =7

t¡ 5

t3

= 7t¡1 ¡ 5t¡3f 0(t) = 7(¡1t¡1¡1)¡ 5(¡3t¡3¡1)

= ¡7t¡2 + 15t¡4 or ¡7t2+15

t4

13. f(t) =14

t+12

t4+p2

= 14t¡1 + 12t¡4 +p2

f 0(t) = 14(¡1t¡1¡1) + 12(¡4t¡4¡1) + 0= ¡14t¡2 ¡ 48t¡5 or ¡14

t2¡ 48t5 239

240 Chapter 4 CALCULATING THE DERIVATIVE

14. y =6

x4¡ 7

x3+3

x+p5

= 6x¡4 ¡ 7x¡3 + 3x¡1 +p5dy

dx= 6(¡4x¡4¡1)¡ 7(¡3x¡3¡1) + 3(¡1x¡1¡1) + 0= ¡24x¡5 + 21x¡4 ¡ 3x¡2

or¡24x5

+21

x4¡ 3

x2

15. y =3

x6+1

x5¡ 7

x2

= 3x¡6 + x¡5 ¡ 7x¡2dy

dx= 3(¡6x¡7) + (¡5x¡6)¡ 7(¡2x¡3)= ¡18x¡7 ¡ 5x¡6 + 14x¡3

or¡18x7

¡ 5

x6+14

x3

16. p(x) = ¡10x¡1=2 + 8x¡3=2

p0(x) = ¡10μ¡12x¡3=2

¶+ 8

μ¡32x¡5=2

¶

= 5x¡3=2 ¡ 12x¡5=2

or5

x3=2¡ 12

x5=2

17. h(x) = x¡1=2 ¡ 14x¡3=2

h0(x) = ¡12x¡3=2 ¡ 14

μ¡32x¡5=2

¶

=¡x¡3=22

+ 21x¡5=2

or¡12x3=2

+21

x5=2

18. y =6

4px= 6x¡1=4

dy

dx= 6

μ¡14

¶x¡5=4

= ¡32x¡5=4

or¡32x5=4

19. y =¡23px

=¡2x1=3

= ¡2x¡1=3

dy

dx= ¡2

μ¡13x¡4=3

¶

=2x¡4=3

3or

2

3x4=3

20. f(x) =x3 + 5

x

= x2 + 5x¡1

f 0(x) = 2x2¡1 + 5(¡1x¡1¡1)= 2x¡ 5x¡2

or 2x¡ 5

x2

21. g(x) =x3 ¡ 4xp

x

=x3 ¡ 4xx1=2

= x5=2 ¡ 4x1=2

g0(x) =5

2x5=2¡1 ¡ 4

μ1

2x1=2¡1

¶

=5

2x3=2 ¡ 2x¡1=2

or5

2x3=2 ¡ 2p

x

22. g(x) = (8x2 ¡ 4x)2= 64x4 ¡ 64x3 + 16x2

g0(x) = 64(4x4¡1)¡ 64(3x3¡1) + 16(2x2¡1)= 256x3 ¡ 192x2 + 32x

23. h(x) = (x2 ¡ 1)3= x6 ¡ 3x4 + 3x2 ¡ 1

h0(x) = 6x6¡1 ¡ 3(4x4¡1) + 3(2x2¡1)¡ 0= 6x5 ¡ 12x3 + 6x

24. A quadratic function has degree 2.When the derivative is taken, the power will de-crease by 1 and the derivative function will be lin-ear, so the correct choice is (b).

26.d

dx(4x3 ¡ 6x¡2)

= 4(3x2)¡ 6(¡2x¡3)= 12x2 + 12x¡3

= 12x2 +12

x3choice (c)

=12x2(x3) + 12

x3

=12x5 + 12

x3choice (b)

Neither choice (a) nor choice (d) equals

d

dx(4x3 ¡ 6x¡2):

Section 4.1 Techniques for Finding Derivatives 241

27. Dx

·9x¡1=2 +

2

x3=2

¸

= Dx[9x¡1=2 + 2x¡3=2]

= 9

μ¡12x¡3=2

¶+ 2

μ¡32x¡5=2

¶

= ¡92x¡3=2 ¡ 3x¡5=2

or¡92x3=2

¡ 3

x5=2

28. Dx

·84px¡ 3p

x3

¸

= Dx [8x¡1=4 ¡ 3x¡3=2]

= 8

μ¡14x¡5=4

¶¡ 3

μ¡32x¡5=2

¶

= ¡2x¡5=4 + 9x¡5=2

2

or¡2x5=4

+9

2x5=2

29. f(x) =x4

6¡ 3x

=1

6x4 ¡ 3x

f 0(x) =1

6(4x3)¡ 3

=2

3x3 ¡ 3

f 0(¡2) = 2

3(¡2)3 ¡ 3

= ¡163¡ 3

= ¡253

30. f(x) =x3

9¡ 7x2

=1

9x3 ¡ 7x2

f 0(x) =1

9(3x2)¡ 7(2x)

=1

3x2 ¡ 14x

f 0(3) =1

3(3)2 ¡ 14(3)

= 3¡ 42 = ¡39

31. y = x4 ¡ 5x3 + 2; x = 2y0 = 4x3 ¡ 15x2

y0(2) = 4(2)3 ¡ 15(2)2= ¡28

The slope of tangent line at x = 2 is ¡28:Use m = ¡28 and (x1; _y1) = (2;¡22) to obtainthe equation.

y ¡ (¡22) = ¡28(x¡ 2)y = ¡28x+ 34

32. y = ¡3x5 ¡ 8x3 + 4x2y0 = ¡3(5x4)¡ 8(3x2) + 4(2x)= ¡15x4 ¡ 24x2 + 8x

y0(1) = ¡15(1)4 ¡ 24(1)2 + 8(1)= ¡15(1)4 ¡ 24(1)2 + 8(1)= ¡15¡ 24 + 8= ¡31

y(1) = ¡3(1)5 ¡ 8(1)3 + 4(1)2 = ¡7

The slope of the tangent line at x = 1 is ¡31:Use m = ¡31 and (x1; y1) = (1;¡7) to obtain theequation.

y ¡ (¡7) = ¡31(x¡ 1)y + 7 = ¡31x+ 31

y = ¡31x+ 24

33. y = ¡2x1=2 + x3=2

y0 = ¡2μ1

2x¡1=2

¶+3

2x1=2

= ¡x¡1=2 + 32x1=2

= ¡ 1

x1=2+3x1=2

2

y0(9) = ¡ 1

(9)1=2+3(9)1=2

2

= ¡13+9

2

=25

6

The slope of the tangent line at x = 9 is 256 :


34. y = ¡x¡3 + x¡2y0 = ¡(¡3x¡4) + (¡2x¡3)= 3x¡4 ¡ 2x¡3

=3

x4¡ 2

x3

y0(2) =3

(2)4¡ 2

(2)3

=3

16¡ 28

= ¡ 1

16

The slope of the tangent line at x = 2 is ¡ 116 .

35. f(x) = 9x2 ¡ 8x+ 4f 0(x) = 18x¡ 8Let f 0(x) = 0 to …nd the point where the slope ofthe tangent line is zero.

18x¡ 8 = 018x = 8

x =8

18=4

9Find the y-coordinate.

f(x) = 9x2 ¡ 8x+ 4

f

μ4

9

¶= 9

μ4

9

¶2¡ 8

μ4

9

¶+ 4

= 9

μ16

81

¶¡ 329+ 4

=16

9¡ 329+36

9=20

9

The slope of the tangent line is zero at one point,¡49 ;

209

¢.

36. f(x) = x3 + 9x2 + 19x¡ 10f 0(x) = 3x2 + 18x+ 19

If the slope of the tangent line is ¡5, then thef 0(x) = ¡5:3x2 + 18x+ 19 = ¡53x2 + 18x+ 24 = 0

3(x2 + 6x+ 8) = 0

3(x+ 2)(x+ 4) = 0

x = ¡2 or x = ¡4f(¡2) = (¡2)3 + 9(¡2)2 + 19(¡2)¡ 10

= ¡8 + 36¡ 38¡ 10= ¡20

f(¡4) = (¡4)3 + 9(¡4)2 + 19(¡4)¡ 10= ¡64 + 144¡ 76¡ 10= ¡6

Thus, the points where the slope of the tangentline is ¡5 are (¡2;¡20) and (¡4;¡6):

37. f(x) = 2x3 + 9x2 ¡ 60x+ 4f 0(x) = 6x2 + 18x¡ 60If the tangent line is horizontal, then its slope iszero and f 0(x) = 0:

6x2 + 18x¡ 60 = 06(x2 + 3x¡ 10) = 06(x+ 5)(x¡ 2) = 0x = ¡5 or x = 2

Thus, the tangent line is horizontal at x = ¡5 andx = 2:

38. f(x) = x3 + 15x2 + 63x¡ 10f 0(x) = 3x2 + 30x+ 63

If the tangent line is horizontal, then its slope iszero and f 0(x) = 0:

3x2 + 30x+ 63 = 0

3(x2 + 10x+ 21) = 0

3(x+ 3)(x+ 7) = 0

x = ¡3 or x = ¡7

Therefore, the tangent line is horizontal at x = ¡3or x = ¡7:

39. f(x) = x3 ¡ 4x2 ¡ 7x+ 8f 0(x) = 3x2 ¡ 8x¡ 7If the tangent line is horizontal, then its slope iszero and f 0(x) = 0:

3x2 ¡ 8x¡ 7 = 0

x =8§p64 + 84

6

x =8§p148

6

x =8§ 2p37

6

x =2(4§p37)

6

x =4§p37

3

Thus, the tangent line is horizontal at x = 4§p373 :


40. f(x) = x3 ¡ 5x2 + 6x+ 3f 0(x) = 3x2 ¡ 10x+ 6If the tangent line is horizontal, then its slope iszero and f 0(x) = 0:

3x2 ¡ 10x+ 6 = 0

x =10§p100¡ 72

6

x =10§p28

6

x =10§ 2p7

6

x =5§p73

The tangent line is horizontal at x = 5§p73 :

41. f(x) = 6x2 + 4x¡ 9f 0(x) = 12x+ 4

If the slope of the tangent line is ¡2; f 0(x) = ¡2:12x+ 4 = ¡2

12x = ¡6

x = ¡12

f

μ¡12

¶= ¡19

2

The slope of the tangent line is ¡2 at ¡¡12 ;¡19

2

¢:

42. f(x) = 2x3 ¡ 9x2 ¡ 12x+ 5f 0(x) = 6x2 ¡ 18x¡ 12If the slope of the tangent line is 12, f 0(x) = 12:

6x2 ¡ 18x¡ 12 = 126x2 ¡ 18x¡ 24 = 06(x2 ¡ 3x¡ 4) = 06(x¡ 4)(x+ 1) = 0x = 4 or x = ¡1

f(4) = ¡59 and f(¡1) = 6The slope of the tangent line is ¡12 at (4;¡59)and (¡1; 6):

43. f(x) = x3 + 6x2 + 21x+ 2

f 0(x) = 3x2 + 12x+ 21

If the slope of the tangent line is 9, f 0(x) = 9:

3x2 + 12x+ 21 = 9

3x2 + 12x+ 12 = 0

3(x2 + 4x+ 4) = 0

3(x+ 2)2 = 0

x = ¡2f(¡2) = ¡24

The slope of the tangent line is 9 at (¡2;¡24):44. f(x) = 3g(x)¡ 2h(x) + 3

f 0(x) = 3g0(x)¡ 2h0(x)f 0(5) = 3g0(5)¡ 2h0(5)

= 3(12)¡ 2(¡3) = 42

45. f(x) =1

2g(x) +

1

4h(x)

f 0(x) =1

2g0(x) +

1

4h0(x)

f 0(2) =1

2g0(2) +

1

4h0(2)

=1

2(7) +

1

4(14) = 7

46. (a) From the graph, f(1) = 2; because the curvegoes through (1; 2):

(b) f 0(1) gives the slope of the tangent line to fat 1. The line goes through (¡1; 1) and (1; 2):

m =2¡ 1

1¡ (¡1) =1

2;

sof 0(1) =

1

2:

(c) The domain of f is [¡1;1) because the x-coordinates of the points of f start at x = ¡1and continue in…nitely through the positive realnumbers.

(d) The range of f is [0;1) because the y-coordinatesof the points on f start at y = 0 and continue in-…nitely through the positive real numbers.

49.f(x)

k=1

k¢ f(x)

Use the rule for the derivative of a constant timesa function.

d

dx

·f(x)

k

¸=d

dx

·1

k¢ f(x)

¸

=1

kf 0(x)

=f 0(x)k

50. Graph the numerical derivative of

f(x) = 1:25x3 + 0:01x2 ¡ 2:9x+ 1for x ranging from ¡5 to 5.(a) When x = 4, the derivative equals 57.18.

(b) The derivative crosses the x-axis at approxi-mately ¡0:88 and 0:88.


51. The demand is given by q = 5000¡ 100p:Solve for p:

p =5000¡ q100

R(q) = q

μ5000¡ q100

¶

=5000q ¡ q2

100

R0(q) =5000¡ 2q100

(a) R0(1000) =5000¡ 2(1000)

100

= 30

(b) R0(2500) =5000¡ 2(2500)

100

= 0

(c) R0(3000) =5000¡ 2(3000)

100

= ¡1052. C(q) = 3000¡ 20q + 0:03q2

R(q) = qp = q

μ5000¡ q100

¶

= q³50¡ q

100

´(from Exercise 51)The pro…t equation is found as follows.

P = R¡C= q

³50¡ q

100

´¡ (3000¡ 20q + 0:03q2)

= 50q ¡ q2

100¡ 3000 + 20q ¡ 0:03q2

= 50q + 20q ¡ q2

100¡ 0:03q2 ¡ 3000

= 70q ¡ 0:01q2 ¡ 0:03q2 ¡ 3000Therefore, P = 70q ¡ 0:04q2 ¡ 3000:Now, the marginal pro…t is

P 0(q) = 70¡ 0:04(2q)¡ 0= 70¡ 0:08q:

(a) P 0(500) = 70¡ 0:08(500)= 70¡ 40= 30

(b) P 0(815) = 70¡ 0:08(815)= 70¡ 65:2= 4:8

(c) P 0(1000) = 70¡ 0:08(1000) = 70¡ 80 = ¡10

53. S(t) = 100¡ 100t¡1S0(t) = ¡100(¡1t¡2)

= 100t¡2

=100

t2

(a) S0(1) =100

(1)2=100

1= 100

(b) S0(10) =100

(10)2=100

100= 1

54. p(q) =1000

q2+ 1000

If R is the revenue function, R(q) = qp(q):

R(q) = q

μ1000

q2+ 1000

¶

R(q) = 1000q¡1 + 1000qR0(q) = ¡1000q¡2 + 1000

R0(q) = 1000¡ 1000q2

R0(q) = 1000μ1¡ 1

q2

¶

R0(10) = 1000μ1¡ 1

102

¶

R0(10) = 1000μ99

100

¶

= 990

The marginal revenue is $990.

55. Pro…t = Revenue ¡ CostP (q) = qp(q)¡C(q)

P (q) = q

μ1000

q2+ 1000

¶¡ (0:2q2 + 6q + 50)

=1000

q+ 1000q ¡ 0:2q2 ¡ 6q ¡ 50

= 1000q¡1 + 994q ¡ 0:2x2 ¡ 50P 0(q) = ¡1000q¡2 + 994¡ 0:4q

= 994¡ 0:4q ¡ 1000q2

P 0(10) = 994¡ 0:4(10)¡ 1000

(10)2

= 994¡ 4¡ 10= 980

The marginal pro…t is $980.


56. C(x) = 2x; R(x) = 6x¡ x2

1000

(a) C0(x) = 2

(b) R0(x) = 6¡ 2x

1000= 6¡ x

500

(c) P (x) = R(x)¡C(x)

=

μ6x¡ x2

1000

¶¡ 2x

= 4x¡ x2

1000

P 0(x) = 4¡ 2x

1000= 4¡ x

500

(d) P 0(x) = 4¡ x

500= 0

4 =x

500

x = 2000

(e) Using part (d), we see that marginal pro…t is0 when x = 2000 units.

P (2000) = 4(2000)¡ (2000)2

1000

= 8000¡ 4000= 4000

The pro…t for 2000 units produced is $4000.

57. (a) 1982 when t = 50:

C(50) = 0:00875(50)2 ¡ 0:108(50) + 1:42= 17:895 ¼ 17:9 cents

2002 when t = 70:

C(70) = 0:00875(70)2 ¡ 0:108(70) + 1:42= 36:735 ¼ 36:7 cents

(b) C 0(t) = 0:00875(2t)¡ 0:108(1)= 0:0175t¡ 0:108

1982 when t = 50:

C 0(50) = 0:0175(50)¡ 0:108= 0:767 cents/year

2002 when t = 70:

C 0(70) = 0:0175(70)¡ 0:108¼ 1:12 cents/year

(c) Using a graphing calculator, a cubic functionthat models the data is

C(t) = (¡1:790£ 10¡4)t3 + 0:02947t2

¡ 0:7105t+ 3:291:

Using the values from the calculator, the rate ofchange in 1982 is C0(50) ¼ 0:894 cents/year.Using the values from the calculator, the rate ofchange in 2002 is C0(70) ¼ 0:784 cents/year.

58. M(t) = 3:044t3 ¡ 379:6t2 + 14,274.5t¡ 139,433

M 0(t) = 3:044(3t3¡1)¡ 379:6(2t2¡1)+ 14,274.5t1¡1 ¡ 0

= 9:132t2 ¡ 759:2t+ 14,274.5(a) 1920¡ 1900 = 20

M 0(20) ¼ 2743(b) 1960¡ 1900 = 60

M 0(60) ¼ 1598(c) 1980¡ 1900 = 80

M 0(80) ¼ 11,983(d) 2000¡ 1900 = 100

M 0(100) ¼ 29,675(e) The amount of money in circulation was

increasing at the rate of $2743 million/yearin 1920 but then began to decrease. Thedecrease of money continued through theearly 1950’s. Since then, the amount hasagain been increasing.

59. N(t) = 0:00437t3:2

N 0(t) = 0:013984t2:2

(a) N 0(5) ¼ 0:4824(b) N 0(10) ¼ 2:216

60. G(x) = ¡0:2x2 + 450(a) G(0) = ¡0:2(0)2 + 450 = 450(b) G(25) = ¡0:2(25)2 + 450

= ¡125 + 450= 325

G0(x) = ¡2(0:2)x = ¡0:4x(c) G0(10) = ¡0:4(10) = ¡4After 10 units of insulin are injected, the bloodsugar level is decreasing at a rate of 4 points perunit of insulin.

(d) G0(25) = ¡0:4(25)= ¡10

After 25 units of insulin are injected, the bloodsugar level is decreasing at a rate of 10 points perunit of insulin.


61. V (t) = ¡2159 + 1313t¡ 60:82t2

(a) V (3) = ¡2159 + 1313(3)¡ 60:82(3)2= 1232.62 cm3

(b) V 0(t) = 1313¡ 121:64tV 0(3) = 1313¡ 121:64(3)

= 948:08 cm3/yr

62. w(c) =c3

100¡ 1500

c

(a) When c = 30; w = 220 g or 0.22 kg.

(b)dw

dc=3c2

100+1500c2

When c = 30; dwdc = 2823 g/cm. When the cir-cumference of the brain is 30 cm, it is increasingby 2823 g with every centimeter the circumferenceincreases.

63. v = 2:69l1:86

dv

dl= (1:86)2:69l1:86¡1 ¼ 5:00l0:86

64. l(x) = ¡2:318 + 0:2356x¡ 0:002674x2

(a) The problem states that a fetus this formulaconcerns is at least 18 weeks old. So, the minimumx value should be 18. Considering the gestationtime of a cow in general, a meaninful range forthis funcion is 18 · x · 44:(b) l0(x) = 0:2356¡ (2)0:002674x

= 0:2356¡ 0:005348x

(c) l0(25) = 0:2356¡ 0:005348(25)= 0:1019 cm/week

65. t = 0:0588s1:125

(a)When s = 1609; t ¼ 238:1 seconds, or 3 minutes,58.1 seconds.

(b)dt

ds= 0:0588(1:125s1:125¡1)

= 0:06615s0:125

When s = 100; dtds ¼ 0:118 sec/m. At 100 meters,the fastest possible time increases by 0.118 sec-onds for each additional meter.

(c) Yes, they have been surpassed. In 2000, theworld record in the mile stood at 3:43.13. (Ref:www.runnersworld.com)

66. V = C(R0 ¡R)R2= CR0R

2 ¡CR3dV

dR= 2CR0R¡ 3CR2 = 0

CR(2R0 ¡ 3R) = 0CR = 0 or 2R0 ¡ 3R = 0

R = 0 R =2

3R0

Discard R = 0, since a closed windpipe producesno air‡ow. Velocity is maximized when R = 2

3R0.

