CALCULATING THE DERIVATIVE - Grosse Pointe Public Schools

[Pages:58]Chapter 4

CALCULATING THE DERIVATIVE

4.1 Techniques for Finding Derivatives

1. y = 12x3 ? 8x2 + 7x + 5

dy = 12(3x3?1) ? 8(2x2?1) + 7x1?1 + 0 dx

= 36x2 ? 16x + 7

2.

y

=

8x3

?

5x2

?

x 12

dy dx

= =

8(3x3?1) ? 5(2x2?1) ?

24x2

?

10x

?

1 12

1 12

x1?1

3.

y

=

3x4

?

6x3

+

x2 8

+

5

dy dx

=

3(4x4?1)

?

6(3x3?1) +

1 8

(2x2?1)

+

0

=

12x3

?

18x2

+

1 4x

4. y = 5x4 + 9x3 + 12x2 ? 7x

dy = 5(4x4?1) + 9(3x3?1) dx

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

5. y = 6x3:5 ? 10x0:5

dy = 6(3:5x3:5?1) ? 10(0:5x0:5?1)

dx

=

21x2:5

? 5x?0:5

or

21x2:5

?

5 x0:5

6. f (x) = ?2x1:5 + 12x0:5 f 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

=

p 8x

+

6x3=4

= 8x1=2 + 6x3=4

?

?

dy = 8 dx

1 2

x1=2?1

+6

3 4

x3=4?1

=

4x?1=2

+

9 2

x?1=4

4

9

or x1=2 + 2x1=4

8. y = ?100px ? 11x2=3

= ?100x1=2 ? 11x2=3

?

?

dy = ?100 dx

1 2

x1=2?1

? 11

2 3

x2=3?1

=

?50x?1=2

?

22x?1=3 3

or

?50 x1=2

?

22 3x1=3

9. g(x) = 6x?5 ? x?1 g0(x) = 6(?5)x?5?1 ? (?1)x?1?1 = ?30x?6 + x?2

or

?30 x6

+

1 x2

10. y = 10x?3 + 5x?4 ? 8x

dy = 10(?3x?3?1) + 5(?4x?4?1) ? 8x1?1 dx

=

?30x?4

?

20x?5

?

8

or

?30 x4

?

20 x5

?

8

11. y = 5x?5 ? 6x?2 + 13x?1

dy = 5(?5x?5?1) ? 6(?2x?2?1) + 13(?1x?1?1) dx

= ?25x?6 + 12x?3 ? 13x?2

or

?25 x6

+

12 x3

?

13 x2

75 12. f(t) = t ? t3

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

=

?7t?2

+

15t?4

or

?7 t2

+

15 t4

14 12 p

13. f(t) = t + t4 + 2

=

14t?1

+

12t?4

+

p 2

f 0(t) = 14(?1t?1?1) + 12(?4t?4?1) + 0

=

?14t?2

?

48t?5

or

?14 t2

?

48 t5

239

240

Chapter 4 CALCULATING THE DERIVATIVE

6 7 3p

14. y = x4 ? x3 + x + 5

=

6x?4

?

7x?3

+

3x?1

+

p 5

dy = 6(?4x?4?1) ? 7(?3x?3?1) + 3(?1x?1?1) + 0 dx

= ?24x?5 + 21x?4 ? 3x?2

or

?24 x5

+

21 x4

?

3 x2

317 15. y = x6 + x5 ? x2

= 3x?6 + x?5 ? 7x?2 dy = 3(?6x?7) + (?5x?6) ? 7(?2x?3) dx

= ?18x?7 ? 5x?6 + 14x?3

or

?18 x7

?

5 x6

+

14 x3

16. p(x) = ?10x?1=2 + 8x?3=2

?

?

p0(x) = ?10

?

1 2

x?3=2

+8

?

3 2

x?5=2

= 5x?3=2 ? 12x?5=2

or

5 x3=2

?

12 x5=2

17. h(x) = x?1=2 ? 14x?3=2

?

h0(x)

=

?

1 2

x?3=2

?

14

?

3 2

x?5=2

=

?x?3=2 2

+ 21x?5=2

or

?1 2x3=2

+

21 x5=2

18. y = 4p6 x = 6x?1=4

?

dy = 6 dx

?

1 4

x?5=4

=

?

3 2

x?5=4

?3 or 2x5=4

19. y = p?2 3x

=

?2 x1=3

=

?2x?1=3 ?

dy = ?2 dx

?

1 3

x?4=3

=

2x?4=3 3

or

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) = x3p? 4x x

=

x3 ? 4x x1=2

= x5=2 ? 4x1=2

?

g0(x)

=

5 2

x5=2?1

?

4

1 2

x1=2?1

=

5 2

x3=2

? 2x?1=2

or

5 2

x3=2

?

p2 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 decrease by 1 and the derivative function will be linear, so the correct choice is (b).

26. d (4x3 ? 6x?2) dx

= 4(3x2) ? 6(?2x?3) = 12x2 + 12x?3

=

12x2

+

12 x3

choice (c)

=

12x2(x3) x3

+

12

=

12x5 + x3

12

choice (b)

Neither choice (a) nor choice (d) equals

d (4x3 ? 6x?2): dx

Section 4.1 Techniques for Finding Derivatives

?

?

27. Dx

9x?1=2

+

2 x3=2

= Dx[9x?1=2 + 2x?3=2]

?

?

=9

?

1 2

x?3=2

+2

?

3 2

x?5=2

=

?

9 2

x?3=2

?

3x?5=2

or

?9 2x3=2

?

3 x5=2

?

?

28. Dx

p8 ? p3

4x

x3

= Dx [8x?1=4 ? 3x?3=2]

?