67. BMI =703w

h2

(a) 60200 = 74 in.

BMI =703(220)

742¼ 28

(b) BMI =703w

742= 24:9 implies

w =24:9(74)2

703¼ 194:

A 220-lb person needs to lose 26 pounds to getdown to 194 lbs.

(c) If f(h) =703(125)

h2= 87,875h¡2, then

f 0(h) = 87,875(¡2h¡2¡1)= ¡175,750h¡3 = ¡175,750

h3

(d) f 0(65) = ¡175,750653

¼ ¡0:64

For a 125-lb female with a height of 65 in. (50500),the BMI decreases by 0.64 for each additional inchof height.

(e) Sample Chart

ht=wt 140 160 180 200

60 27 31 35 3965 23 27 30 3370 20 23 26 2975 17 20 22 25

68. s(t) = 11t2 + 4t+ 2

(a) v(t) = s0(t) = 22t+ 4

(b) v(0) = 22(0) + 4 = 4

v(5) = 22(5) + 4 = 114

v(10) = 22(10) + 4 = 224


69. s(t) = 18t2 ¡ 13t+ 8(a) v(t) = s0(t) = 18(2t)¡ 13 + 0

= 36t¡ 13(b) v(0) = 36(0)¡ 13 = ¡13

v(5) = 36(5)¡ 13 = 167v(10) = 36(10)¡ 13 = 347

70. s(t) = 4t3 + 8t2 + t

(a) v(t) = s0(t) = 4(3t2) + 8(2t) + 1= 12t2 + 16t+ 1

(b) v(0) = 12(0)2 + 16(0) + 1 = 1

v(5) = 12(5)2 + 16(5) + 1

= 300 + 80 + 1 = 381

v(10) = 12(10)2 + 16(10) + 1

= 1200 + 160 + 1 = 1361

71. s(t) = ¡3t3 + 4t2 ¡ 10t+ 5(a) v(t) = s0(t) = ¡3(3t2) + 4(2t)¡ 10 + 0

= ¡9t2 + 8t¡ 10(b) v(0) = ¡9(0)2 + 8(0)¡ 10 = ¡10

v(5) = ¡9(5)2 + 8(5)¡ 10= ¡225 + 40¡ 10 = ¡195

v(10) = ¡9(10)2 + 8(10)¡ 10= ¡900 + 80¡ 10 = ¡830

72. s(t) = ¡16t2 + 144velocity = s0(t)

= ¡32t(a) s0(1) = ¡32 ¢ 1 = ¡32

s0(2) = ¡32 ¢ 2= ¡64

The rock’s velocity is ¡32 ft/sec after 1 secondand ¡64 ft/sec after 2 seconds.(b) The rock will hit the ground when s(t) = 0:

¡16t2 + 144 = 0

t =

r144

16

= 3

The rock will hit the ground after 3 seconds.

(c) The velocity at impact is the velocity at 3seconds.

s0(3) = ¡32 ¢ 3= ¡96

Its velocity on impact is ¡96 ft/sec.

73. s(t) = ¡16t2 + 64t

(a) v(t) = s0(t) = ¡16(2t) + 64= ¡32t+ 64

v(2) = ¡32(2) + 64 = ¡64 + 64 = 0v(3) = ¡32(3) + 64 = ¡96 + 64 = ¡32

The ball’s velocity is 0 ft/sec after 2 seconds and¡32 ft/sec after 3 seconds.

(b) As the ball travels upward, its speed decreasesbecause of the force of gravity until, at maximumheight, its speed is 0 ft/sec.In part (a), we found that v(2) = 0:

It takes 2 seconds for the ball to reach its maxi-mum height.

(c) s(2) = ¡16(2)2 + 64(2)= ¡16(4) + 128= ¡64 + 128= 64

It will go 64 ft high.

74. (a) d(x) = 1:66¡ 0:90x+ 0:47x2d(0:5) = 1:66¡ 0:90(0:5) + 0:47(0:5)2

= 1:3275 g/cm3

(b) d0(x) = ¡0:90 + 0:47(2x)= ¡0:90 + 0:94x

d0(0:5) = ¡0:90 + 0:94(0:5)= ¡0:43 g/cm3

When the level of the Dead Sea decreases to 50%of the current level, the density of the brine isdecreasing at the rate of 0.43 gram per cm3.

75. y1 = 4:13x+ 14:63y2 = ¡0:033x2 + 4:647x+ 13:347(a) When x = 5; y1 ¼ 35 and y2 ¼ 36.

(b)dy1dx

= 4:13

dy2dx

= 0:033(2x) + 4:647

= ¡0:066x+ 4:647

When x = 5; dy1dx = 4:13 and dy2dx ¼ 4:32. These

values are fairly close and represent the rate ofchange of four years for a dog for one year of ahuman, for a dog that is actually 5 years old.


(c)With the …rst three points eliminated, the dogage increases in 2-year steps and the human ageincreases in 8-year steps, for a slope of 4. Theequation has the form y = 4x + b. A value of16 for b makes the numbers come out right. y =4x+ b. For a dog of age x = 5 years or more, theequivalent human age is given by y = 4x+ 16.

4.2 Derivatives of Products andQuotients

1. y = (3x2 + 2)(2x¡ 1)dy

dx= (3x2 + 2)(2) + (2x¡ 1)(6x)= 6x2 + 4+ 12x2 ¡ 6x= 18x2 ¡ 6x+ 4

2. y = (5x2 ¡ 1)(4x+ 3)dy

dx= (5x2 ¡ 1)(4) + (10x)(4x+ 3)= 20x2 ¡ 4 + 40x2 + 30x= 60x2 + 30x¡ 4

3. y = (2x¡ 5)2

= (2x¡ 5)(2x¡ 5)dy

dx= (2x¡ 5)(2) + (2x¡ 5)(2)= 4x¡ 10 + 4x¡ 10= 8x¡ 20

4. y = (7x¡ 6)2= (7x¡ 6)(7x¡ 6)

dy

dx= (7x¡ 6)(7) + (7)(7x¡ 6)= 49x¡ 42 + 49x¡ 42= 98x¡ 84

5. k(t) = (t2 ¡ 1)2 = (t2 ¡ 1)(t2 ¡ 1)k0(t) = (t2 ¡ 1)(2t) + (t2 ¡ 1)(2t)

= 2t3 ¡ 2t+ 2t3 ¡ 2t= 4t3 ¡ 4t

6. g(t) = (3t2 + 2)2

= (3t2 + 2)(3t2 + 2)

g0(t) = (3t2 + 2)(6t) + (6t)(3t2 + 2)= 18t3 + 12t+ 18t3 + 12t

= 36t3 + 24t

7. y = (x+ 1)(px+ 2)

= (x+ 1)(x1=2 + 2)

dy

dx= (x+ 1)

μ1

2x¡1=2

¶+ (x1=2 + 2)(1)

=1

2x1=2 +

1

2x¡1=2 + x1=2 + 2

=3

2x1=2 +

1

2x¡1=2 + 2

or3x1=2

2+

1

2x1=2+ 2

8. y = (2x¡ 3)(px¡ 1)= (2x¡ 3)(x1=2 ¡ 1)

dy

dx= (2x¡ 3)

μ1

2x¡1=2

¶+ 2(x1=2 ¡ 1)

= x1=2 ¡ 32x¡1=2 + 2x1=2 ¡ 2

= 3x1=2 ¡ 3x¡1=2

2¡ 2

or 3x1=2 ¡ 3

2x1=2¡ 2

9. p(y) = (y¡1 + y¡2)(2y¡3 ¡ 5y¡4)p0(y) = (y¡1 + y¡2)(¡6y¡4 + 20y¡5)

+ (¡y2 ¡ 2y¡3)(2y¡3 ¡ 5y¡4)= ¡6y¡5 + 20y¡6 ¡ 6y¡6 + 20y¡7

¡ 2y¡5 + 5y¡6 ¡ 4y¡6 + 10y¡7= ¡8y¡5 + 15y¡6 + 30y¡7

10. q(x) = (x¡2 ¡ x¡3)(3x¡1 + 4x¡4)q0(x) = (x¡2 ¡ x¡3)(¡3x¡2 ¡ 16x¡5)

+ (¡2x¡3 + 3x¡4)(3x¡1 + 4x¡4)q0(x) = ¡3x¡4 ¡ 16x¡7 + 3x¡5 + 16x¡8 ¡ 6x¡4

¡ 8x¡7 + 9x¡5 + 12x¡8q0(x) = ¡9x¡4 + 12x¡5 ¡ 24x¡7 + 28x¡8

11. f(x) =6x+ 1

3x+ 10

f 0(x) =(3x+ 10)(6)¡ (6x+ 1)(3)

(3x+ 10)2

=18x+ 60¡ 18x¡ 3

(3x+ 10)2

=57

(3x+ 10)2

Section 4.2 Derivatives of Products and Quotients 249

12. f(x) =8x¡ 117x+ 3

f 0(x) =(7x+ 3)(8)¡ (8x¡ 11)(7)

(7x+ 3)2

=56x+ 24¡ 56x+ 77

(7x+ 3)2

=101

(7x+ 3)2

13. y =5¡ 3t4 + t

dy

dx=(4 + t)(¡3)¡ (5¡ 3t)(1)

(4 + t)2

=¡12¡ 3t¡ 5 + 3t

(4 + t)2

=¡17

(4 + t)2

14. y =9¡ 7t1¡ t

dy

dx=(1¡ t)(¡7)¡ (9¡ 7t)(¡1)

(1¡ t)2

=¡7 + 7t+ 9¡ 7t

(1¡ t)2

=2

(1¡ t)2

15. y =x2 + x

x¡ 1dy

dx=(x¡ 1)(2x+ 1)¡ (x2 + x)(1)

(x¡ 1)2

=2x2 + x¡ 2x¡ 1¡ x2 ¡ x

(x¡ 1)2

=x2 ¡ 2x¡ 1(x¡ 1)2

16. y =x2 ¡ 4xx+ 3

dy

dx=(x+ 3)(2x¡ 4)¡ (x2 ¡ 4x)(1)

(x+ 3)2

=2x2 + 6x¡ 4x¡ 12¡ x2 + 4x

(x+ 3)2

=x2 + 6x¡ 12(x+ 3)2

17. f(t) =4t2 + 11

t2 + 3

f 0(t) =(t2 + 3)(8t)¡ (4t2 + 11)(2t)

(t2 + 3)2

=8t3 + 24t¡ 8t3 ¡ 22t

(t2 + 3)2

=2t

(t2 + 3)2

18. y =¡x2 + 8x4x2 ¡ 5

dy

dx=(4x2 ¡ 5)(¡2x+ 8)¡ (¡x2 + 8x)(8x)

(4x2 ¡ 5)2=¡8x3 + 32x2 + 10x¡ 40 + 8x3 ¡ 64x2

(4x2 ¡ 5)2=¡32x2 + 10x¡ 40

(4x2 ¡ 5)2

19. g(x) =x2 ¡ 4x+ 2x2 + 3

g0(x) =(x2 + 3)(2x¡ 4)¡ (x2 ¡ 4x+ 2)(2x)

(x2 + 3)2

=2x3 ¡ 4x2 + 6x¡ 12¡ 2x3 + 8x2 ¡ 4x

(x2 + 3)2

=4x2 + 2x¡ 12(x2 + 3)2

20. k(x) =x2 + 7x¡ 2x2 ¡ 2

k0(x) =(x2 ¡ 2)(2x+ 7)¡ (x2 + 7x¡ 2)(2x)

(x2 ¡ 2)2

=2x3 + 7x2 ¡ 4x¡ 14¡ 2x3 ¡ 14x2 + 4x

(x2 ¡ 2)2

=¡7x2 ¡ 14(x2 ¡ 2)2

21. p(t) =

pt

t¡ 1

=t1=2

t¡ 1

p0(t) =(t¡ 1) ¡12 t¡1=2¢¡ t1=2(1)

(t¡ 1)2

=12 t1=2 ¡ 1

2 t¡1=2 ¡ t1=2

(t¡ 1)2

=¡12 t1=2 ¡ 1

2 t¡1=2

(t¡ 1)2

=¡pt2 ¡ 1

2pt

(t¡ 1)2 or¡t¡ 1

2pt(t¡ 1)2


22. r(t) =

pt

2t+ 3=

t1=2

2t+ 3

r0(t) =(2t+ 3)

¡12 t¡1=2¢¡ (t1=2)(2)

(2t+ 3)2=t1=2 + 3

2 t¡1=2 ¡ 2t1=2

(2t+ 3)2=¡t1=2 + 3

2t1=2

(2t+ 3)2=¡pt+ 3

2pt

(2t+ 3)2or

¡2t+ 32pt(2t+ 3)2

23. y =5x+ 6p

x=5x+ 6

x1=2

dy

dx=(x1=2)(5)¡ (5x+ 6) ¡12x¡1=2¢

(x1=2)2=5x1=2 ¡ 5

2x1=2 ¡ 3x¡1=2x

=52x

1=2 ¡ 3x¡1=2x

=

5px2 ¡ 3p

x

xor

5x¡ 62xpx

24. h(z) =z2:2

z3:2 + 5

h0(z) =(z3:2 + 5)(2:2z1:2)¡ z2:2(3:2z2:2)

(z3:2 + 5)2=2:2z4:4 + 11z1:2 ¡ 3:2z4:4

(z3:2 + 5)2=¡z4:4 + 11z1:2(z3:2 + 5)2

25. g(y) =y1:4 + 1

y2:5 + 2

g0(y) =(y2:5 + 2)(1:4y0:4)¡ (y1:4 + 1)(2:5y1:5)

(y2:5 + 2)2=1:4y2:9 + 2:8y0:4 ¡ 2:5y2:9 ¡ 2:5y1:5

(y2:5 + 2)2

=¡1:1y2:9 ¡ 2:5y1:5 + 2:8y0:4

(y2:5 + 2)2

26. f(x) =(3x2 + 1)(2x¡ 1)

5x+ 4

f 0(x) =(5x+ 4)[(3x2 + 1)(2) + (6x)(2x¡ 1)]¡ (3x2 + 1)(2x¡ 1)(5)

(5x+ 4)2

=(5x+ 4)(18x2 ¡ 6x+ 2)¡ (3x2 + 1)(10x¡ 5)

(5x+ 4)2

=90x3 ¡ 30x2 + 10x+ 72x2 ¡ 24x+ 8¡ 30x3 + 15x2 ¡ 10x+ 5

(5x+ 4)2

=60x3 + 57x2 ¡ 24x+ 13

(5x+ 4)2

27. g(x) =(2x2 + 3)(5x+ 2)

6x¡ 7g0(x) =

(6x¡ 7)[(2x2 + 3)(5) + (4x)(5x+ 2)]¡ (2x2 + 3)(5x+ 2)(6)(6x¡ 7)2

=(6x¡ 7)(30x2 + 8x+ 15)¡ (2x2 + 3)(30x+ 12)

(6x¡ 7)2

=180x3 + 48x2 + 90x¡ 210x2 ¡ 56x¡ 105¡ 60x3 ¡ 24x2 ¡ 90x¡ 36

(6x¡ 7)2

=120x3 ¡ 186x2 ¡ 56x¡ 141

(6x¡ 7)2

28. h(x) = f(x)g(x)

h0(x) = f(x)g0(x) + g(x)f 0(x)h0(3) = f(3)g0(3) + g(3)f 0(3) = 9(5) + 4(8) = 77


29. h(x) =f(x)

g(x)

h0(x) =g(x)f 0(x)¡ f(x)g0(x)

[g(x)]2

h0(3) =g(3)f 0(3)¡ f(3)g0(3)

[g(3)]2=4(8)¡ 9(5)

42= ¡13

16

30. In the …rst step, the two terms in the numerator are reversed. The correct work follows.

Dx

μ2x+ 5

x2 ¡ 1¶

=(x2 ¡ 1)(2)¡ (2x+ 5)(2x)

(x2 ¡ 1)2

=2x2 ¡ 2¡ 4x2 ¡ 10x

(x2 ¡ 1)2

=¡2x2 ¡ 10x¡ 2(x2 ¡ 1)2

31. In the …rst step, the denominator, (x3)2 = x6; was omitted. The correct work follows.

Dx

μx2 ¡ 4x3

¶=x3(2x)¡ (x2 ¡ 4)(3x2)

(x3)2=2x4 ¡ 3x4 + 12x2

x6=¡x4 + 12x2

x6=x2(¡x2 + 12)x2(x4)

=¡x2 + 12x4

32. f(x) =x

x¡ 2 ; at (3; 3)

m = f 0(x) =(x¡ 2)(1)¡ x(1)

(x¡ 2)2 = ¡ 2

(x¡ 2)2

At (3; 3);

m = ¡ 2

(3¡ 2)2 = ¡2:

Use the point-slope form.

y ¡ 3 = ¡2(x¡ 3)y = ¡2x+ 9

33. (a) f(x) =3x3 + 6

x2=3

f 0(x) =(x2=3)(9x2)¡ (3x3 + 6)(23x¡1=3)

(x2=3)2=9x8=3 ¡ 2x8=3 ¡ 4x¡1=3

x4=3=7x8=3 ¡ 4

x1=3

x4=3=7x3 ¡ 4x5=3

(b) f(x) = 3x7=3 + 6x¡2=3

f 0(x) = 3μ7

3x4=3

¶+ 6

μ¡23x¡5=3

¶= 7x4=3 ¡ 4x¡5=3

(c) The derivatives are equivalent.

34. f 0(x) = kg0(x) + g(x) ¢ 0 = kg0(x)The result is the same as applying the rule for di¤erentiating a constant times a function.


35. f(x) =u(x)

v(x)

f 0(x) = limh!0

f(x+ h)¡ f(x)h

= limh!0

u(x+h)v(x+h) ¡ u(x)

v(x)

h= limh!0

u(x+ h)v(x)¡ u(x)v(x+ h)hv(x+ h)v(x)

= limh!0

u(x+h)v(x)¡u(x)v(x)+u(x)v(x)¡u(x)v(x+ h)hv(x+ h)v(x)

= limh!0

v(x)[u(x+h)¡u(x)]¡u(x)[v(x+h)¡v(x)]hv(x+ h)v(x)

= limh!0

v(x)u(x+h)¡u(x)h ¡ u(x)v(x+h)¡v(x)h

v(x+ h)v(x)=v(x) ¢ u0(x)¡ u(x)v0(x)

[v(x)]2

36. Given f(x) = u(x)v(x) ; multiply both sides by v(x) to obtain

f(x)v(x) = u(x):

Using the product rule, we havef(x) ¢ v0(x) + v(x) ¢ f 0(x) = u0(x):

Substitute u(x)v(x) for f(x) and solve for

hu(x)v(x)

i0:

u(x)

v(x)¢ v0(x)+v(x)

·u(x)

v(x)

¸0= u0(x)

u(x) ¢ v0(x)+[v(x)]2·u(x)

v(x)

¸0= v(x) ¢ u0(x)

[v(x)]2·u(x)

v(x)

¸0= v(x) ¢ u0(x)¡u(x) ¢ v0(x)

·u(x)

v(x)

¸0=v(x) ¢ u0(x)¡u(x) ¢ v0(x)

[v(x)]2

37. Graph the numerical derivative of f(x) = (x2¡ 2)(x2¡p2) for x ranging from ¡2 to 2. The derivative crossesthe x-axis at 0 and at approximately ¡1:307 and 1.307.

38. Graph the numerical derivative of f(x) = x¡2x2+4 for x ranging from ¡5 to 10. The derivative crosses the x-axis

at approximately ¡0:828 and 4.828.

39. C(x) =3x+ 2

x+ 4

C(x) =C(x)

x=3x+ 2

x2 + 4x

(a) C(10) =3(10) + 2

102 + 4(10)=32

140¼ 0.2286 hundreds of dollars or $22:86 per unit

(b) C(20) =3(20) + 2

(20)2 + 4(20)=62

480¼ 0:1292 hundreds of dollars or $12:92 per unit

(c) C(x) =3x+ 2

x2 + 4xper unit

(d) C0(x) =

(x2 + 4x)(3)¡ (3x+ 2)(2x+ 4)(x2 + 4x)2

=3x2 + 12x¡ 6x2 ¡ 12x¡ 4x¡ 8

(x2 + 4x)2=¡3x2 ¡ 4x¡ 8(x2 + 4x)2


40. P (x) =5x¡ 62x+ 3

P (x) =P (x)

x=

5x¡ 62x2 + 3x

(a) P (8) =5(8)¡ 6

2(8)2 + 3(8)=34

152

¼ 0.224 tens of dollars or $2.24 per book

(b) P (15) =5(15)¡ 6

2(15)2 + 3(15)=69

495

¼ 0.139 tens of dollars or $1.39 per book

(c) P (x) =5x¡ 62x2 + 3x

per book

(d) P0(x) =

(2x2 + 3x)(5)¡ (5x¡ 6)(4x+ 3)(2x2 + 3x)2

=10x2 + 15x¡ 20x2 ¡ 15x+ 24x+ 18

(2x2 + 3x)2=¡10x2 + 24x+ 18(2x2 + 3x)2

41. M(d) =100d2

3d2 + 10

(a) M 0(d) =(3d2 + 10)(200d)¡ (100d2)(6d)

(3d2 + 10)2=600d3 + 2000d¡ 600d3

(3d2 + 10)2=

2000d

(3d2 + 10)2

(b) M 0(2) =2000(2)

[3(2)2 + 10]2=4000

484¼ 8:3

This means the new employee can assemble about 8.3 additional bicycles per day after 2 days of training.

M 0(5) =2000(5)

[3(5)2 + 10]2=10,0007225

¼ 1:4

This means the new employee can assemble about 1.4 additional bicycles per day after 5 days of training.

42. The revenue R(q) is given by R(q) = pq = D(q)q:Using the product rule, the derivative of R(q); which is the marginal revenue, is

R0(q) = D(q) ¢ 1 + qD0(q) = D(q) + qD0(q):

43. C(x) =C(x)

x

Let u(x) = C(x); with u0(x) = C0(x):Let v(x) = x with v0(x) = 1: Then, by the quotient rule,

C(x) =v(x) ¢ u0(x)¡ u(x) ¢ v0(x)

[v(x)]2=x ¢C0(x)¡C(x) ¢ 1

x2=xC0(x)¡C(x)

x2

44. Let d(t) be the demand as a function of time and p(t) be the price as a function of time.Then R(t) = d(t)p(t) is the revenue as a function of time. Let t = t1 represent the beginning of the year.

R0(t) = d(t)p0(t) + d0(t)p(t)R0(t1) = d(t1)p0(t1) + d0(t1)p(t1) = (500)(0:5) + (30)(15) = 250 + 450 = 700

Revenue is increasing $700 per month.


45. Let C(t) be the cost as a function of time and q(t)be the quantity as a function of time.Then C(t) = C(t)

q(t) is the revenue as a function oftime. Let t = t1 represent last month.

C0(t) =

q(t)C0(t)¡C(t)q0(t)[g(t)]2

C0(t1) =

q(t1)C0(t1)¡C(t1)q0(t1)[g(t1)]2

=(12,500)(1200)¡ (27,000)(350)

(12,500)2

= 0:03552

The average cost is increasing at a rate of $0.03552per gallon per month.

46. s(x) =x

m+ nx; m and n constants

(a) s0(x) =(m+ nx)1¡ x(n)

(m+ nx)2

=m+ nx¡ nx(m+ nx)2

=m

(m+ nx)2

(b) x = 50; m = 10; n = 3

s0(50) =m

(m+ 50n)2

=10

[10 + 50(3)]2

=1

2560

¼ 0:000391 mm per ml

47. f(x) =Kx

A+ x

(a) f 0(x) =(A+ x)K ¡Kx(1)

(A+ x)2

f 0(x) =AK

(A+ x)2

(b) f 0(A) =AK

(A+A)2

=AK

4A2

=K

4A

48. N(t) = 3t(t¡ 10)2 + 40

(a) N(t) = 3t(t2 ¡ 20t+ 100) + 40= 3t3 ¡ 60t2 + 300t+ 40

N 0(t) = 9t2 ¡ 120t+ 300

(b) N 0(8) = 9(8)2 ¡ 120(8) + 300= 9(64)¡ 960 + 300= ¡84

The rate of change is ¡84 million per hour.

(c) N 0(11) = 9(11)2 ¡ 120(11) + 300= 9(121)¡ 1320 + 300= 69

The rate of change is 69 million per hour.

(d) The population …rst declines, and then in-creases.

49. R(w) =30(w ¡ 4)w ¡ 1:5

(a) R(5) =30(5¡ 4)5¡ 1:5

¼ 8:57 min

(b) R(7) =30(7¡ 4)7¡ 1:5

¼ 16:36 min

(c) R0(w) =(w ¡ 1:5)(30)¡ 30(w ¡ 4)(1)

(w ¡ 1:5)2

=30w ¡ 45¡ 30w + 120

(w ¡ 1:5)2

=75

(w ¡ 1:5)2

R0(5) =75

(5¡ 1:5)2

¼ 6:12min2

kcal

R0(7) =75

(7¡ 1:5)2

¼ 2:48min2

kcal

Section 4.3 The Chain Rule 255

50. W =

μ1 +

20

H ¡ 0:93¶H = H +

20H

H ¡ 0:93

(a)dW

dH= 1+

(H ¡ 0:93)(20)¡ 20H(1)(H ¡ 0:93)2

=(H ¡ 0:93)2(H ¡ 0:93)2 +

20H ¡ 18:6¡ 20H(H ¡ 0:93)2

=H2 ¡ 1:86H + 0:8649

(H ¡ 0:93)2 +¡18:6

(H ¡ 0:93)2

=H2 ¡ 1:86H ¡ 17:7351

(H ¡ 0:93)2

(b)dW

dH= 0 when H2 ¡ 1:86H ¡ 17:7351 = 0

By the quadratic formula,

H =1:86§p1:862 + 4(17:7351)

2

¼ ¡3:38 or 5:24

Discarding the negative solution, W is minimizedat H = 5:24 m.

(c) Crows apply optimal foraging techniques.

51. f(t) =90t

99t¡ 90f 0(t) =

(99t¡ 90)(90)¡ (90t)(99)(99t¡ 90)2

=¡8100

(99t¡ 90)2

(a) f 0(1) =¡8100

(99¡ 90)2

=¡810092

=¡810081

= ¡100

(b) f 0(10) =¡8100

[99(10)¡ 90]2

=¡8100(900)2

=¡8100810; 000

= ¡ 1

100or ¡ 0:01

52. f(x) =x2

2(1¡ x)

f 0(x) =2(1¡ x)(2x)¡ x2(¡2)

[2(1¡ x)]2

=4x¡ 4x2 + 2x24(1¡ x)2

=4x¡ 2x24(1¡ x)2

=2x(2¡ x)4(1¡ x)2

=x(2¡ x)2(1¡ x)2

(a) f 0(0:1) =0:1(2¡ 0:1)2(1¡ 0:1)2 ¼ 0:1173

(b) f 0(0:6) =0:6(2¡ 0:6)2(1¡ 0:6)2 = 2:625

4.3 The Chain Rule

In Exercises 1-6, f(x) = 5x2 ¡ 2x and g(x) = 8x+ 3:

1. g(2) = 8(2) + 3 = 19

f [g(2)] = f [19]

= 5(19)2 ¡ 2(19)= 1805¡ 38 = 1767

2. g(¡5) = 8(¡5) + 3 = ¡37f [g(¡37)] = f [¡37]

= 5(¡37)2 ¡ 2(¡37)= 6845 + 74 = 6919

3. f(2) = 5(2)2 ¡ 2(2)= 20¡ 4 = 16

g[f(2)] = g[16]

= 8(16) + 3

= 128 + 3 = 131

4. f(¡5) = 5(¡5)2 ¡ 2(¡5)= 125 + 10 = 135

g[f(¡5)] = f [135]= 8(135) + 3

= 1080 + 3 = 1083

5. g(k) = 8k + 3

f [g(k)] = f [8k + 3]

= 5(8k + 3)2 ¡ 2(8k + 3)= 5(64k2 + 48k + 9)¡ 16k ¡ 6= 320k2 + 224k + 39


6. f(5z) = 5(5z)2 ¡ 2(5z)= 125z2 ¡ 10z

g[f(5z)] = g(125z2 ¡ 10z)= 8(125z2 ¡ 10z) + 3= 1000z2 ¡ 80z + 3

7. f(x) =x

8+ 7; g(x) = 6x¡ 1

f [g(x)] =6x¡ 18

+ 7

=6x¡ 18

+56

8

=6x+ 55

8

g[f(x)] = 6hx8+ 7i¡ 1

=6x

8+ 42¡ 1

=3x

4+ 41

=3x

4+164

4

=3x+ 164

4

8. f(x) = ¡8x+ 9; g(x) = x

5+ 4

f [g(x)] = ¡8hx5+ 4i+ 9

=¡8x5

¡ 32 + 9

=¡8x5

¡ 23

=¡8x¡ 115

5

g[f(x)] =¡8x+ 9

5+ 4

=¡8x+ 9

5+20

5

=¡8x+ 29

5

9. f(x) =1

x; g(x) = x2

f [g(x)] =1

x2

g[f(x)] =

μ1

x

¶2

=1

x2

10. f(x) =2

x4; g(x) = 2¡ x

f [g(x)] =2

(2¡ x)4

g[f(x)] = 2¡μ2

x4

¶= 2¡ 2

x4

11. f(x) =px+ 2; g(x) = 8x2 ¡ 6

f [g(x)] =p(8x2 ¡ 6) + 2

=p8x2 ¡ 4

g[f(x)] = 8(px+ 2)2 ¡ 6

= 8x+ 16¡ 6= 8x+ 10

12. f(x) = 9x2 ¡ 11x; g(x) = 2px+ 2

f [g(x)] = 9(2px+ 2)2 ¡ 11(2px+ 2)

= 9[4(x+ 2)]¡ 22px+ 2= 36(x+ 2)¡ 22px+ 2= 36x+ 72¡ 22px+ 2

g[f(x)] = 2p(9x2 ¡ 11x) + 2

= 2p9x2 ¡ 11x+ 2

13. f(x) =px+ 1; g(x) =

¡1x

f [g(x)] =

r¡1x+ 1

=

rx¡ 1x

g[f(x)] =¡1px+ 1

14. f(x) =8

x; g(x) =

p3¡ x

f [g(x)] =8p3¡ x

=8p3¡ x ¢

p3¡ xp3¡ x

=8p3¡ x3¡ x

g[f(x)] =

r3¡ 8

x=

r3x¡ 8x

=

p3x¡ 8px

¢pxpx

=

p3x2 ¡ 8xx


16. y = (3x2 ¡ 7)2=3

If f(x) = x2=3 and g(x) = 3x2 ¡ 7; then

y = f [g(x)] = (3x2 ¡ 7)2=3:

17. y = (5¡ x2)3=5

If f(x) = x3=5 and g(x) = 5¡ x2, then

y = f [g(x)] = (5¡ x2)3=5:

18. y =p9¡ 4x

If f(x) =px and g(x) = 9¡ 4x; then

y = f [g(x)] =p9¡ 4x:

19. y = ¡p13 + 7xIf f(x) = ¡px andg(x) = 13 + 7x;

then y = f [g(x)] = ¡p13 + 7x:

20. y = (x1=2 ¡ 3)2 + (x1=2 ¡ 3) + 5If f(x) = x2 + x+ 5 and

g(x) = x1=2 ¡ 3; theny = f [g(x)]

= (x1=2 ¡ 3)2 + (x1=2 ¡ 3) + 5:

21. y = (x2 + 5x)1=3 ¡ 2(x2 + 5x)2=3 + 7If f(x) = x1=3 ¡ 2x2=3 + 7 andg(x) = x2 + 5x;

then

y = f [g(x)] = (x2 + 5x)1=3

¡ 2(x2 + 5x)2=3 + 7:

22. y = (2x3 + 9x)5

Let f(x) = x5 and g(x) = 2x3 + 9x: Then(2x3 + 9x)5 = f [g(x)]:

dy

dx= f 0[g(x)] ¢ g0(x)

f 0(x) = 5x4

f 0[g(x)] = 5[g(x)]4

= 5(2x3 + 9x)4

g0(x) = 6x2 + 9

dy

dx= 5(2x3 + 9x)4(6x2 + 9)

23. y = (8x4 ¡ 5x2 + 1)4

Let f(x) = x4 and g(x) = 8x4 ¡ 5x2 + 1: Then

(8x4 ¡ 5x2 + 1)4 = f [g(x)]:

Use the alternate form of the chain rule.dy

dx= f 0[g(x)] ¢ g0(x)

f 0(x) = 4x3

f 0[g(x)] = 4[g(x)]3

= 4(8x4 ¡ 5x2 + 1)3g0(x) = 32x3 ¡ 10x

dy

dx= 4(8x4 ¡ 5x2 + 1)3(32x3 ¡ 10x)

24. f(x) = ¡7(3x4 + 2)¡4

Use the generalized power rule withu = 3x4 + 2; n = ¡4 and u0 = 12x3:

f 0(x) = ¡7[¡4(3x4 + 2)¡4¡1 ¢ 12x3]= ¡7[¡48x3(3x4 + 2)¡5]= 336x3(3x4 + 2)¡5

25. f(x) = ¡2(12x2 + 5)¡6

Use the generalized power rule withu = 12x2 + 5; n = ¡6; and u0 = 24x:

f 0(x) = ¡2[¡6(12x2 + 5)¡6¡1 ¢ 24x]= ¡2[¡144x(12x2 + 5)¡7]= 288x(12x2 + 5)¡7

26. s(t) = 12(2t4 + 5)3=2

Use generalized power rule with u = 2t4 + 5;n = 3

2 ; and u0 = 8t3:

s0(t) = 12·3

2(2t4 + 5)1=2 ¢ 8t3

¸

= 12[12t3(2t4 + 5)1=2]

= 144t3(2t4 + 5)1=2

27. s(t) = 45(3t3 ¡ 8)3=2

Use the generalized power rule withu = 3t3 ¡ 8; n = 3

2 ; and u0 = 9t2:

s0(t) = 45·3

2(3t3 ¡ 8)1=2 ¢ 9t2

¸

= 45

·27

2t2(3t3 ¡ 8)1=2

¸

=1215

2t2(3t3 ¡ 8)1=2


28. f(t) = 8p4t2 + 7

= 8(4t2 + 7)1=2

Use generalized power rule with u = 4t2 + 7;n = 1

2 ; and u0 = 8t:

f 0(t) = 8·1

2(4t2 + 7)¡1=2 ¢ 8t

¸

= 8[4t(4t2 + 7)¡1=2]= 32t(4t2 + 7)¡1=2

=32t

(4t2 + 7)1=2

=32tp4t2 + 7

29. g(t) = ¡3p7t3 ¡ 1= ¡3(7t3 ¡ 1)1=2

Use generalized power rule withu = 7t3 ¡ 1; n = 1

2 ; and u0 = 21t2:

g0(t) = ¡3·1

2(7t3 ¡ 1)¡1=2 ¢ 21t2

¸

= ¡3·21

2t2(7t3 ¡ 1)¡1=2

¸

=¡632t2 ¢ 1

(7t3 ¡ 1)1=2

=¡63t2

2p7t3 ¡ 1

30. r(t) = 4t(2t5 + 3)4

Use the product rule and the power rule.

r0(t) = 4t[4(2t5 + 3)3 ¢ 10t4]+ (2t5 + 3)4 ¢ 4

= 160t5(2t5 + 3)3 + 4(2t5 + 3)4

= 4(2t5 + 3)3[40t5 + (2t5 + 3)]

= 4(2t5 + 3)3(42t5 + 3)

31. m(t) = ¡6t(5t4 ¡ 1)4


m0(t) = ¡6t[4(5t4 ¡ 1)3 ¢ 20t3] + (5t4 ¡ 1)4(¡6)= ¡480t4(5t4 ¡ 1)3 ¡ 6(5t4 ¡ 1)4= ¡6(5t4 ¡ 1)3[80t4 + (5t4 ¡ 1)]= ¡6(5t4 ¡ 1)3(85t4 ¡ 1)

32. y = (x3 + 2)(x2 ¡ 1)4


dy

dx= (x3 + 2)[4(x2 ¡ 1)3 ¢ 2x]

+ (x2 ¡ 1)4(3x2)= 8x(x3 + 2)(x2 ¡ 1)3 + 3x2(x2 ¡ 1)4= x(x2 ¡ 1)3[8(x3 + 2) + 3x(x2 ¡ 1)]= x(x2 ¡ 1)3(8x3 + 16 + 3x3 ¡ 3x)= x(x2 ¡ 1)3(11x3 ¡ 3x+ 16)

33. y = (3x4 + 1)4(x3 + 4)


dy

dx= (3x4 + 1)4(3x2) + (x3 + 4)[4(3x4 + 1)3

¢ 12x3]= 3x2(3x4 + 1)4 + 48x3(x3 + 4)(3x4 + 1)3

= 3x2(3x4 + 1)3[3x4 + 1+ 16x(x3 + 4)]

= 3x2(3x4 + 1)3(3x4 + 1+ 16x4 + 64x)

= 3x2(3x4 + 1)3(19x4 + 64x+ 1)

34. p(z) = z(6z + 1)4=3


p0(z) = z ¢ 43(6z + 1)1=3 ¢ 6

+ 1 ¢ (6z + 1)4=3= 8z(6z + 1)1=3 + (6z + 1)4=3

= (6z + 1)1=3[8z + (6z + 1)1]

= (6z + 1)1=3(14z + 1)

35. q(y) = 4y2(y2 + 1)5=4


q0(y) = 4y2 ¢ 54(y2 + 1)1=4(2y) + 8y(y2 + 1)5=4

= 10y3(y2 + 1)1=4 + 8y(y2 + 1)5=4

= 2y(y2 + 1)1=4[5y2 + 4(y2 + 1)4=4]

= 2y(y2 + 1)1=4(9y2 + 4)


36. y =1

(3x2 ¡ 4)5 = (3x2 ¡ 4)¡5

dy

dx= ¡5(3x2 ¡ 4)¡6 ¢ 6x= ¡30x(3x2 ¡ 4)¡6

=¡30x

(3x2 ¡ 4)6

37. y =¡5

(2x3 + 1)2= ¡5(2x3 + 1)¡2

dy

dx= ¡5[¡2(2x3 + 1)¡3 ¢ 6x2]= ¡5[¡12x2(2x3 + 1)¡3]= 60x2(2x3 + 1)¡3

=60x2

(2x3 + 1)3

38. p(t) =(2t+ 3)3

4t2 ¡ 1p0(t) =

(4t2¡1)[3(2t+3)2 ¢ 2]¡(2t+3)3(8t)(4t2 ¡ 1)2

=6(4t2 ¡ 1)(2t+ 3)2 ¡ 8t(2t+ 3)3

(4t2 ¡ 1)2

=(2t+ 3)2[6(4t2 ¡ 1)¡ 8t(2t+ 3)]

(4t2 ¡ 1)2

=(2t+ 3)2[24t2 ¡ 6¡ 16t2 ¡ 24t]

(4t2 ¡ 1)2

=(2t+ 3)2[8t2 ¡ 24t¡ 6]

(4t2 ¡ 1)2

=2(2t+ 3)2(4t2 ¡ 12t¡ 3)

(4t2 ¡ 1)2

39. r(t) =(5t¡ 6)43t2 + 4

r0(t)

=(3t2 + 4)[4(5t¡ 6)3 ¢ 5]¡ (5t¡ 6)4(6t)

(3t2 + 4)2

=20(3t2 + 4)(5t¡ 6)3 ¡ 6t(5t¡ 6)4

(3t2 + 4)2

=2(5t¡ 6)3[10(3t2 + 4)¡ 3t(5t¡ 6)]

(3t2 + 4)2

=2(5t¡ 6)3(30t2 + 40¡ 15t2 + 18t)

(3t2 + 4)2

=2(5t¡ 6)3(15t2 + 18t+ 40)

(3t2 + 4)2

40. y =x2 + 4x

(3x3 + 2)4

dy

dx=(3x3+2)4(2x+4)¡(x2+4x)[4(3x3+2)3 ¢9x2]

[(3x3 + 2)4]2

=(3x3+2)4(2x+ 4)¡ 36x2(x2 + 4x)(3x3 + 2)3

(3x3 + 2)8

=2(3x3 + 2)3[(3x3 + 2)(x+ 2)¡ 18x2(x2 + 4x)]

(3x3 + 2)8

=2(3x4 + 6x3 + 2x+ 4¡ 18x4 ¡ 72x3)

(3x3 + 2)5

=¡30x4 ¡ 132x3 + 4x+ 8

(3x3 + 2)5

41. y =3x2 ¡ x(2x¡ 1)5

dy

dx=(2x¡ 1)5(6x¡ 1)¡ (3x2 ¡ x)[5(2x¡ 1)4 ¢ 2]

[(2x¡ 1)5]2

=(2x¡ 1)5(6x¡ 1)¡ 10(3x2 ¡ x)(2x¡ 1)4

(2x¡ 1)10

=(2x¡ 1)4[(2x¡ 1)(6x¡ 1)¡ 10(3x2 ¡ x)]

(2x¡ 1)10

=12x2 ¡ 2x¡ 6x+ 1¡ 30x2 + 10x

(2x¡ 1)6

=¡18x2 + 2x+ 1(2x¡ 1)6

42. Let f(x) = xn: Then y = f(g(x)) = [g(x)]n:Using the chain rule,

dy

dx= f 0[g(x)] ¢ g0(x):

Then, using the power rule, f 0(x) = nxn¡1:

dy

dx= n[g(x)]n¡1 ¢ g0(x):

43. (a) Dx(f [g(x)]) at x = 1

= f 0[g(1)] ¢ g0(1)

= f 0(2) ¢μ2

7

¶

= ¡7μ2

7

¶

= ¡2


(b) Dx(f [g(x)]) at x = 2

= f 0[g(2)] ¢ g0(2)

= f 0(3) ¢μ3

7

¶

= ¡8μ3

7

¶

= ¡247

44. (a) Dx (g[f(x)]) at x = 1= g0[f(1)] ¢ f 0(1)= g0(2) ¢ (¡6)=3

7(¡6) = ¡18

7

(b) Dx (g[f(x)]) at x = 2= g0[f(2)] ¢ f 0(2)= g0(4) ¢ (¡7)=5

7(¡7) = ¡5

45. f(x) =px2 + 16;x = 3

f(x) = (x2 + 16)1=2

f 0(x) =1

2(x2 + 16)¡1=2(2x)

f 0(x) =xp

x2 + 16

f 0(3) =3p

32 + 16=3

5

f(3) =p32 + 16 = 5

We use m = 35 and the point P (3; 5) in the point-

slope form.

y ¡ 5 = 3

5(x¡ 3)

y ¡ 5 = 3

5x¡ 9

5

y =3

5x+

16

5

46. f(x) = (x3 + 7)2=3;x = 1

f 0(x) =2

3(x3 + 7)¡1=3(3x2)

f 0(x) =2x2

(x3 + 7)1=3

f 0(1) =2

2= 1

f 0(1) = 82=3 = 4

We use m = 1 and the point P (1; 4):

y ¡ 4 = 1(x¡ 1)y = x+ 3

47. f(x) = x(x2 ¡ 4x+ 5)4;x = 2f 0(x) = x ¢ 4(x2 ¡ 4x+ 5)3 ¢ (2x¡ 4)

+ 1 ¢ (x2 ¡ 4x+ 5)4= (x2 ¡ 4x+ 5)3

¢ [4x(2x¡ 4) + (x2 ¡ 4x+ 5)]= (x2 ¡ 4x+ 5)3(9x2 ¡ 20x+ 5)

f 0(2) = (1)3(1) = 1f(2) = 2(1)4 = 2


y ¡ 2 = 1(x¡ 2)y ¡ 2 = x¡ 2

y = x

48. f(x) = x2px4 ¡ 12;x = 2

f(x) = x2(x4 ¡ 12)1=2

f 0(x) = x2 ¢ 12(x4 ¡ 12)¡1=2(4x3)

+ 2x(x4 ¡ 12)1=2

f 0(x) =2x5

(x4 ¡ 12)1=2 + 2x(x4 ¡ 12)1=2

f 0(2) =64

41=2+ 4(4)1=2

f 0(2) = 32 + 8 = 40f(2) = 4

p4 = 8


y ¡ 8 = 40(x¡ 2)y = 40x¡ 72

49. f(x) =px3 ¡ 6x2 + 9x+ 1

f(x) = (x3 ¡ 6x2 + 9x+ 1)1=2

f 0(x) =1

2(x3 ¡ 6x2 + 9x+ 1)¡1=2¢ (3x2 ¡ 12x+ 9)

f 0(x) =3(x2 ¡ 4x+ 3)

2px3 ¡ 6x2 + 9x+ 1

If the tangent line is horizontal, its slope is zeroand f 0(x) = 0:

3(x2 ¡ 4x+ 3)2px3 ¡ 6x2 + 9x+ 1 = 0

3(x2 ¡ 4x+ 3) = 03(x¡ 1)(x¡ 3) = 0

x = 1 or x = 3

The tangent line is horizontal at x = 1 and x = 3:


50. f(x) =x

(x2 + 4)4

f 0(x) =(x2 + 4)4 ¢ 1¡ x ¢ 4(x2 + 4)3(2x)

(x2 + 4)8

=(x2 + 4)3[(x2 + 4)¡ 8x2]

(x2 + 4)8=

4¡ 7x2(x2 + 4)5

If the tangent line is horizontal, its slope is zeroand f 0(x) = 0:

4¡ 7x2(x2 + 4)5

= 0

4¡ 7x2 = 0

x2 =4

7

x = § 2p7

The tangent line is horizontal at x = § 2p7:

53. D(p) =¡p2100

+ 500; p(c) = 2c¡ 10

The demand in terms of the cost is

D(c) = D[p(c)]

=¡(2c¡ 10)2

100+ 500

=¡4(c¡ 5)2

100+ 500

=¡c2 + 10c¡ 25

25+ 500

=¡c2 + 10c¡ 25 + 12,500

25

=¡c2 + 10c+ 12,475

25:

54. R(x) = 24(x2 + x)2=3

R0(x) = 24·2

3(x2 + x)¡1=3 ¢ (2x+ 1)

¸

= 16(x2 + x)¡1=3(2x+ 1)

(a) R0(100) = 16(1002 + 100)¡1=3(2 ¢ 100 + 1)¼ 148:78

The marginal revenue is $148.78.

(b) R0(200) = 16(2002 + 200)¡1=3(2 ¢ 200 + 1)¼ 187:29


(c) R0(300) = 16(3002 + 300)¡1=3(2 ¢ 300 + 1)¼ 214:34


(d) R(x) =R(x)

x=24(x2 + x)2=3

x

(e) R0(x)

=x ¢ 16(x2 + x)¡1=3(2x+ 1)¡24(x2 + x)2=3 ¢ 1

x2

=16x(x2 + x)¡1=3(2x+ 1)¡ 24(x2 + x)2=3

x2

=16x(2x+ 1)¡ 24(x2 + x)

x2(x2 + x)1=3

=32x2 + 16x¡ 24x2 ¡ 24x

x2(x2 + x)1=3

=8x2 ¡ 8x

x2(x2 + x)1=3

=8(x¡ 1)

x(x2 + x)1=3

55. A = 1500μ1 +

r

36,500

¶1825

dAdr is the rate of change of A with respect to r:

dA

dr= 1500(1825)

μ1 +

r

36,500

¶1824μ1

36,500

¶

= 75

μ1 +

r

36,500

¶1824

(a) For r = 6%;

dA

dr= 75

μ1 +

6

36,500

¶1824= $101:22:

(b) For r = 8%;

dA

dr= 75

μ1 +

8

36,500

¶1824= $111:86:

(c) For r = 9%;

dA

dr= 75

μ1 +

9

36,500

¶1824= $117:59:


56. q = D(p)

= 30

Ã5¡ pp

p2 + 1

!

= 150¡ 30ppp2 + 1

= 150¡ 30p

(p2 + 1)1=2

dq

dp= 0¡

·(p2+1)1=2Dp(30p)¡(30p)(Dp(p2+1)1=2)

[(p2 + 1)1=2]2

¸

= ¡"(p2+ 1)1=2(30)¡(30p) ¡12¢ (p2+1)¡1=2(2p)

(p2 + 1)

#

= ¡·(p2+ 1)1=2(30)¡(30p)(p)(p2+ 1)¡1=2

(p2 + 1)

¸

= ¡·(30)(p2+ 1)¡1=2([p2+1]¡p2)

(p2 + 1)

¸

=¡30(p2 + 1)¡1=2(1)

(p2 + 1)

= ¡ 30

(p2 + 1)3=2

57. V =60; 000

1 + 0:3t+ 0:1t2

The rate of change of the value is

V 0(t)

=(1 + 0:3t+ 0:1t2)(0)¡ 60; 000(0:3 + 0:2t)

(1 + 0:3t+ 0:1t2)2

=¡60; 000(0:3 + 0:2t)(1 + 0:3t+ 0:1t2)2

:

(a) 2 years after purchase, the rate of change inthe value is

V 0(2) =¡60; 000[0:3 + 0:2(2)][1 + 0:3(2) + 0:1(2)2]2

=¡60; 000(0:3 + 0:4)(1 + 0:6 + 0:4)2

=¡42; 000

4

= ¡$10; 500:(b) 4 years after purchase, the rate of change inthe value is

V 0(4) =¡60; 000[0:3 + 0:2(4)][1 + 0:3(4) + 0:1(4)2]2

=¡66; 00014:44

= ¡$4570:64:

58. C = 2000q + 3500q =

p15; 000¡ 1:5p

Solve for p.

q =p15,000¡ 1:5p

q2 = 15; 000¡ 1:5pq2 ¡ 15,000

¡1:5 = p

q2

¡1:5 +15,0001:5

= p

¡2q23

+ 10,000 = p

(a) R(q) = qp = qμ¡2q2

3+ 10,000

¶

=¡2q33

+ 10,000q

=¡2q3 + 30,000q

3

=30,000q ¡ 2q3

3

(b) P (q) = R(q)¡C(q)

=30,000q ¡ 2q3

3

¡ (2000q + 3500)

=

μ¡2q33

+ 10,000q¶

¡ (2000q + 3500)

=¡2q33

+ 8000q ¡ 3500

= 8000q ¡ 2q3

3¡ 3500

P (q) is the pro…t function.

(c)dP

dq= Dq (8000q ¡ 2q

3

3¡ 3500)

= 8000¡ 2q2

dP

dqor P 0(q) gives the marginal pro…t.

(d) If p = $5000 and

q =p15,000 ¡ 1:5p;

then q =p15,000 ¡ 1:5(5000)

=p15,000 ¡ 7500

=p7500:


Thus, q2 = 7500:

P 0(q) = 8000¡ 2q2= 8000¡ 2(7500)= 8000¡ 15,000= ¡7000

When the price is $5000, the marginal pro…t is¡$7000:

59. P (x) = 2x2 + 1; x = f(a) = 3a+ 2

P [f(a)] = 2(3a+ 2)2 + 1

= 2(9a2 + 12a+ 4) + 1

= 18a2 + 24a+ 9

60. (a) A(r) = ¼r2

r(t) = t2

A[r(t)] = ¼[t2]2

= ¼(t4)

= ¼t4

This function represents the area of the oil slickas a function of time t after the beginning of theleak.

(b) DtA[r(t)] = 4¼t3

DtA[r(100)] = 4¼(100)3

= 4,000,000¼

At 100 minutes the area of the spill is changing atthe rate of 4,000,000¼ ft2/min.

61. (a) r(t) = 2t; A(r) = ¼r2

A[r(t)] = ¼(2t)2

= 4¼t2

A = 4¼t2 gives the area of the pollution in termsof the time since the pollutants were …rst emitted.

(b) DtA[r(t)] = 8¼tDtA[r(4)] = 8¼(4) = 32¼

At 12 P.M., the area of pollution is changing atthe rate of 32¼ mi2/hr.

62. N(t) = 2t(5t+ 9)1=2 + 12

N 0(t) = (2t)·1

2(5t+ 9)¡1=2(5)

¸

+ 2(5t+ 9)1=2 + 0

= 5t(5t+ 9)¡1=2 + 2(5t+ 9)1=2

= (5t+ 9)¡1=2[5t+ 2(5t+ 9)]= (5t+ 9)¡1=2(15t+ 18)

=15t+ 18

(5t+ 9)1=2

(a) N 0(0) =15(0) + 18

[5(0) + 9]1=2

=18

91=2= 6

(b) N 0μ7

5

¶=

15¡75

¢+ 18£

5¡75

¢+ 9¤1=2

=21 + 8

(7 + 9)1=2

=39

(16)1=2

=39

4= 9:75

(c) N 0(8) =15(8) + 18

[5(8) + 9]1=2

=120 + 18

(49)1=2

=138

7¼ 19:71

63. C(t) =1

2(2t+ 1)¡1=2

C 0(t) =1

2

μ¡12

¶(2t+ 1)¡3=2(2)

= ¡12(2t+ 1)¡3=2

(a) C0(0) = ¡12[2(0) + 1]¡3=2

= ¡12

= ¡0:5

(b) C0(4) = ¡12[2(4) + 1]¡3=2

= ¡12(9)¡3=2

=¡12¢ 1

(p9)3

= ¡ 1

54

¼ ¡0:02

(c) C0(7:5) = ¡12[2(7:5) + 1]¡3=2

= ¡12(16)¡3=2

= ¡12

μ1

(p16)3

¶

= ¡ 1

128

¼ ¡0:008


(d) C is always decreasing because

C 0 = ¡12(2t+ 1)

¡3=2

is always negative for t ¸ 0:(The amount of calcium in the bloodstream willcontinue to decrease over time.)

64. (a) R(Q) = Q

μC ¡ Q

3

¶1=2

R0(Q) = Q

"1

2

μC ¡ Q

3

¶¡1=2μ¡13

¶#

+

μC ¡ Q

3

¶1=2(1)

= ¡16Q

μC ¡ Q

3

¶¡1=2+

μC ¡ Q

3

¶1=2

= ¡ Q

6³C ¡ Q

3

´1=2 +μC ¡ Q

3

¶1=2

(b) R0(Q) = ¡ Q

6³C ¡ Q

3

´1=2 +μC ¡ Q

3

¶1=2

If Q = 87 and C = 59; then

R0(Q) =μ59¡ 87

3

¶1=2¡ 87

6¡59¡ 87

3

¢1=2= (30)1=2 ¡ 87

6(30)1=2

= 5:48¡ 87

32:88

= 5:48¡ 2:65= 2:83:

(c) Because R0(Q) is positive, the patient’s sensi-tivity to the drug is increasing.

65. V (r) =4

3¼r3; S(r) = 4¼r2; r(t) = 6¡ 3

17t

(a) r(t) = 0 when 6¡ 3

17t = 0;

t =17(6)

3= 34 min.

(b)dV

dr= 4¼r2;

dS

dr= 8¼r;

dr

dt= ¡ 3

17

dV

dt=dV

dr¢ drdt= ¡12

17¼r2

= ¡1217¼

μ6¡ 3

17t

¶2dS

dt=dS

dr¢ drdt= ¡24

17¼r

= ¡2417¼

μ6¡ 3

17t

¶

When t = 17,

dV

dt= ¡12

17¼

·6¡ 3

17(17)

¸2

= ¡10817¼ mm3/min

dS

dt= ¡24

17¼

·6¡ 3

17(17)

¸

= ¡7217¼ mm2/min

At t = 17 minutes, the volume is decreasing by10817 ¼ mm

3 per minute and the surface area is de-creasing by 72

17¼ mm2 per minute.

4.4 Derivatives of ExponentialFunctions

1. y = e4x

Let g(x) = 4x;

with g0(x) = 4:

dy

dx= 4e4x

2. y = e¡2x

Let g(x) = ¡2x;with g0(x) = ¡2:

dy

dx= ¡2e¡2x

3. y = ¡8e3xdy

dx= ¡8(3e3x) = ¡24e3x

4. y = 1:2e5x

dy

dx= 1:2(5e5x) = 6e5x

Section 4.4 Derivatives of Exponential Functions 265

5. y = ¡16e2x+1g(x) = 2x+ 1

g0(x) = 2

dy

dx= ¡16(2e2x+1) = ¡32e2x+1

6. y = ¡4e¡0:3xdy

dx= ¡4(¡0:3e¡0:3x)= 1:2e¡0:3x

7. y = ex2

g(x) = x2

g0(x) = 2x

dy

dx= 2xex

2

8. y = e¡x2

g(x) = ¡x2g0(x) = ¡2xdy

dx= ¡2xe¡x2

9. y = 3e2x2

g(x) = 2x2

g0(x) = 4x

dy

dx= 3(4xe2x

2)

= 12xe2x2

10. y = ¡5e4x3g(x) = 4x3

g0(x) = 12x2

dy

dx= (¡5)(12x2)e4x3

= ¡60x2e4x3

11. y = 4e2x2¡4

g(x) = 2x2 ¡ 4g0(x) = 4x

dy

dx= 4[(4x)e2x

2¡4]

= 16xe2x2¡4

12. y = ¡3e3x2+5g(x) = 3x2 + 5

g0(x) = 6xy0 = (¡3)(6x)e3x2+5= ¡18xe3x2+5

13. y = xex

Use the product rule.

dy

dx= xex + ex ¢ 1= ex(x+ 1)

14. y = x2e¡2x


dy

dx= x2(¡2e¡2x) + 2xe¡2x= ¡2x2e¡2x + 2xe¡2x= 2x(1¡ x)e¡2x

15. y = (x+ 3)2e4x


dy

dx= (x+ 3)2(4)e4x + e4x ¢ 2(x+ 3)= 4(x+ 3)2e4x + 2(x+ 3)e4x

= 2(x+ 3)e4x[2(x+ 3) + 1]

= 2(x+ 3)(2x+ 7)e4x

16. y = (3x3 ¡ 4x)e¡5x


dy

dx= (3x3 ¡ 4x)(¡5e¡5x) + e¡5x(9x2 ¡ 4)= (¡15x3 + 20x)e¡5x + (9x2 ¡ 4)e¡5x= (¡15x3 + 9x2 + 20x¡ 4)e¡5x

17. y =x2

ex

Use the quotient rule.

dy

dx=ex(2x)¡ x2ex

(ex)2

=xex(2¡ x)

e2x

=x(2¡ x)ex

18. y =ex

2x+ 1

dy

dx=ex(2x+ 1)¡ (2)(ex)

(2x+ 1)2

=ex(2x+ 1¡ 2)(2x+ 1)2

=ex(2x¡ 1)(2x+ 1)2

19. y =ex + e¡x

x

dy

dx=x(ex ¡ e¡x)¡ (ex + e¡x)

x2


20. y =ex ¡ e¡x

x

dy

dx=x(ex ¡ (¡1)e¡x)¡ (ex ¡ e¡x)(1)

x2

=xex + xe¡x ¡ ex + e¡x

x2

=ex(x¡ 1) + e¡x(x+ 1)

x2

21. p =10,000

9 + 4e¡0:2t

dp

dt=(9 + 4e¡0:2t) ¢ 0¡10,000[0 + 4(¡0:2)e¡0:2t]

(9 + 4e¡0:2t)2

=8000e¡0:2t

(9 + 4e¡0:2t)2

22. p =500

12 + 5e¡0:5t

dp

dt=(12+5e¡0:5t) ¢ 0¡500[0+5(¡0:5)e¡0:5t]

(12 + 5e¡0:5t)2

=1250e¡0:5t

(12 + 5e¡0:5t)2

23. f(z) = (2z + e¡z2)2

f 0(z) = 2(2z + e¡z2

)1(2¡ 2ze¡z2)= 4(2z + e¡z

2

)(1¡ ze¡z2)24. y = 73x+1

Let g(x) = 3x+ 1; with g0(x) = 3: Then

dy

dx= (ln 7)(73x+1) ¢ 3= 3(ln 7)73x+1

25. y = 4¡5x+2

Let g(x) = ¡5x+ 2; with g0(x) = ¡5: Thendy

dx= (ln 4)(4¡5x+2) ¢ (¡5)= ¡5(ln 4)4¡5x+2

26. y = 3 ¢ 4x2+2

Let g(x) = x2 + 2; with g0(x) = 2x: Then

dy

dx= 3(ln 4)4x

2+2 ¢ 2x= 6x(ln 4)4x

2+2

27. y = ¡103x2¡4

Let g(x) = 3x2 ¡ 4; with g0(x) = 6x:dy

dx= ¡(ln 10)103x2¡4 ¢ 6x= ¡6x(103x2¡4) ln 10

28. s = 2 ¢ 3pt

Let g(t) =pt; with g0(t) =

1

2pt: Then

ds

dt= 2(ln 3)3

pt ¢ 1

2pt

=(ln 3)3

pt

pt

29. s = 5 ¢ 2pt¡2

Let g(t) = t¡ 2, with g0(t) = 12pt¡2 . Then

ds

dt= 5(ln 2)(2

pt¡2) ¢ 1

2pt¡ 2

=(5 ln 2)2

pt¡2

2pt¡ 2

30. y =tet + 2

e2t + 1

Use the quotient rule and product rule.

dy

dt=(e2t + 1)(tet + et ¢ 1)¡ (tet + 2)(2e2t)

(e2t + 1)2

=(e2t + 1)(tet + et)¡ (tet + 2)(2e2t)

(e2t + 1)2

=te3t + e3t + tet + et ¡ 2te3t ¡ 4e2t

(e2t + 1)2

=¡te3t + e3t + tet + et ¡ 4e2t

(e2t + 1)2

=(1¡ t)e3t ¡ 4e2t + (1 + t)et

(e2t + 1)2

31. y =t2e2t

t+ e3t

Use the quotient rule and product rule.

dy

dt=(t+ e3t)(2te2t + t2 ¢ 2e2t)¡ t2e2t(1 + 3e3t)

(t+ e3t)2

=(t+ e3t)(2te2t + 2t2e2t)¡ t2e2t(1 + 3e3t)

(t+ e3t)2

=(2t2e2t + 2t3e2t + 2te5t + 2t2e5t)¡ (t2e2t + 3t2e5t)

(t+ e3t)2

=t2e2t + 2t3e2t + 2te5t ¡ t2e5t

(t+ e3t)2

=(2t3 + t2)e2t + (2t¡ t2)e5t

(t+ e3t)2


32. f(x) = exp3x+2

Let g(x) = xp3x+ 2:

g0(x) = 1 ¢ p3x+ 2+ xμ

3

2p3x+ 2

¶

=p3x+ 2+

3x

2p3x+ 2

=2(3x+ 2)

2p3x+ 2

+3x

2p3x+ 2

=9x+ 4

2p3x+ 2

f 0(x) = exp3x+2 ¢

μ9x+ 4

2p3x+ 2

¶

33. f(x) = ex2=(x3+2)

Let g(x) =x2

x3 + 2

g0(x) =(x3 + 2)(2x)¡ x2(3x2)

(x3 + 2)2

=2x4 + 4x¡ 3x4(x3 + 2)2

=4x¡ x4(x3 + 2)2

=x(4¡ x3)(x3 + 2)2

f 0(x) = ex2=(x3+2) ¢

·x(4¡ x3)(x3 + 2)2

¸

=x(4¡ x3)ex2=(x3+2)

(x3 + 2)2

34. y = yoekt

dy

dt=d

dt[yoe

kt] = yokekt = k(yoe

kt) = ky

35. Graph

y =ex+0:0001 ¡ ex

0:0001on a graphing calculator. A good choice for theviewing window is [¡1; 4] by [¡1; 16] with Xscl =1, Yscl = 2.

If we graph y = ex on the same screen, we see thatthe two graphs coincide. They are close enough tobeing identical that they are indistinguishable.By the de…nition of the derivative, if f(x) = ex;

f 0(x) = limh!0

f(x+ h)¡ f(x)h

= limh!0

ex+h ¡ exh

;

and h = 0:0001 is very close to 0.Comparing the two graphs provides graphical ev-idence that

f 0(x) = ex:

36. Graph the function y = ex:

Sketch the lines tangent to the graph at x = ¡1;0, 1, 2.

Estimate the slopes of the tangent lines at thesepoints.At x = ¡1 the slope is a little steeper than 1

3 orapproximately 0.3̄.At x = 0 the slope is 1.At x = 1 the slope is a little steeper than 5

2 or 2.5.At x = 2 the slope is a little steeper than 713 or7.3̄.Note that e¡1 ¼ 0:36787944; e0 = 1;e1 = e ¼ 2:7182812; and e2 ¼ 7:3890561: Thevalues are close enough to the slopes of the tangentlines to convince us that de

x

dx = ex:


37. S(t) = 100¡ 90e¡0:3tS0(t) = ¡90(¡0:3)e¡0:3t

= 27e¡0:3t

(a) S0(1) = 27e¡0:3(1)

= 27e¡0:3

¼ 20(b) S0(5) = 27e¡0:3(5)

= 27e¡1:5

¼ 6(c) As time goes on, the rate of change of sales isdecreasing.

(d) S0(t) = 27e¡0:3t 6= 0, but

limt!1 S

0(t) = limt!1 27e¡0:3t = 0:

Although the rate of change of sales never equalszero, it gets closer and closer to zero as t increases.

38. C(x) =p900¡ 800 ¢ 1:1¡x

C(x) = [900¡ 800(1:1¡x)]1=2

=1

2[900¡ 800(1:1¡x)]¡1=2

¢ [¡800(ln 1:1)(1:1¡x)(¡1)]

C0(x) =(400 ln 1:1)(1:1¡x)p900¡ 800(1:1¡x)

(a) C 0(0) =400 ln 1:1p

100¼ 3:81

The marginal cost is $3.81.

(b) C0(20) =(400 ln 1:1)(1:1¡20)p900¡ 800(1:1¡20) ¼ 0:20

The marginal cost is $.20.

(c) As x becomes larger and larger, C0(x) ap-proaches zero.

39. A(t) = 10t2 2¡t

A0(t) = 10t2(ln 2)2¡t(¡1) + 20t 2¡tA0(t) = 10t 2¡t(¡t ln 2 + 2)(a) A0(2) = 10(2)(2¡2)(¡2 ln 2 + 2)

¼ 3:07(b) A0(4) = 10(4)(2¡4)(¡4 ln 2 + 2)

¼ ¡1:93(c) Public awareness increased at …rst and thendecreased.

40. f(t) = 0:028(3:824)t¡2001

f 0(t) = 0:028(ln 3:824)(3:824)t¡2001(1)= 0:028(ln 3:824)(3:824)t¡2001

(a) f 0(2002) = 0:028(ln 3:824)(3:824)2002¡2001

¼ 0:144The instantaneous rate of change in 2002 is 0.144million or 144,000 subscribers per year.

(b) f 0(2006) = 0:028(ln 3:824)(3:824)2006¡2001

¼ 30:7The instantaneous rate of change in 2006 is 30.7million or 30,700,000 subscribers per year.

41. y = 100e¡0:03045t

(a) For t = 0;

y = 100e¡0:03045(0)

= 100e0

= 100%:

(b) For t = 2;

y = 100e¡0:03045(2)

= 100e¡0:0609

¼ 94%:

(c) For t = 4;

y = 100e¡0:03045(4)

¼ 89%:

(d) For t = 6;

y0 = 100e¡0:03045(6)

¼ 83%:

(e) y0 = 100(¡0:03045)e¡0:03045t= ¡3:045e¡0:03045t

For t = 0;

y0 = ¡3:045e¡0:03045(0)= ¡3:045:

(f) For t = 2;

y0 = ¡3:045e¡0:03045(2)¼ ¡2:865:

(g) The percent of these cars on the road is de-creasing, but at a slower rate as they age.


42. S(t) = 5000e0:1(e0:25t)

Let g(t) = 0:1e0:25t; withg0(t) = 0:1(e0:25t)(0:25) = 0:025e0:25t

S0(t) = 5000e0:1(e0:25t)(0:025e0:25t)

= 125e0:1(e0:25t)e0:25t

= 125e0:1(e0:25t)+0:25t

S0(8) = 125e0:1(e0:25¢8)+0:25(8)

= 125e0:1(e2)+2 ¼ 1934

The answer is b.

43. (a) G0 = 0:7;m = 10:3

G(t) =10:3

1 +¡10:30:7 ¡ 1

¢e¡0:03036(10:3)t

=10:3

1 + 13:71e¡0:3127t

(b) G0(t) = ¡10:3(1 + 13:71e¡0:3127t)¡2¢ 13:71e¡0:3127t(¡0:3127)

=44:1573051e¡0:3127t

(1 + 13:71e¡0:3127t)2

1990 when t = 5:

G(5) =10:3

1 + 13:71e¡0:3127(5)

¼ 2:66

G0(5) =44:1573051e¡0:3127(5)

[1 + 13:71e¡0:3127(5)]2

¼ 0:617The population in 1990 is 2.66 million and thegrowth rate is 0.617 million per year.

(c) 1995 when t = 10:

G(10) =10:3

1 + 13:71e¡0:3127(10)

¼ 6:43G0(10) =

44:1573051e¡0:3127(10)

[1 + 13:71e¡0:3127(10)]2


(d) 2000 when t = 15:

G(15) =10:3

1 + 13:71e¡0:3127(15)

¼ 9:15

G0(15) =44:1573051e¡0:3127(15)

[1 + 13:71e¡0:3127(15)]2

¼ 0:320

The population in 2000 is 9.15 million and thegrowth rate is 0.320 million per year.

(e) The rate of growth over time increases for awhile and then gradually decreases to 0.

44. (a) G0 = 0:7;m = 13:7

G(t) =13:7

1 +¡13:70:7 ¡ 1

¢e¡0:02352(13:7)t

=13:7

1 + 18:57e¡0:3222t

(b) G0(t) = ¡13:7(1 + 18:57e¡0:3222t)¡2¢ 18:57e¡0:3222t(¡0:3222)

=81:9705798e¡0:3222t

(1 + 18:57e¡0:3222t)2

1990 when t = 5:

G(5) =13:7

1 + 18:57e¡0:3222(5)

¼ 2:91

G0(5) =81:9705798e¡0:3222(5)

[1 + 18:57e¡0:3222(5)]2


(c) 1995 when t = 10:

G(10) =13:7

1 + 18:57e¡0:3222(10)

¼ 7:87

G0(10) =81:9705798e¡0:3222(10)

[1 + 18:57e¡0:3222(10)]2


(d) 2000 when t = 15:

G(15) =13:7

1 + 18:57e¡0:3222(15)

¼ 11:94

G0(15) =81:9705798e¡0:3222(15)

[1 + 18:57e¡0:3222(15)]2


(e) The rate of growth over time increases for awhile and then gradually decreases to 0.


45. p(t) = 9:865(1:025)t

p0(t) = 9:865(ln 1:025)(1:025)t

(a) For 1998, t = 18:

p0(18) = 9:865(ln 1:025)(1:025)18

= 0:380

The instantaneous rate of growth is 380,000 peopleper year.

(b) For 2006, t = 26:

p0(26) = 9:865(ln 1:025)(1:025)26

= 0:463

The instantaneous rate of growth is 463,000 peopleper year.

46. G(t) =m G0

G0 + (m¡G0)e¡kmt ;

where G0 = 200;m = 10,000; and k = 0:00001:

(a)G(t) =10,000(200)

200+(10,000¡200)e¡(0:00001)(10;000)t

G(t) =10,000

1 + 49e¡0:1t

(b) G(t) = 10,000(1 + 49e¡0:1t)¡1

G0(t) = ¡10,000(1 + 49e¡0:1t)¡2(¡4:9e¡0:1t)

=49,000e¡0:1t

(1 + 49e¡0:1t)2

G(6) =10,000

1 + 49e¡0:6¼ 359

G0(6) =49,000e¡0:6

(1 + 49e¡0:6)2¼ 34:6

(c) After 3 years, t = 36:

G(36) =10,000

1 + 49e¡3:6¼ 4276

G0(36) =49,000e¡3:6

(1 + 49e¡3:6)2¼ 245

(d) After 7 years, t = 84:

G(84) =10,000

1 + 49e¡8:4¼ 9891

G0(84) =49,000e¡8:4

(1 + 49e¡8:4)2¼ 10:8

(e) It increases for a while and then gradually de-creases to 0.

47. G(t) =mGo

Go + (m¡Go)e¡kmt ; where Go = 400;m = 5200; and k = 0:0001:

(a) G(t) =(5200)(400)

400 + (5200¡ 400)e(¡0:0001)(5200)t

=(400)(5200)

400 + 4800e¡0:52t

=5200

1 + 12e¡0:52t

(b) G(t) = 5200(1 + 12e¡0:52t)¡1

G0(t) = ¡5200(1 + 12e¡0:52t)¡2(¡6:24e¡0:52t)

=32,448e¡0:52t

(1 + 12e¡0:52t)2

G(1) =5200

1 + 12e¡0:52¼ 639

G0(1) =32,448e¡0:52

(1 + 12e¡0:52)2¼ 292

(c) G(4) =5200

1 + 12e¡2:08¼ 2081

G0(4) =32,448e¡2:08

(1 + 12e¡2:08)2¼ 649

(d) G(10) =5200

1 + 12e¡5:2¼ 4877

G0(10) =34; 448e¡5:2

(1 + 12e¡5:2)2¼ 167

(e) It increases for a while and then gradually de-creases to 0.

48. P (x) = 0:04e¡4x

(a) P (0:5) = 0:04e¡4(0:5)

= 0:04e¡2

¼ 0:005(b) P (1) = 0:04e¡4(1)

= 0:04e¡4

¼ 0:0007(c) P (2) = 0:04e¡4(2)

= 0:04e¡8

¼ 0:000013P 0(x) = 0:04(¡4)e¡4x

= ¡0:16e¡4x

(d) P 0(0:5) = ¡0:16e¡4(0:5)= ¡0:16e¡2¼ ¡0:022

(e) P 0(1) = ¡0:16e¡4(1)= ¡0:16e¡4¼ ¡0:0029


(f) P 0(2) = ¡0:16e¡4(2)= ¡0:16e¡8¼ ¡0:000054

49. V (t) = 1100[1023e¡0:02415t + 1]¡4

(a) V (240) = 1100[1023e¡0:02415(240) + 1]¡4

¼ 3:857 cm3

(b) V =4

3¼r3, so r(V ) = 3

r3V

4¼

r(3:857) =3

r3(3:857)

4¼¼ 0:973 cm

(c) V (t) = 1100[1023e¡0:02415t + 1]¡4 = 0:5

[1023e¡0:02415t + 1]¡4 =1

2200

(1023e¡0:02415t + 1)4 = 2200

1023e¡0:02415t + 1 = 22001=4

1023e¡0:02415t = 22001=4 ¡ 1

e¡0:02415t =22001=4 ¡ 11023

¡0:02415t = lnμ22001=4 ¡ 11023

¶

t =1

¡0:02415 lnμ22001=4 ¡ 11023

¶¼ 214 months

The tumor has been growing for almost 18 years.

(d) As t goes to in…nity, e¡0:02415t goes to zero,and V (t) = 1100[1023e¡0:02415t+1]¡4 goes to 1100cm3, which corresponds to a sphere with a radius

of 3

q3(1100)4¼ ¼ 6:4 cm. It makes sense that a tu-

mor growing in a person’s body reaches a maxi-mum volume of this size.

(e) By the chain rule,

dV

dt= 1100(¡4)[1023e¡0:02415t + 1]¡5

¢ (1023)(e¡0:02415t)(¡0:02415)= 108,703.98[1023e¡0:02415t + 1]¡5e¡0:02415t

At t = 240,dV

dt¼ 0:282.

At 240 months old, the tumor is increasing in vol-ume at the instantaneous rate of 0.282 cm3/month.

50. P (t) = 0:00239e0:0957t

(a) P (25) = 0:00239e0:0957(25)

¼ 0:026%P (50) = 0:00239e0:0957(50)

¼ 0:286%P (75) = 0:00239e0:0957(75)

¼ 3:130%(b) P 0(t) = 0:00239e0:0957t(0:0957)

= 0:000228723e0:0957t

P 0(25) = 0:000228723e0:0957(25)

¼ 0:0025%=yearP 0(50) = 0:000228723e0:0957(50)

¼ 0:0274%=yearP 0(75) = 0:000228723e0:0957(75)

¼ 0:300%=year

51. URR=1¡½(0:96)0:14t¡1+

8t

126t+900[1¡(0:96)0:14t¡1]

¾(a)When t = 180, URR ¼ 0:589. The patient hasnot received adequate dialysis.

(b) When t = 240, URR ¼ 0:690. The patienthas received adequate dialysis.

(c) DtURR= ¡©(ln 0:96)(0:96)0:14t¡1(0:14)+

8t

126t+ 900(¡ ln 0:96)(0:96)0:14t¡1(0:14)

+(126t+ 900)(8)¡ 8t(126)

(126t+ 900)2[1¡ (0:96)0:14t¡1]

¾When t = 240;DtURR ¼ 0:001. The URR is in-creasing instantaneously by 0.001 units per minutewhen t = 240 minutes.The rate of increase is low, and it will take a signif-icant increase in time on dialysis to increase URRsigni…cantly.

52. p(x) = 0:001131e0:1268x

(a) p(25) = 0:001131e0:1268(25) ¼ 0:027(b) When p(x) = 1,

0:001131e0:1268x = 1

e0:1268x =1

0:001131

0:1268x = ln1

0:001131

x =1

0:1268ln

1

0:001131

¼ 54This represents the year 2024.


(c) p0(x) = 0:001131e0:1268x(0:1268)= 0:0001434108e0:1268x

p0(32) = 0:0001434108e0:1268(32)

¼ 0:008The marginal increase in the proportion per yearin 2002 is approximately 0.008.

53. M(t) = 3102e¡e¡0:022(t¡56)

(a)M(200) = 3102e¡e¡0:022(200¡56) ¼ 2974:15 grams,

or about 3 kilograms.

(b) As t gets very large, ¡e¡0:022(t¡56) goes tozero, e¡e

¡0:022(t¡56)goes to 1, andM(t) approaches

3102 grams or about 3.1 kilograms.

(c) 80% of 3102 is 2481.6.

2481:6 = 3102e¡e¡0:022(t¡56)

¡ ln 2481:63102

= e¡0:022(t¡56)

ln

μln3102

2481:6

¶= ¡0:022(t¡ 56)

t = ¡ 1

0:022ln

μln3102

2481:6

¶+ 56

¼ 124 days

(d) DtM(t) = 3102e¡e¡0:022(t¡56)

Dt(¡e¡0:022(t¡56))= 3102e¡e

¡:0022(t¡56)(¡e¡0:022(t¡56))(¡0:022)

= 68:244e¡e¡0:022(t¡56)

e¡0:022(t¡56)

When t = 200, DtM(t) ¼ 2:75 g/day.(e)

Growth is initially rapid, then tapers o¤.

(f)Day Weight Rate

50 991 24.88100 2122 17.73150 2734 7.60200 2974 2.75250 3059 0.94300 3088 0.32

54. L(t) = 589[1¡ e¡0:168(t+2:682)]

(a) Over time, the length of the average cutlass-…sh approaches 589 mm assymptotically.

(b) (0:95)589 = 589[1¡ e¡0:168(t+2:682)]0:95 = 1¡ e¡0:168(t+2:682)

ln(0:05) = ¡0:168(t+ 2:682)t = 15:1497 : : : ¼ 15 years old

(c) L0(t) = 589[¡e¡0:168(t+2:682)](¡0:168)= 98:952e¡0:168(t+2:682)

L0(4) = 98:952e¡0:168(4+2:682)

¼ 32:2 mm/yearWhen a cutlass…sh is 4 years old, it is growing inlength at a rate of 32.2 millimeters per year.

(d)

55. W1(t) = 509:7(1¡ 0:941e¡0:00181t)W2(t) = 498:4(1¡ 0:889e¡0:00219t)1:25

(a) Both W1 and W2 are strictly increasing func-tions, so they approach their maximum values ast approaches 1.

limt!1W1(t) = lim

t!1 509:7(1¡ 0:941e¡0:00181t)

= 509:7(1¡ 0) = 509:7limt!1W2(t) = lim

t!1 498:4(1¡ 0:889e¡0:00219t)1:25

= 498:4(1¡ 0)1:25 = 498:4So, the maximum values of W1 and W2 are 509.7kg and 498.4 kg respectively.

(b) 0:9(509:7) = 509:7(1¡ 0:941e¡0:00181t)0:9 = 1¡ 0:941e¡0:00181t0:1

0:941= e¡0:00181t

1239 ¼ t0:9(498:4) = 498:4(1¡ 0:889e¡0:00219t)1:25

0:9 = (1¡ 0:889e¡0:00219t)1:251¡ 0:90:80:889

= e¡0:00219t

1095 ¼ t


Respectively, it will take the average beef cowabout 1239 days or 1095 days to reach 90% ofits maximum.

(c) W 01(t) = (509:7)(¡0:941)(¡0:00181)e¡0:00181t

¼ 0:868126e¡0:00181tW 01(750) ¼ 0:868126e¡0:00181(750)

¼ 0:22 kg/dayW 02(t) = (498:4)(1:25)(1¡ 0:889e¡0:00219t)0:25

¢ (¡0:889)(¡0:00219)e¡0:00219t¼ 1:21292e¡0:00219t(1¡ 0:889e¡0:00219t)0:25

W 02(750) ¼ 1:12192e¡0:00219(750)

¢ (1¡ 0:889e¡0:00219(750))0:25¼ 0:22 kg/day

Both functions yield a rate of change of about 0.22kg per day.

(d) Looking at the graph, the growth patternsof the two functions are very similar.

(e) The graphs of the rates of change of thetwo functions are also very similar.

56. R(c) = 3:19(1:006c)

dR

dt= 3:19(ln 1:006)(1:006c)

dc

dt

If c = 180 and dcdt = 15; then

dR

dt= 3:19(ln 1:006)(1:006180)(15)

¼ 0:840

57. (a) G0 = 0:00369;m = 1; k = 3:5

G(t) =1

1 +¡

10:00369 ¡ 1

¢e¡3:5(1)t

=1

1 + 270e¡3:5t

(b) G0(t) = ¡(1 + 270e¡3:5t)¡2 ¢ 270e¡3:5t(¡3:5)

=945e¡3:5t

(1 + 270e¡3:5t)2

G(1) =1

1 + 270e¡3:5(1)

¼ 0:109

G0(1) =945e¡3:5(1)

[1 + 270e¡3:5(1)]2

¼ 0:341The proportion is 0.109 and the rate of growth is0.341 per century.

(c) G(2) =1

1 + 270e¡3:5(2)

¼ 0:802

G0(2) =945e¡3:5(2)

[1 + 270e¡3:5(2)]2

¼ 0:555The proportion is 0.802 and the rate of growth is0.555 per century.

(d) G(3) =1

1 + 270e¡3:5(3)

¼ 0:993

G0(3) =945e¡3:5(2)

[1 + 270e¡3:5(2)]2¼ 0:0256

The proportion is 0.993 and the rate of growth is0.0256 per century.

(e) The rate of growth increases for a while andthen gradually decreases to 0.

58. H(N) = 1000(1¡ e¡kN); k = 0:1H 0 = ¡1000e¡0:1N(¡0:1)= 100e¡0:1N

(a) H 0(10) = 100e¡0:1(10)

¼ 36:8(b) H0(100) = 100e¡0:1(100)

¼ :00454(c) H 0(1000) = 100e¡0:1(1000)

¼ 0(d) 100e¡0:1N is always positive since powers ofe are never negative. This means that repetitionalways makes a habit stronger.


59. P (t) = 37:79(1:012)t

(a) P (10) = 37:79(1:021)10 ¼ 46:5So, the U.S. Latino-American population in 2010was approximately 46,500,000.

(b) P 0(t) = 37:79(ln 1:021)(1:021)t

¼ 0:7854(1:021)tP 0(10) = 0:7854(1:021)10

¼ 0:967The Latino-American population was increasingat the rate of 0.967 million/year at the end of theyear 2010.

60. A(t) = 500e¡0:31t

A0(t) = 500(¡0:31)e¡0:31t= ¡155e¡0:31t

(a) A0(4) = ¡155e¡0:31(4)= ¡155e¡1:24¼ ¡44:9 grams per year

(b) A0(6) = ¡155e¡0:31(6)= ¡155e¡1:86¼ ¡24:1 grams per year

(c) A0(10) = ¡155e¡0:31(10)= ¡155e¡3:1¼ ¡6:98 grams per year

(d) limt!1 A0(t) = lim

t!1 ¡155e¡0:31t = 0

As the number of years increases, the rate of changeapproaches zero.

(e) A0(t) will never equal zero, although as t in-creases, it approaches zero. Powers of e are alwayspositive.

61. Q(t) = CV (1¡ e¡t=RC)

(a) Ic =dQ

dt= CV

·0¡ e¡t=RC

μ¡ 1

RC

¶¸

= CV

μ1

RC

¶e¡t=RC

=V

Re¡t=RC

(b) When C = 10¡5 farads, R = 107 ohms,and V = 10 volts, after 200 seconds

Ic =10107 e

¡200=(107¢10¡5) ¼ 1:35£ 10¡7 amps

62. t(r) = 218 + 31(0:933)n

(a) t(55) = 218 + 31(0:933)55

¼ 218:7 sec(b) t0(n) = (31 ln 0:933)(0:933)n

t0(55) = (31 ln 0:933)(0:933)55

¼ ¡0:047The record is decreasing by 0.047 seconds per yearat the end of 2005.

(c) As n!1, (0:933)n ! 0 andt(n) ! 218. If the estimate is correct, then thisis the least amount of time that it will ever takea human to run a mile. At the end of 2005, theworld record stood at 3:43.13, or 223.13 seconds.

63. T (h) = 80e¡0:000065h

dT

dt= 80e¡0:000065h

μ¡0:000065dh

dt

¶

= ¡0:0052e¡0:000065hdhdt

If h = 1000 anddh

dt= 800; then

dT

dt= ¡0:0052e¡0:000065(1000)(800)¼ ¡3:90

The temperature is decreasing at 3.90 degrees/hr.

4.5 Derivatives of LogarithmicFunctions

1. y = ln (8x)

dy

dx=d

dx(ln 8x)

=d

dx(ln 8 + ln x)

=d

dx(ln 8) +

d

dx(ln x)

= 0 +1

x

=1

x

2. y = ln (¡4x)dy

dx=d

dx[ln (¡4x)] = d

dx[ln 4 + ln(¡x)]

=d

dxln 4 +

d

dxln(¡x) = 0 + ¡1

¡x =1

x

Section 4.5 Derivatives of Logarithmic Functions 275

3. y = ln (8¡ 3x)g(x) = 8¡ 3xg0(x) = ¡3dy

dx=g0(x)g(x)

=¡3

8¡ 3x or3

3x¡ 8

4. y = ln (1 + x3)

g(x) = 1 + x3

g0(x) = 3x2

dy

dx=g0(x)g(x)

=3x2

1 + x3

5. y = ln¯̄4x2 ¡ 9x¯̄

g(x) = 4x2 ¡ 9xg0(x) = 8x¡ 9dy

dx=g0(x)g(x)

=8x¡ 94x2 ¡ 9x

6. y = ln¯̄¡8x3 + 2x¯̄

g(x) = ¡8x3 + 2xg0(x) = ¡24x2 + 2dy

dx=¡24x2 + 2¡8x3 + 2x

7. y = lnpx+ 5

g(x) =px+ 5

= (x+ 5)1=2

g0(x) =1

2(x+ 5)¡1=2

dy

dx=

12(x+ 5)

¡1=2

(x+ 5)1=2

=1

2(x+ 5)

8. y = lnp2x+ 1 = ln (2x+ 1)1=2

g(x) = (2x+ 1)1=2

g0(x) =1

2(2x+ 1)¡1=2(2) = (2x+ 1)¡1=2

dy

dx=(2x+ 1)¡1=2

(2x+ 1)1=2=

1

2x+ 1

9. y = ln (x4 + 5x2)3=2

=3

2ln (x4 + 5x2)

dy

dx=3

2Dx [ln (x

4 + 5x2)]

g(x) = x4 + 5x2

g0(x) = 4x3 + 10x

dy

dx=3

2

μ4x3 + 10x

x4 + 5x2

¶

=3

2

·2x(2x2 + 5)

x2(x2 + 5)

¸

=3(2x2 + 5)

x(x2 + 5)

10. y = ln (5x3 ¡ 2x)3=2

=3

2ln (5x3 ¡ 2x)

dy

dx=3

2Dx ln (5x3 ¡ 2x)

g(x) = 5x3 ¡ 2xg0(x) = 15x2 ¡ 2dy

dx=3

2

μ15x2 ¡ 25x3 ¡ 2x

¶

=3(15x2 ¡ 2)2(5x3 ¡ 2x)

11. y = ¡5x ln(3x+ 2)


dy

dx= ¡5x

·d

dxln(3x+ 2)

¸

+ ln(3x+ 2)

·d

dx(¡5x)

¸

= ¡5xμ

3

3x+ 2

¶+ [ln(3x+ 2)](¡5)

= ¡ 15x

3x+ 2¡ 5 ln(3x+ 2)

12. y = (3x+ 7) ln (2x¡ 1)


dy

dx= (3x+ 7)

μ2

2x¡ 1¶+ [ln(2x¡ 1)](3)

=2(3x+ 7)

2x¡ 1 + 3 ln (2x¡ 1)


13. s = t2 ln jtjds

dt= t2 ¢ 1

t+ 2t ln jtj

= t+ 2t ln jtj= t(1 + 2 ln jtj)

14. y = x ln¯̄2¡ x2¯̄


dy

dx= x

μ1

2¡ x2¶(¡2x) + ln ¯̄2¡ x2¯̄

= ¡ 2x2

2¡ x2 + ln¯̄2¡ x2¯̄

15. y =2 ln (x+ 3)

x2


dy

dx=x2³

2x+3

´¡ 2 ln (x+ 3) ¢ 2x(x2)2

=2x2

x+3 ¡ 4x ln (x+ 3)x4

=2x2 ¡ 4x(x+ 3) ln (x+ 3)

x4(x+ 3)

=x[2x¡ 4(x+ 3) ln (x+ 3)]

x4(x+ 3)

=2x¡ 4(x+ 3) ln (x+ 3)

x3(x+ 3)

16. v =ln u

u3


dv

du=u3¡1u

¢¡ (ln u)(3u2)(u3)2

=u2 ¡ 3u2 ln u

u6

=u2(1¡ 3 ln u)

u6

=1¡ 3 ln u

u4

17. y =ln x

4x+ 7


dy

dx=(4x+ 7)

¡1x

¢¡ (ln x)(4)(4x+ 7)2

=4x+7x ¡ 4 ln x(4x+ 7)2

=4x+ 7¡ 4x ln xx(4x+ 7)2

18. y =¡2 ln x3x¡ 1


dy

dx=(3x¡ 1)(¡2) ¡ 1x¢¡ (¡2 ln x)(3)

(3x¡ 1)2

=¡2(3x¡1)

x + 6 ln x

(3x¡ 1)2

=¡2(3x¡ 1) + 6x ln x

x(3x¡ 1)2

=¡2(3x¡ 1¡ 3x ln x)

x(3x¡ 1)2

19. y =3x2

ln x

dy

dx=(ln x)(6x)¡ 3x2 ¡ 1x¢

(ln x)2

=6x ln x¡ 3x(ln x)2

20. y =x3 ¡ 12 ln x

dy

dx=(2 ln x)(3x2)¡ (x3 ¡ 1)(2) ¡ 1x¢

(2 ln x)2

=6x2 ln x¡ 2

x(x3 ¡ 1)

(2 ln x)2¢ xx

=6x3 ln x¡ 2(x3 ¡ 1)

4x (ln x)2

=3x3 ln x¡ (x3 ¡ 1)

2x (ln x)2

21. y = (ln jx+ 1j)4dy

dx= 4(ln jx+ 1j3

μ1

x+ 1

¶

=4(ln jx+ 1j)3

x+ 1

22. y =pln jx¡ 3j = (ln jx¡ 3j)1=2

dy

dx=1

2(ln jx¡ 3j)¡1=2 ¢ d

dx(ln jx¡ 3j)

=1

2(ln jx¡ 3j)1=2μ

1

x¡ 3¶

=1

2(x¡ 3)(ln jx¡ 3j)1=2

=1

2(x¡ 3)pln jx¡ 3j


23. y = ln jln xjg(x) = ln x

g0(x) =1

x

dy

dx=g0(x)g(x)

=1x

ln x

=1

x ln x

24. y = (ln 4)(ln j3xj)dy

dx= (ln 4)

μ1

3x

¶(3)

=3 ln 4

3x

=ln 4

x

(Recall that ln 4 is a constant.)

25. y = ex2

ln x; x > 0

dy

dx= ex

2

μ1

x

¶+ (ln x)(2x)ex

2

=ex

2

x+ 2xex

2

ln x

26. y = e2x¡1 ln (2x¡ 1)dy

dx= (2)e2x¡1 ln (2x¡ 1)

+ (e2x¡1)μ

1

2x¡ 1¶(2)

= 2e2x¡1 ln (2x¡ 1) + 2e2x¡1

2x¡ 1

27. y =ex

ln x; x > 0


dy

dx=(ln x)ex ¡ ex ¡ 1x¢

(ln x)2¢ xx

=xex ln x¡ exx (ln x)2

28. p(y) =ln y

ey

p0(y) =ey ¢ 1y ¡ (ln y)ey

(ey)2

=ey ¡ y(ln y)ey

ye2y

=ey(1¡ y ln y)

ye2y

=1¡ y ln y

yey

29. g(z) = (e2z + ln z)3

g0(z) = 3(e2z + ln z)2μe2z ¢ 2 + 1

z

¶

= 3(e2z + ln z)2μ2ze2z + 1

z

¶

30. y = log (6x)

g(x) = 6x and g0(x) = 6:

dy

dx=

1

ln 10

μ6

6x

¶

=1

x ln 10

31. y = log(4x¡ 3)g(x) = 4x¡ 3g0(x) = 4

dy

dx=

1

ln 10¢ 4

4x¡ 3=

4

(ln 10)(4x¡ 3)32. y = log j1¡ xj

g(x) = 1¡ x and g0(x) = ¡1:dy

dx=

1

ln 10¢ ¡11¡ x

= ¡ 1

(ln 10)(1¡ x)or

1

(ln 10)(x¡ 1)33. y = log j3xj

g(x) = 3x and g0(x) = 3:

dy

dx=

1

ln 10¢ 33x

=1

x ln 10


34. y = log5p5x+ 2

g(x) =p5x+ 2 and g0(x) =

5

2p5x+ 2

:

dy

dx=

1

ln 5¢

52p5x+2p5x+ 2

=5

2 ln 5 (5x+ 2)

35. y = log7p4x¡ 3

g(x) =p4x¡ 3

g0(x) =4

2p4x¡ 3 =

2p4x¡ 3

dy

dx=

1

ln 7¢

2p4x¡3p4x¡ 3

=2

(ln 7)(4x¡ 3)

36. y = log3 (x2 + 2x)3=2

g(x) = (x2 + 2x)3=2 and

g0(x) =3

2(x2 + 2x)1=2 ¢ (2x+ 2)

= 3(x+ 1)(x2 + 2x)1=2

dy

dx=

1

ln 3¢ 3(x+ 1)(x

2 + 2x)1=2

(x2 + 2x)3=2

=3(x+ 1)

(ln 3)(x2 + 2x)

37. y = log2 (2x2 ¡ x)5=2

g(x) = (2x2 ¡ x)5=2 and

g0(x) =5

2(2x2 ¡ x)3=2 ¢ (4x¡ 1):

dy

dx=

1

ln 2¢52(2x

2 ¡ x)3=2 ¢ (4x¡ 1)(2x2 ¡ x)5=2

=5(4x¡ 1)

(2 ln 2)(2x2 ¡ x)

38. w = log8 (2p ¡ 1)g(p) = 2p ¡ 1g0(x) = (ln 2)2pdw

dp=

1

ln 8¢ (ln 2) 2

p

2p ¡ 1

=(ln 2)2p

(ln 8)(2p ¡ 1)39. z = 10y log y

g(y) = 10y and g0(y) = (ln 10)10y:

dz

dy= 10y ¢ 1

(ln 10)y+ log y ¢ (ln 10)10y

=10y

(ln 10)y+ (log y)(ln 10)10y

40. f(x) = epx ln(

px+ 5)


f 0(x)= epx

Ã1

2pxp

x+ 5

!+ [ln(

px+ 5)]e

px

μ1

2px

¶

=epx

2

·1p

x(px+ 5)

+ln(px+ 5)px

¸

41. f(x) = ln(xepx + 2)

g(x) = xepx + 2

g0(x) = x·epx

μ1

2px

¶¸+ e

px(1)

=epxpx

2+ e

px

=epx

2(px+ 2)

f 0(x) =g0(x)g(x)

=epx

2 (px+ 2)

xepx + 2

=epx(px+ 2)

2(xepx + 2)


42. f(t) =ln(t2 + 1) + t

ln(t2 + 1) + 1


u(t) = ln(t2 + 1) + t; u0(t) =2t

t2 + 1+ 1

v(t) = ln(t2 + 1) + 1; v0(t) =2t

t2 + 1

f 0(t) =[ln(t2 + 1) + 1]

³2tt2+1 + 1

´¡ [ln(t2 + 1) + t]

³2tt2+1

´[ln(t2 + 1) + 1]2

=

2t ln(t2+1)t2+1 + 2t

t2+1 + ln(t2 + 1) + 1¡ 2t ln(t2+1)

t2+1 ¡ 2t2

t2+1

[ln(t2 + 1) + 1]2

=2t¡2t2t2+1 + ln(t2 + 1) + 1

[ln(t2 + 1) + 1]2

=2t¡ 2t2 + (t2 + 1) ln(t2 + 1) + t2 + 1

(t2 + 1)[ln(t2 + 1) + 1]2

=¡t2 + 2t+ 1+ (t2 + 1) ln(t2 + 1)

(t2 + 1)[ln(t2 + 1) + 1]2

43. f(t) =2t3=2

ln(2t3=2 + 1)


u(t) = 2t3=2; u0(t) = 3t1=2

v(t) = ln(2t3=2 + 1); v0(t) =3t1=2

2t3=2 + 1

f 0(t) =ln(2t3=2 + 1)(3t1=2)¡ 2t3=2

h3t1=2

2t3=2+1

i[ln(2t3=2 + 1)]2

=(3t1=2) ln(2t3=2 + 1)¡ 6t2

2t3=2+1

[ln(2t3=2 + 1)]2

=(6t2 + 3t1=2) ln(2t3=2 + 1)¡ 6t2(2t3=2 + 1)[ln(2t3=2 + 1)]2

45. Note that a is a constant.d

dxln jaxj = d

dx(ln jaj+ ln jxj)

=d

dxln jaj+ d

dxln x

= 0 +d

dxln jxj

=d

dxln jxj

Therefore,d

dxln jaxj = d

dxln jxj :


47. Graph

y =ln jx+ 0:0001j ¡ ln jxj

0:0001

on a graphing calculator. A good choice for the viewing window is [¡3; 3]:

If we graph y = 1x on the same screen, we see that the two graphs coincide.

By the de…nition of the derivative, if f(x) = ln jxj ;

f 0(x) = limh!0

f(x+ h)¡ f(x)h

= limh!0

ln jx+ hj ¡ ln jxjh

;

and h = 0:0001 is very close to 0.Comparing the two graphs provides graphical evidence that

f 0(x) =1

x:

48. Use the derivative of lnx:d ln[u(x)v(x)]

dx=

1

u(x)v(x)¢ d[u(x)v(x)]

dx

d lnu(x)

dx=

1

u(x)¢ d[u(x)]

dx

d ln v(x)

dx=

1

v(x)¢ d[v(x)]

dx

Then since ln[u(x)v(x)] = lnu(x) + ln v(x);

1

u(x)v(x)¢ d[u(x)v(x)]

dx=

1

u(x)¢ d[u(x)]

dx+

1

v(x)¢ d[v(x)]

dx:

Multiply both sides of this equation by u(x)v(x): Then

d[u(x)v(x)]

dx= v(x)

d[u(x)]

dx+ u(x)

d[v(x)]

dx:

This is the product rule.

49. Use the derivative of lnx:

d ln u(x)v(x)

dx=

1u(x)v(x)

¢dhu(x)v(x)

idx

=v(x)

u(x)¢dhu(x)v(x)

idx

d lnu(x) =1

u(x)¢ d[u(x)]

dx

d ln v(x) =1

v(x)¢ d[v(x)]

dx


Then, since ln u(x)v(x) = lnu(x)¡ ln v(x);

v(x)

u(x)¢dhu(x)v(x)

idx

=1

u(x)¢ d[u(x)]

dx¡ 1

v(x)¢ d[v(x)]

dx:

Multiply both sides of this equation by u(x)v(x) : Then

dhu(x)v(x)

idx

=1

v(x)¢ d[u(x)]

dx¡ u(x)

[v(x)]2¢ d[v(x)]

dx

=v(x)

[v(x)]2¢ d[u(x)]

dx¡ u(x)

[v(x)]2¢ d[v(x)]

dx

=v(x) ¢ d[u(x)]

dx¡ u(x) ¢ d[v(x)]

dx[v(x)]2

This is the quotient rule.

50. Graph the function y = lnx: Sketch lines tangent to the graph at x = 12 ; 1, 2, 3, 4.

Estimate the slopes of the tangent lines at these points

x slope of tangent12 2

1 1

2 12

3 13

4 14

The values of the slopes at x are 1x :

Thus we see that d lnxdx = 1x :

51. The change-of-base theorem for logarithms states loga x =lnxlna : Find the derivative of each side.

d loga x

dx=lna ¢ d lnxdx ¡ lnx ¢ d lnadx

(lna)2

=lna ¢ 1x ¡ lnx ¢ 0

(lna)2

=1x

lna

=1x

x lna


52. (a) h(x) = u(x)v(x)

d

dxlnh(x) =

d

dxln[u(x)v(x)]

=d

dx[v(x) lnu(x)]

= v(x)d

dxlnu(x) + (lnu(x))v0(x)

= v(x)u0(x)u(x)

+ (lnu(x))v0(x)

=v(x)u0(x)u(x)

+ (lnu(x))v0(x)

(b) Since ddx lnh(x) =

h0(x)h(x) ;

h0(x) = h(x)d

dxlnh(x)

= h(x)

·v(x)u0(x)u(x)

+ (lnu(x))v0(x)¸

= u(x)v(x)·v(x)u0(x)u(x)

+ (lnu(x))v0(x)¸

53. h(x) = xx

u(x) = x; u0(x) = 1v(x) = x; v0(x) = 1

h0(x) = xx·x(1)

x+ (lnx)(1)

¸

= xx(1 + lnx)

54. h(x) = (x2 + 1)5x

u(x) = x2 + 1; u0(x) = 2xv(x) = 5x; v0(x) = 5

h0(x) = (x2 + 1)5x·5x(2x)

x2 + 1+ ln(x2 + 1) ¢ (5)

¸

= (x2 + 1)5x·10x2

x2 + 1+ 5 ln(x2 + 1)

¸

55. R(x) = 30 ln (2x+ 1)

C(x) =x

2

P (x) = R(x)¡C(x) = 30 ln (2x+ 1)¡ x2

The pro…t will be a maximum when the derivativeof the pro…t function is equal to 0.

P 0(x) = 30μ

1

2x+ 1

¶(2)¡ 1

2=

60

2x+ 1¡ 12

Now, P 0(x) =60

2x+ 1¡ 12= 0

when60

2x+ 1=1

2

120 = 2x+ 1

119

2= x:

Thus, a maximum pro…t occurs when x = 1192

or, in a practical sense, when 59 or 60 items aremanufactured. (Both 59 and 60 give the samepro…t.)

56. p = 100 +50

ln q; q > 1

(a) R = pq

R = 100q +50q

ln q

The marginal revenue is

dR

dq= 100 +

(ln q)(50)¡ 50q³1q

´(ln q)2

= 100 +50(ln q ¡ 1)(ln q)2

:

(b) The revenue from one more unit is dRdq for

q = 8:

100 +50(ln 8¡ 1)(ln 8)2

= $112:48

(c) The manager can use the information from (b)to decide if it is reasonable to sell additional items.If the revenue does not exceed the cost, there willbe no pro…t.

57. C(q) = 100q + 100

(a) The marginal cost is given by C 0(q):

C0(q) = 100

(b) P (q) = R(q)¡C(q)

= q

μ100 +

50

ln q

¶

¡ (100q + 100)

= 100q +50q

ln q¡ 100q ¡ 100

=50q

ln q¡ 100


(c) The pro…t from one more unit is is dPdq for

q = 8:

dP

dq=(ln q)(50)¡ 50q

³1q

´(ln q)2

=50 ln q ¡ 50(ln q)2

=50(ln q ¡ 10)(ln q)2

When q = 8; the pro…t from one more unit is

50(ln 8¡ 1)(ln 8)2

= $12:48:

(d) The manager can use the information frompart (c) to decide whether it is pro…table to makeand sell additional items.

58. C(x) = 5 log2 x+ 10

C(x) =C(x)

x=5 log2 x+ 10

x

C0(x) =

x ¢ 5 ¢ 1ln 2 ¢ 1x¡(5 log2 x+10) ¢ 1

x2

=5¡ (ln 2)(5 log2 x+ 10)

x2 ln 2

(a) C0(10) =

5¡ (ln 2)(5 log2 10 + 10)(102) ln 2

¼ ¡0:19396

(b) C0(20) =

5¡ (ln 2)(5 log2 20 + 10)(202) ln 2

¼ ¡0:06099

59. A(w) = 4.688w0:8168¡0:0154 log10 w

(a) A(4000) = 4:688(4000)0:8168¡0:0154 log10 4000

¼ 2590 cm2

(b)A(w)

4:688= w0:8166¡0:0154 log10 w

lnA(w)¡ln 4:688 = (lnw)μ0:8168¡ 0:0154 lnw

ln 10

¶

A0(w)A(w)

=1

w

μ0:8168¡ 0:0154 lnw

ln 10

¶

+ lnw

μ¡0:0154ln 10

¶1

w

=0:8168

w¡ 0:0308ln 10

lnw

w

=1

w

μ0:8168¡ 0:0308

ln 10lnw

¶

A0(w) =1

w

μ0:8168¡ 0:0308

ln 10lnw

¶

¢ (4:688w0:8168¡0:0154 log10 w)A0(4000) ¼ 0:4571 ¼ 0:46 g/cm2When the infant weighs 4000 g, it is gaining0.46 square centimeters per gram of weightincrease.

(c)

60. lnμN(t)

N0

¶= 9:8901e¡e

2:54197¡0:2167t

(a)N(t)

1000= e9:8901e

¡e2:54197¡0:2167t

N(t) = 1000e9:8901e¡e2:54197¡0:2167t

(b) N 0(20) ¼ 1,307,416 bacteria/hourTwenty hours into the experiment, the number ofbacteria are increasing at a rate of 1,307,416 perhour.

(c) S(t) = ln

μN(t)

N0

¶

(d)


The two graphs have the same general shape, butN(t) is scaled much larger.

(e) limt!1 S(t) = lim

t!1 9:8901e¡e2:54197¡0:2167t

= 9:8901

S(t) = ln

μN(t)

N0

¶

N(t) = N0eS(t)

limt!1 N(t) = N0e

limt!1

S(t)= 1000e9:8901

¼ 19,734,033 bacteria

61. F (x) = 0:774 + 0:727 log(x)

(a) F (25,000) = 0:774 + 0:727 log(25,000)= 3:9713 : : :

¼ 4 kJ/day

(b) F 0(x) = 0:7271

x ln 10

=0:727

ln 10x¡1

F 0(25,000) =0:727

ln 1025,000¡1

¼ 0:000012629 : : :¼ 1:3£ 10¡5

When a fawn is 25 kg in size, the rate of changeof the energy expenditure of the fawn is about1:3£ 10¡5 kJ/day per gram.

(c)

62. log y = 1:54¡ 0:008x¡ 0:658 log x

(a) y(x) = 10(1:54¡0:008x¡0:658 log x)

= 101:54(10¡0:008x)(10¡0:658 logx)= 101:54(100:008)¡x(10logx)¡0:658

¼ 34:7(1:0186)¡xx¡0:658

(b) (i) y(20) = 34:7(1:0186)¡2020¡0:658

¼ 3:343(ii) y(40) = 34:7(1:0186)¡4040¡0:658

¼ 1:466

(c)dy

dx= ¡34:7(1:0186)¡xx¡0:658

μln(1:0186)+

0:658

x

¶

(i)dy

dx(20) = ¡0:17160 : : : ¼ ¡0:172

(ii)dy

dx(40) = ¡0:051120 : : : ¼ ¡0:0511

63. M(t) = (0:1t+ 1) lnpt

(a) M(15) = [0:1(15) + 1] lnp15

¼ 3:385

When the temperature is 15±C, the number ofmatings is about 3.

(b) M(25) = [0:1(25) + 1] lnp25

¼ 5:663

When the temperature is 25±C, the number ofmatings is about 6.

(c) M(t) = (0:1t+ 1) lnpt

= (0:1t+ 1) ln t1=2

M 0(t) = (0:1t+ 1)μ1

2¢ 1t

¶

+ (ln t1=2)(0:1)

= 0:1 lnpt+

1

2t(0:1t+ 1)

M 0(15) = 0:1 lnp15 +

1

2 ¢ 15[(0:1)(15) + 1]

¼ 0:22

When the temperature is 15±C, the rate of changeof the number of matings is about 0.22.

64. P (t) = (t+ 100) ln (t+ 2)

P 0(t) = (t+ 100)μ

1

t+ 2

¶+ ln (t+ 2)

P 0(2) = (102)μ1

4

¶+ ln (4)

¼ 26:9

P 0(8) = (108)μ1

10

¶+ ln (10)

¼ 13:1

Chapter 4 Review Exercises 285

65. M =2

3log

E

0:007

(a) 8:9 =2

3log

E

0:007

13:35 = logE

0:007

1013:35 =E

0:007

E = 0:007(1013:35)

¼ 1:567£ 1011 kWh(b) 10,000,000£ 247 kWh/month

= 2,470,000,000 kWh/month

1:567£ 1011 kWh2,470,000,000 kWh/month

¼ 63:4 months

(c) M =2

3logE ¡ 2

3log 0:007

dM

dE=2

3

μ1

(ln 10)E

¶

=2

(3 ln 10)E

When E = 70,000,

dM

dE=

2

(3 ln 10)70,000

¼ 4:14£ 10¡6

(d) dMdE varies inversely with E, so as E increases,dMdE decreases and approaches zero.

66. f(x) =29,000(2:322¡ logx)

x

f 0(x) =x

·29,000

μ¡ 1

(ln 10)x

¶̧¡29,000(2:322¡ logx)(1)x2

= 29,000 ¢ ¡1

ln 10 ¡ (2:322¡ log x)x2

= ¡29,000 ¢ 1 + (ln 10)(2:322¡ log x)x2 ln 10

(a) f(30) =29,000(2:322¡ log 30)

30

¼ 817

f 0(30) = ¡29,0001 + (ln 10)(2:322¡ log 30)(30)2 ln 10

¼ ¡41:2For a street 30 ft wide, the maximum tra¢c ‡owis 817 vehicles/hr, and the rate of change is ¡41:2vehicles/hr per foot.

(b) f(40) =29; 000(2:322¡ log 40)

40

¼ 522

f 0(40) = ¡29; 000 ¢ 1 + (ln 10)(2:322¡ log 40)(40)2 ln 10

¼ ¡20:9For a street 40 ft wide, the maximum tra¢c ‡owis 522 vehicles/hr, and the rate of change is ¡20:9vehicles/hr per foot.

Chapter 4 Review Exercises

1. y = 5x3 ¡ 7x2 ¡ 9x+p5dy

dx= 5(3x2)¡ 7(2x)¡ 9 + 0= 15x2 ¡ 14x¡ 9

2. y = 7x3 ¡ 4x2 ¡ 5x+p2dy

dx= 7(3x2)¡ 4(2x)¡ 5 + 0

= 21x2 ¡ 8x¡ 5

3. y = 9x8=3

dy

dx= 9

μ8

3x5=3

¶

= 24x5=3

4. y = ¡4x¡3dy

dx= ¡4(¡3x¡4)

= 12x¡4 or12

x4

5. f(x) = 3x¡4 + 6px

= 3x¡4 + 6x1=2

f 0(x) = 3(¡4x¡5) + 6μ1

2x¡1=2

¶

= ¡12x¡5 + 3x¡1=2 or ¡ 12x5+

3

x1=2

6. f(x) = 19x¡1 ¡ 8px= 19x¡1 ¡ 8x1=2

f 0(x) = 19(¡x¡2)¡ 8μ1

2x¡1=2

¶

= ¡19x¡2 ¡ 4x¡1=2 or ¡ 19x2¡ 4

x1=2


7. k(x) =3x

4x+ 7

k0(x) =(4x+ 7)(3)¡ (3x)(4)

(4x+ 7)2

=12x+ 21¡ 12x(4x+ 7)2

=21

(4x+ 7)2

8. r(x) =¡8x2x+ 1

r0(x) =(2x+ 1)(¡8)¡ (¡8x)(2)

(2x+ 1)2

=¡16x¡ 8 + 16x(2x+ 1)2

=¡8

(2x+ 1)2

9. y =x2 ¡ x+ 1x¡ 1

dy

dx=(x¡ 1)(2x¡ 1)¡ (x2 ¡ x+ 1)(1)

(x¡ 1)2

=2x2 ¡ 3x+ 1¡ x2 + x¡ 1

(x¡ 1)2

=x2 ¡ 2x(x¡ 1)2

10. y =2x3 ¡ 5x2x+ 2

dy

dx=(x+ 2)(6x2 ¡ 10x)¡ (2x3 ¡ 5x2)(1)

(x+ 2)2

=6x3 + 12x2 ¡ 10x2 ¡ 20x¡ 2x3 + 5x2

(x+ 2)2

=4x3 + 7x2 ¡ 20x

(x+ 2)2

11. f(x) = (3x2 ¡ 2)4f 0(x) = 4(3x2 ¡ 2)3[3(2x)]

= 24x(3x2 ¡ 2)3

12. k(x) = (5x3 ¡ 1)6k0(x) = 6(5x3 ¡ 1)5[5(3x2)]

= 90x2(5x3 ¡ 1)5

13. y =p2t7 ¡ 5

= (2t7 ¡ 5)1=2dy

dx=1

2(2t7 ¡ 5)¡1=2[2(7t6)]

= 7t6(2t7 ¡ 5)¡1=2 or 7t6

(2t7 ¡ 5)1=2

14. y = ¡3p8t4 ¡ 1 = ¡3(8t4 ¡ 1)1=2dy

dx= ¡3

μ1

2

¶(8t4 ¡ 1)¡1=2[8(4t3)]

= ¡48t3(8t4 ¡ 1)¡1=2

or¡48t3

(8t4 ¡ 1)1=2

15. y = 3x(2x+ 1)3

dy

dx= 3x(3)(2x+ 1)2(2) + (2x+ 1)3(3)

= (18x)(2x+ 1)2 + 3(2x+ 1)3

= 3(2x+ 1)2[6x+ (2x+ 1)]

= 3(2x+ 1)2(8x+ 1)

16. y = 4x2(3x¡ 2)5dy

dx= (4x2)[5(3x¡ 2)4(3)] + (3x¡ 2)5(8x)= 60x2(3x¡ 2)4 + 8x(3x¡ 2)5= 4x(3x¡ 2)4[15x+ 2(3x¡ 2)]= 4x(3x¡ 2)4(15x+ 6x¡ 4)= 4x(3x¡ 2)4(21x¡ 4)

17. r(t) =5t2 ¡ 7t(3t+ 1)3

r0(t) =(3t+1)3(10t¡7)¡(5t2¡7t)(3)(3t+1)2(3)

[(3t+ 1)3]2

=(3t+ 1)3(10t¡ 7)¡ 9(5t2 ¡ 7t)(3t+ 1)2

(3t+ 1)6

=(3t+ 1)(10t¡ 7)¡ 9(5t2 ¡ 7t)

(3t+ 1)4

=30t2 ¡ 11t¡ 7¡ 45t2 + 63t

(3t+ 1)4

=¡15t2 + 52t¡ 7

(3t+ 1)4

18. s(t) =t3 ¡ 2t(4t¡ 3)4

s0(t) =(4t¡3)4(3t2¡2)¡(t3¡2t)(4)(4t¡3)3(4)

[(4t¡ 3)4]2

=(4t¡ 3)4(3t2 ¡ 2)¡ 16(t3 ¡ 2t)(4t¡ 3)3

(4t¡ 3)8

=(4t¡ 3)3[(4t¡ 3)(3t2 ¡ 2)¡ 16(t3 ¡ 2t)]

(4t¡ 3)8

=(4t¡ 3)3(12t3 ¡ 9t2 ¡ 8t+ 6¡ 16t3 + 32t)

(4t¡ 3)8

=¡4t3 ¡ 9t2 + 24t+ 6

(4t¡ 3)5


19. p(t) = t2(t2 + 1)5=2

p0(t) = t2 ¢ 52(t2 + 1)3=2 ¢ 2t+ 2t(t2 + 1)5=2

= 5t3(t2 + 1)3=2 + 2t(t2 + 1)5=2

= t(t2 + 1)3=2[5t2 + 2(t2 + 1)1]

= t(t2 + 1)3=2(7t2 + 2)

20. g(t) = t3(t4 + 5)7=2

g0(t) = t3 ¢ 72(t4 + 5)5=2(4t3) + 3t2 ¢ (t4 + 5)7=2

= 14t6(t4 + 5)5=2 + 3t2(t4 + 5)7=2

= t2(t4 + 5)5=2[14t4 + 3(t4 + 5)]

= t2(t4 + 5)5=2(17t4 + 15)

21. y = ¡6e2xdy

dx= ¡6(2e2x) = ¡12e2x

22. y = 8e0:5x

dy

dx= 8(0:5e0:5x) = 4e0:5x

23. y = e¡2x3

g(x) = ¡2x3g0(x) = ¡6x2y0 = ¡6x2e¡2x3

24. y = ¡4ex2g(x) = x2

g0(x) = 2x

dy

dx= (2x)(¡4ex2)= ¡8xex2

25. y = 5x ¢ e2x


dy

dx= 5x(2e2x) + e2x(5)

= 10xe2x + 5e2x

= 5e2x(2x+ 1)

26. y = ¡7x2e¡3x


dy

dx= (¡7x2)(¡3e¡3x) + e¡3x(¡14x)= 21x2e¡3x ¡ 14xe¡3xor 7xe¡3x(3x¡ 2)

27. y = ln (2 + x2)

g(x) = 2 + x2

g0(x) = 2x

dy

dx=

2x

2 + x2

28. y = ln (5x+ 3)

g(x) = 5x+ 3

g0(x) = 5

dy

dx=

5

5x+ 3

29. y =ln j3xjx¡ 3

dy

dx=(x¡ 3) ¡ 13x¢ (3)¡ (ln j3xj)(1)

(x¡ 3)2

=x¡3x ¡ ln j3xj(x¡ 3)2 ¢ x

x

=x¡ 3¡ x ln j3xj

x(x¡ 3)2

30. y =ln j2x¡ 1jx+ 3

dy

dx=(x+ 3)

³2

2x¡1´¡ (ln j2x¡ 1j)(1)

(x+ 3)2¢ 2x¡ 12x¡ 1

=2(x+ 3)¡ (2x¡ 1) ln j2x¡ 1j

(2x¡ 1)(x+ 3)2

31. y =xex

ln (x2 ¡ 1)

dy

dx=ln (x2 ¡ 1)[xex + ex]¡ xex

³1

x2¡1´(2x)

[ln (x2 ¡ 1)]2

=ex(x+ 1) ln (x2 ¡ 1)¡ 2x2ex

x2¡1[ln (x2 ¡ 1)]2 ¢ x

2 ¡ 1x2 ¡ 1

=ex(x+ 1)(x2 ¡ 1) ln (x2 ¡ 1)¡ 2x2ex

(x2 ¡ 1)[ln (x2 ¡ 1)]2

32. y =(x2 + 1)e2x

ln x

dy

dx=lnx[(x2+1)(2e2x)+(e2x)(2x)]¡(x2+1)e2x ¡1x¢

(ln x)2¢ xx

=x ln x[2e2x(x2 + 1) + 2xe2x]¡ (x2 + 1)e2x

x(ln x)2

=e2x[2x (ln x)(x2 + 1+ x)¡ (x2 + 1)]

x(ln x)2


33. s = (t2 + et)2

s0 = 2(t2 + et)(2t+ et)

34. q = (e2p+1 ¡ 2)4

Use the chain rule.

dq

dp= 4(e2p+1 ¡ 2)3[2e2p+1]= 8e2p+1(e2p+1 ¡ 2)3

35. y = 3 ¢ 10¡x2

dy

dx= 3 ¢ (ln 10)10¡x2(¡2x)= ¡6x(ln 10) ¢ 10¡x2

36. y = 10 ¢ 2pxdy

dx= 10 ¢ (ln 2) ¢ 2

px ¢ 12x¡1=2

=5(ln 2)2

px

x1=2

37. g(z) = log2 (z3 + z + 1)

g0(z) =1

ln 2¢ 3z2 + 1

z3 + z + 1

=3z2 + 1

(ln 2)(z3 + z + 1)

38. h(z) = log (1 + ez)

h0(z) =1

ln 10¢ ez

1 + ez

=ez

(ln 10)(1 + ez)

39. f(x) = e2x ln(xex + 1)


f 0(x) = e2xμex + xex

xex + 1

¶+ [ln(xex + 1)](2e2x)

=(1 + x)e3x

xex + 1+ 2e2x ln(xex + 1)

40. f(x) =epx

ln(px+ 1)


u(x) = epx; u0(x) =

epx

2px

v(x) = ln(px+ 1); v0(x) =

1

2px(px+ 1)

f 0(x) =[ln(px+ 1)]

³epx

2px

´¡ epx

μ1

2px(px+ 1)

¶[ln(px+ 1)]2

=epx[(px+ 1) ln(

px+ 1)¡ 1]

2px(px+ 1)[ln(

px+ 1)]2

41. (a) Dx(f [g(x)]) at x = 2

= f 0[g(2)]g0(2)

= f 0(1)μ3

10

¶

= ¡5μ3

10

¶

= ¡32

(b) Dx(f [g(x)]) at x = 3

= f 0[g(3)]g0(3)

= f 0(2)μ4

11

¶

= ¡6μ4

11

¶

= ¡2411

42. (a) Dx(g[f(x)]) at x = 2

= g0[f(2)]f 0(2)

= g0(4)(¡6)

=6

13(¡6)

= ¡3613

(b) Dx(g[f(x)]) at x = 3

= g0[f(3)]f 0(3)

= g0(2)(¡7)

=3

10(¡7)

= ¡2110


44. y = x2 ¡ 6x; tangent at x = 2dy

dx= 2x¡ 6

Slope = y0(2) = 2(2)¡ 6 = ¡2

Use (2;¡8) and point-slope form.

y ¡ (¡8) = ¡2(x¡ 2)y + 8 = ¡2x+ 4y + 2x = ¡4

y = ¡2x¡ 4

45. y = 8¡ x2; x = 1y = 8¡ x2dy

dx= ¡2x

slope = y0(1) = ¡2(1) = ¡2

Use (1; 7) and m = ¡2 in the point-slope form.

y ¡ 7 = ¡2(x¡ 1)y ¡ 7 = ¡2x+ 22x+ y = 9

y = ¡2x+ 9

46. y =3

x¡ 1; tangent at x = ¡1

y =3

x¡ 1 = 3(x¡ 1)¡1

dy

dx= 3(¡1)(x¡ 1)¡2(1)= ¡3(x¡ 1)¡2

Slope = y0(¡1) = ¡3(¡1¡ 1)¡2 = ¡34

Use¡¡1;¡3

2

¢and point-slope form.

y ¡μ¡32

¶= ¡3

4[x¡ (¡1)]

y +3

2= ¡3

4(x+ 1)

y +6

4= ¡3

4x¡ 3

4

y = ¡34x¡ 9

4

47. y =x

x2 ¡ 1; x = 2dy

dx=(x2 ¡ 1) ¢ 1¡ x(2x)

(x2 ¡ 1)2

=¡x2 ¡ 1(x2 ¡ 1)2

The value of dydx when x = 2 is the slope.

m =¡(22)¡ 1(22 ¡ 1)2 =

¡59= ¡5

9

When x = 2;

y =2

4¡ 1 =2

3:

Use m = ¡59 with P

¡2; 23

¢:

y ¡ 23= ¡5

9(x¡ 2)

y ¡ 69= ¡5

9x+

10

9

y = ¡59x+

16

9

48. y =p6x¡ 2; tangent at x = 3

y =p6x¡ 2 = (6x¡ 2)1=2

dy

dx=1

2(6x¡ 2)¡1=2(6)

= 3(6x¡ 2)¡1=2

slope = y0(3) = 3(6 ¢ 3¡ 2)¡1=2= 3(16)¡1=2

=3

161=2

=3

4

Use (3; 4) and point-slope form.

y ¡ 4 = 3

4(x¡ 3)

y ¡ 164=3

4x¡ 9

4

y =3

4x+

7

4


49. y = ¡p8x+ 1; x = 3y = ¡(8x+ 1)1=2dy

dx= ¡1

2(8x+ 1)¡1=2(8)

dy

dx= ¡ 4

(8x+ 1)1=2


m = ¡ 4

(24 + 1)1=2= ¡4

5

When x = 3;

y = ¡p24 + 1 = ¡5:

Use m = ¡45 with P (3;¡5):

y + 5 = ¡45(x¡ 3)

y +25

5= ¡4

5x+

12

5

y = ¡45x¡ 13

5

50. y = ex; x = 0

dy

dx= ex

The value of dydx when x = 0 is the slopem = e0 = 1:

When x = 0; y = e0 = 1: Use m = 1 with P (0; 1):

y ¡ 1 = 1(x¡ 0)y = x+ 1

51. y = xex; x = 1

dy

dx= xex + 1 ¢ ex

= ex(x+ 1)


m = e1(1 + 1) = 2e

When x = 1; y = 1e1 = e: Use m = 2e withP (1; e):

y ¡ e = 2e(x¡ 1)y = 2ex¡ e

52. y = ln x; x = 1

dy

dx=1

x

The value of dydx when x = 1 is the slope m = 11 =

1:

When x = 1; y = ln 1 = 0: Use m = 1 withP (1; 0):

y ¡ 0 = 1(x¡ 1)y = x¡ 1

53. y = x ln x; x = e

dy

dx= x ¢ 1

x+ 1 ¢ ln x

= 1+ ln x

The value of dydx when x = e is the slope

m = 1 + ln e = 1 + 1 = 2:

When x = e; y = e ln e = e ¢ 1 = e: Use m = 2

with P (e; e):

y ¡ e = 2(x¡ e)y = 2x¡ e

54. The slope of the graph of y = x+k is 1. First, we…nd the point on the graph of f(x) =

p2x¡ 1 at

which the slope is also 1.

f(x) = (2x¡ 1)1=2

f 0(x) =1

2(2x¡ 1)¡1=2(2)

f 0(x) =1p2x¡ 1

The slope is 1 when

1p2x¡ 1 = 1

1 =p2x¡ 1

1 = 2x¡ 12x = 2

x = 1;

andf(1) = 1:

Therefore, at P (1; 1) on the graph of f(x) =p2x¡ 1;

the slope is 1. An equation of the tangent line is

y ¡ 1 = 1(x¡ 1)y ¡ 1 = x¡ 1

y = x+ 0:


Any tangent line intersects the curve in exactlyone point.From this we see that if k = 0; there is one pointof intersection.

The graph of f is below the line y = x+0: There-fore, if k > 0; the graph of y = x + k will notintersect the graph.

Consider the point Q(12 ; 0) on the graph. We …ndan equation of the line through Q with slope 1.

y ¡ 0 = 1μx¡ 1

2

¶

y = x¡ 12

The line with a slope of 1 through Q(12 ; 0) willintersect the graph in two points. One is Q andthe other is some point on the graph to the rightof P:The graph of y = x+0 intersects the graph in onepoint, while the graph of y = x¡ 1

2 intersects it intwo points. If we use a value of k in y = x+k with¡12 < k < 0; we will have a line with a y-intercept

between ¡12 and a 0 and a slope of 1 which will

intersect the graph in two points.If k; the y-intercept, is less than ¡1

2 ; the graph ofy = x+ k will be below point Q and will intersectthe graph of f in exactly one point.

To summarize, the graph of y = x+k will intersectthe graph of f(x) =

p2x¡ 1 in

(1) no points if k > 0;

(2) exactly one point if k = 0 or if k < ¡12 ;

(3) exactly two points if ¡12 · k < 0:

55. (a) Use the chain rule.

Let g(x) = lnx: Then g0(x) =1

x:

Let y = g[f(x)]: Thendy

dx= g0[f(x)] ¢ f 0(x):

so

d ln f(x)

dx=

1

f(x)¢ f 0(x) = f 0(x)

f(x):

(b)^f =

f 0(x)f(x)

;^g= f

g0(x)g(x)

^fg =

(fg)0(x)(fg)(x)

=f(x) ¢ g0(x) + g(x) ¢ f 0(x)

f(x)g(x)

=f(x) ¢ g0(x)f(x) ¢ g(x) +

g(x) ¢ f 0(x)f(x) ¢ g(x)

=g0(x)g(x)

+f 0(x)f(x)

=^g +

^f or

^f +

^g

56. Using the result^fg=

^f +

^g; the total amount

of tuition collected goes up by approximately2% + 3% = 5%.Let T = tuition per person before the increase andS = number of students before the increase. Thenthe new tuition is 1.03T and the new numbers ofstudents is 1.02S; so the total amount of tuitioncollected is (1:03T )(1:02S) = 1:0506TS; which isan increase of 5.06%.

57. C(x) =px+ 1

C(x) =C(x)

x=

px+ 1

x

=(x+ 1)1=2

x

C0(x) =

x£12(x+ 1)

¡1=2¤¡ (x+ 1)1=2(1)x2

=12x(x+ 1)

¡1=2 ¡ (x+ 1)1=2x2

=x(x+ 1)¡1=2 ¡ 2(x+ 1)1=2

2x2

=(x+ 1)¡1=2[x¡ 2(x+ 1)]

2x2

=(x+ 1)¡1=2(¡x¡ 2)

2x2

=¡x¡ 2

2x2(x+ 1)1=2


58. C(x) =p3x+ 2

C(x) =C(x)

x=

p3x+ 2

x=(3x+ 2)1=2

x

C0(x) =

x£12(3x+ 2)

¡1=2(3)¤¡ (3x+ 2)1=2(1)

x2

=32x(3x+ 2)

¡1=2 ¡ (3x+ 2)1=2x2

=3x(3x+ 2)¡1=2 ¡ 2(3x+ 2)1=2

2x2

=(3x+ 2)¡1=2[3x¡ 2(3x+ 2)]

2x2

=3x¡ 6x¡ 42x2(3x+ 2)1=2

=¡3x¡ 4

2x2(3x+ 2)1=2

59. C(x) = (x2 + 3)3

C(x) =C(x)

x=(x2 + 3)3

x

C0(x) =

x[3(x2 + 3)2(2x)]¡ (x2 + 3)3(1)x2

=6x2(x2 + 3)2 ¡ (x2 + 3)3

x2

=(x2 + 3)2[6x2 ¡ (x2 + 3)]

x2

=(x2 + 3)2(5x2 ¡ 3)

x2

60. C(x) = (4x+ 3)4

C(x) =C(x)

x=(4x+ 3)4

x

C0(x) =

x[4(4x+ 3)3(4)]¡ (4x+ 3)4(1)x2

=16x(4x+ 3)3 ¡ (4x+ 3)4

x2

=(4x+ 3)3[16x¡ (4x+ 3)]

x2

=(4x+ 3)3(12x¡ 3)

x2

61. C(x) = 10¡ e¡x

C(x) =C(x)

x

C(x) =10¡ e¡x

x

C0(x) =

x(e¡x)¡ (10¡ e¡x) ¢ 1x2

=e¡x(x+ 1)¡ 10

x2

62. C(x) = ln (x+ 5)

C(x) =ln (x+ 5)

x

C0(x) =

x ¢ 1x+5 ¡ ln (x+ 5) ¢ 1

x2

=x¡ (x+ 5) ln (x+ 5)

x2(x+ 5)

63. S(x) = 1000 + 60px+ 12x

= 1000 + 60x1=2 + 12x

dS

dx= 60

μ1

2x¡1=2

¶+ 12

= 30x¡1=2 + 12 =30px+ 12

(a)dS

dx(9) =

30p9+ 12 =

30

3+ 12 = 22

Sales will increase by $22 million when $1000 moreis spent on research.

(b)dS

dx(16) =

30p16+ 12 =

30

4+ 12 = 19:5

Sales will increase by $19.5 million when $1000more is spent on research.

(c)dS

dx(25) =

30p25+ 12 =

30

5+ 12 = 18

Sales will increase by $18 million when $1000 moreis spent on research.

(d) As more money is spent on research, the in-crease in sales is decreasing.

64. P (x) =x2

2x+ 1

P 0(x) =(2x+ 1)(2x)¡ x2(2)

(2x+ 1)2

=4x2 + 2x¡ 2x2(2x+ 1)2

=2x2 + 2x

(2x+ 1)2

(a) P 0(4) =2(4)2 + 2(4)

[2(4) + 1]2

=40

81¼ 0:4938

In dollars, this is 4081(100) ¼ $49:38; which is the

approximate increase in pro…t from selling the …fthunit.


(b) P 0(12) =2(12)2 + 2(12)

[2(12) + 1]2

=312

625= 0:4992

In dollars, this is 312625(100) ¼ $49:92; which is

the approximate increase in pro…t from selling thethirteenth unit.

(c) P 0(20) =2(20)2 + 2(20)

[2(20) + 1]2

=840

1681¼ 0:4997

In dollars, this is 8401681(100) ¼ $49:97; which is

the approximate increase in pro…t from selling thetwenty-…rst unit.

(d) As the number sold increases, the marginalpro…t increases.

(e) The average pro…t is de…ned by

P (x) =P (x)

x=

x2

2x+1

x=

x

2x+ 1

The marginal average pro…t is given by

d

dx(P (x)) =

(2x+ 1)(1)¡ (x)(2)(2x+ 1)2

=2x+ 1¡ 2x(2x+ 1)2

=1

(2x+ 1)2

The marginal average pro…t when 4 units are soldis

d

dx(P (4)) =

1

[2(4) + 1]2

=1

81¼ 0:0123

The average pro…t is going up at a rate of 100¡181

¢ ¼$1:23 per unit when 4 units are sold.

65. T (x) =1000 + 60x

4x+ 5

T 0(x) =(4x+ 5)(60)¡ (1000 + 60x)(4)

(4x+ 5)2

=240x+ 300¡ 4000¡ 240x

(4x+ 5)2

=¡3700(4x+ 5)2

(a) T 0(9) =¡3700

[4(9) + 5]2

=¡37001681

¼ ¡2:201

Costs will decrease $2201 for the next $100 spenton training.

(b) T 0(19) =¡3700

[4(19) + 5]2

=¡37006561

¼ ¡0:564Costs will decrease $564 for the next $100 spenton training.

(c) Costs will always decrease because

T 0(x) =¡3700(4x+ 5)2

will always be negative.

66. A(r) = 1000³1 +

r

400

´48A0(r) = 1000 ¢ 48

³1 +

r

400

´47¢ 1400

= 120³1 +

r

400

´47

A0(5) = 120μ1 +

5

400

¶47¼ 215:15

The balance increases by approximately $215.15for every 1% increase in the interest rate whenthe rate is 5%.

67. A(r) = 1000e12r=100

A0(r) = 1000e12r=100 ¢ 12100

= 120e12r=100

A0(5) = 120e0:6 ¼ 218:65The balance increases by approximately $218.65for every 1% increase in the interest rate whenthe rate is 5%.

68. T (r) =ln 2

ln¡1 + r

100

¢T (r) = ln 2

hln³1 +

r

100

´i¡1T 0(r) = (ln 2)(¡1)

hln³1 +

r

100

´i¡2¢

1100

1 + r100

T 0(r) =¡ ln 2

(100 + r)£ln¡1 + r

100

¢¤2T 0(5) = ¡ ln 2

105 (ln 1:05)2¼ ¡2:77

The doubling time decreases by approximately 2.77years for every 1% increase in the interest ratewhen the interest rate is 5%.


69. f(t) = 1:5207t4 ¡ 19:166t3 + 62:91t2 + 6:0726t+ 1026f 0(t) = 1:5207(4t3)¡ 19:166(3t2) + 62:91(2t) + 6:0726

= 6:0828t3 ¡ 57:498t2 + 125:82t = 7 corresponds to the beginning of 2005.

f 0(7) = 6:0828(7)3 ¡ 57:498(7)2 + 125:82(7) + 6:0726¼ 156

Rents were increasing at the rate of $156 per month per year.

70. (a) Using the regression feature on a graphing calculator, a cubic function that models the data is

y = (7:259£ 10¡5)t3 ¡ 0:01217t2 + 0:01447t+ 66:32:

Using the regression feature on a graphing calculator, a quartic function that models the data is

y = (1:896£ 10¡6)t4 ¡ 3:676£ 10¡4t3 + 0:02002t2 ¡ 0:7448t+ 68:52

(b) Using the cubic function, dydx at x = 90 is about 0:41 percent per year. Using the quartic function,dydx at

x = 90 is about ¡0:55 percent per year.71. (a) Using the regression feature on a graphing calculator, a cubic function that models the data is

y = (2:458£ 10¡5)t3 ¡ (6:767£ 10¡4)t2 ¡ 0:02561t+ 2:031:

Using the regression feature on a graphing calculator, a quartic function that models the data is

y = (¡1:314£ 10¡6)t4 + (3:363£ 10¡4)t3 ¡ 0:02565t2 + 0:7410t¡ 5:070:

(b) Using the cubic function, dydx at x = 95 is about 0:51 dollar per year. Using the quartic function,dydx at

x = 95 is about 0:47 dollar per year.

72. P (t) = ae0:05t

P 0(t) = ae0:05t(0:05)P 0(t) = P (t)(0:05)

If P (t) = 1,000,000; thenP 0(t) = 1,000,000(0:05) = 50,000.

The population is growing at a rate of 50,000 per year.

73. G(t) =m Go

Go + (m¡Go)e¡kmt ; where m = 30,000, Go = 2000; and k = 5 ¢ 10¡6:

(a) G(t) =(30,000)(2000)

2000 + (30,000¡ 2000)e¡5¢10¡6(30;000)t =30; 000

1 + 14e¡0:15t

(b) G(t) = 30; 000(1 + 14e¡0:15t)¡1

G(6) = 30,000(1 + 14e¡0:90)¡1 ¼ 4483G0(t) = ¡30; 000(1 + 14e¡0:15t)¡2(¡2:1e¡0:15t) = 63; 000e¡0:15t

(1 + 14e¡0:15t)2

G0(6) =63; 000e¡0:90

(1 + 14e¡0:90)2¼ 572

The population is 4483, and the rate of growth is 572.


74. L(t) = 71:5(1¡ e¡0:1t) andW (L) = 0:01289 ¢ L2:9

(a) L(5) = 71:5(1¡ e¡0:5) ¼ 28:1The approximate length of a 5-year-old monkey-face is 28.1 cm.

(b) L0(t) = 71:5(0:1e¡0:1t)L0(5) = 71:5(0:1e¡0:5)

¼ 4:34The length is growing by about 4.34 cm/year.

(c) W [L(5)] ¼ 0:01289(28:1)2:9 ¼ 205The approximate weight is 205 grams.

(d) W 0(L) = 0:01289(2:9)L1:9

= 0:037381L1:9

W 0[L(5)] ¼ 0:037381(28:1)1:9¼ 21:2

The rate of change of the weight with respect tolength is 21.2 grams/cm.

(e)dW

dt=dW

dL¢ dLdt

¼ (21:2)(4:34)¼ 92:0

The weight is growing at about 92.0 grams/year.

75. M(t) = 3583e¡e¡0:020(t¡66)

(a) M(250) = 3583e¡e¡0:020(250¡66)

¼ 3493:76 grams,or about 3.5 kilograms

(b) As t!1;¡e¡0:020(t¡66)! 0; e¡e¡0:020(t¡66) ! 1,

and M(t)! 3583 grams or about 3.6 kilograms.

(c) 50% of 3583 is 1791.5.

1791:5 = 3583e¡e¡0:020(t¡66)

ln

μ1791:5

3583

¶= ¡e¡0:020(t¡66)

ln

μln3583

1791:5

¶= ¡0:020(t¡ 66)

t = ¡ 1

0:020ln

μln3583

1791:5

¶+ 66

¼ 84 days

(d) DtM(t)= 3583e¡e

¡0:020(t¡66)Dt(¡e¡0:020(t¡66))

= 3583e¡e¡:0:020(t¡66)

(¡e¡0:020(t¡66))(¡0:020)= 71:66e¡e

¡0:020(t¡66)(e¡0:020(t¡66))

when t = 250, DtM(t) ¼ 1:76 g/day.(e)

Growth is initially rapid, then tapers o¤.

(f) Day Weight Rate

50 904 24.90100 2159 21.87150 2974 11.08200 3346 4.59250 3494 1.76300 3550 0.66

76. h(t) = 37:79(1:021)t

h0(t) = 37:79(ln 1:021)(1:021)t

(a) h0(5) = 37:79(ln 1:021)(1:021)5

¼ 0:871In 2005, the instantaneous rate of change is 871,000per year.

(b) h0(25) = 37:79(ln 1:021)(1:021)25

¼ 1:320In 2025, the instantaneous rate of change will be1,320,000 per year.

77. f(t) =8

t+ 1+

20

t2 + 1

(a) The average velocity from t = 1 to t = 3 isgiven by

average velocity =f(3)¡ f(1)3¡ 1

=

¡84 +

2010

¢¡ ¡82 + 202

¢2

=4¡ 142

= ¡5Belmar’s average velocity between 1 sec and 3 secis ¡5 ft/sec.


(b) f(t) = 8(t+ 1)¡1 + 20(t2 + 1)¡1

f 0(t) = ¡8(t+ 1)¡2 ¢ 1¡ 20(t2 + 1)¡2 ¢ 2t

= ¡ 8

(t+ 1)2¡ 40t

(t2 + 1)2

f 0(3) = ¡ 8

16¡ 120100

= ¡0:5¡ 1:2= ¡1:7

Belmar’s instantaneous velocity at 3 sec is ¡1:7ft/sec.

78. p(x) = 1:757(1:0249)x¡1930

p0(x) = 1:757(ln 1:0249)(1:0249)x¡1930

p0(2000) = 1:757(ln 1:0249)(1:0249)2000¡1930

¼ 0:242The production of corn is increasing at a rate of0.242 billion bushels per year in 2000.

79. (a) N(t) = N0e¡0:217t; where t = 1 and N0 = 210

N(1) = 210e¡0:217(1)

¼ 169

The number of words predicted to be in use in1950 is 169, and the actual number in use was167.

(b) N(2) = 210e¡0:217(2)

¼ 136In 2050 the will be about 136 words still beingused.

(c) N(t) = 210e¡0:217t

N 0(t) = 210e¡0:217t ¢ (¡0:217)= ¡45:57e¡0:217t

N 0(2) = ¡45:57e¡0:217(2)¼ ¡30

In the year 2050 the number of words in use willbe decreasing by 30 words per millenium.

80. f(x) = k(x¡ 49)6 + :8f 0(x) = k ¢ 6(x¡ 49)5

= (3:8£ 10¡9)(6)(x¡ 49)5= (2:28£ 10¡8)(x¡ 49)5

(a) f 0(20) = (2:28£ 10¡8)(20¡ 49)5¼ ¡0:4677

fatalities per 1000 licensed drivers per 100 millionmiles per year.

At the age of 20, each extra year results in a de-crease of 0.4677 fatalities per 1000 licensed driversper 100 million miles.

(b) f 0(60) = (2:28£ 10¡8)(60¡ 49)5¼ 0:003672

fatalities per 1000 licensed drivers per 100 millionmiles per year.

At the age of 60, each extra year results in anincrease of 0.003672 fatalities per 1000 licenseddrivers per 100 million miles.

calculating the derivative - grosse pointe … 4.1 techniques for finding derivatives 243 40....

Documents