?

=8

?

1 4

x?5=4

?3

?

3 2

x?5=2

=

?2x?5=4

+

9x?5=2 2

or

?2 x5=4

+

9 2x5=2

29.

f (x)

=

x4 6

?

3x

=

1 6

x4

?

3x

f 0(x)

=

1 6

(4x3)

?

3

=

2 3

x3

?

3

f 0 (?2)

=

2 3

(?2)3

?

3

=

?

16 3

?

3

=

?

25 3

30.

f (x)

=

x3 9

?

7x2

=

1 9

x3

?

7x2

f 0 (x)

=

1 9

(3x2)

?

7(2x)

=

1 3

x2

?

14x

241

f 0(3)

=

1 (3)2 3

?

14(3)

= 3 ? 42 = ?39

31. y = x4 ? 5x3 + 2; x = 2 y0 = 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; y_1) = (2; ?22) to obtain the equation.

y ? (?22) = ?28(x ? 2) y = ?28x + 34

32. y = ?3x5 ? 8x3 + 4x2 y0 = ?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 the equation.

y ? (?7) = ?31(x ? 1) y + 7 = ?31x + 31 y = ?31x + 24

33.

y = ?2x1=2 + x3=2

?

y0 = ?2

1 2

x?1=2

+

3 2

x1=2

=

?x?1=2

+

3 2

x1=2

=

?

1 x1=2

+

3x1=2 2

y0(9)

=

?

1 (9)1=2

+

3(9)1=2 2

=

?

1 3

+

9 2

=

25 6

The

slope

of

the

tangent

line

at

x

=

9

is

25 6

:

242

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

=

3 x4

?

2 x3

y0(2)

=

3 (2)4

?

2 (2)3

=

3 16

?

2 8

=

?

1 16

The

slope

of

the

tangent

line

at

x

=

2

is

?

1 16

.

35. f (x) = 9x2 ? 8x + 4 f0(x) = 18x ? 8

Let f 0(x) = 0 to ...nd the point where the slope of the tangent line is zero.

18x ? 8 = 0 18x = 8

x

=

8 18

=

4 9

Find the y-coordinate.

f (x) = 9x2 ? 8x + 4

f

4 9

?

=

9

4 9

?2

?

8

4 9

?

+

4

?

=9

16 81

?

32 9

+

4

=

16 9

?

32 9

+

36 9

=

20 9

T? 49h;e29s0l?o.pe of the tangent line is zero at one point,

36. f (x) = x3 + 9x2 + 19x ? 10 f0(x) = 3x2 + 18x + 19

If the slope of the tangent line is ?5, then the f 0(x) = ?5:

3x2 + 18x + 19 = ?5 3x2 + 18x + 24 = 0 3(x2 + 6x + 8) = 0 3(x + 2)(x + 4) = 0

x = ?2 or x = ?4

f(?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

Chapter 4 CALCULATING THE DERIVATIVE

Thus, the points where the slope of the tangent line is ?5 are (?2; ?20) and (?4; ?6):

37. f(x) = 2x3 + 9x2 ? 60x + 4 f 0(x) = 6x2 + 18x ? 60

If the tangent line is horizontal, then its slope is zero and f0(x) = 0:

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

Thus, the tangent line is horizontal at x = ?5 and x = 2:

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

If the tangent line is horizontal, then its slope is zero and f0(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 = ?3 or x = ?7:

39. f(x) = x3 ? 4x2 ? 7x + 8 f 0(x) = 3x2 ? 8x ? 7

If the tangent line is horizontal, then its slope is zero and f0(x) = 0:

3x2 ? 8x ? 7 = 0

p

x= 8?

64 + 84 6

p x = 8 ? 6 148

p

x

=

8

?

2 6

37

p

x

=

2(4

? 6

37)

p

x=

4? 3

37

p

Thus,

the

tangent

line

is

horizontal

at

x

=

4? 3

37 :

Section 4.1 Techniques for Finding Derivatives

40. f (x) = x3 ? 5x2 + 6x + 3 f0(x) = 3x2 ? 10x + 6

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

3x2 ? 10x + 6 = 0

p

x

=

10

?

100 p6

?

72

x

=

10 ? 6

28 p

x

=

10

?2 p6

7

x

=

5

? 3

7

p

The

tangent

line

is

horizontal

at

x

=

5? 3

7:

41. f (x) = 6x2 + 4x ? 9 f0(x) = 12x + 4

If the slope of the tangent line is ?2; f 0(x) = ?2:

12x + 4 = ?2 12x = ?6

x ?

=

?

1 2

f

?

1 2

=

?

19 2

The

slope

of

the

tangent

line

is

?2

at

? ?

1 2

;

?

19 2

?

:

42. f (x) = 2x3 ? 9x2 ? 12x + 5 f0(x) = 6x2 ? 18x ? 12

If the slope of the tangent line is 12, f0(x) = 12:

6x2 ? 18x ? 12 = 12 6x2 ? 18x ? 24 = 0 6(x2 ? 3x ? 4) = 0 6(x ? 4)(x + 1) = 0

x = 4 or x = ?1

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

43. f (x) = x3 + 6x2 + 21x + 2 f0(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 = ?2

f (?2) = ?24

243

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 2

g(x)

+

1 4

h(x)

f 0(x)

=

1 2

g0

(x)

+

1 4

h0(x)

f 0(2)

=

1 2

g0

(2)

+

1 4

h0(2)

=

1 2

(7)

+

1 4

(14)

=

7

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

(b) f0(1) gives the slope of the tangent line to f at 1. The line goes through (?1; 1) and (1; 2):

m

=

1

2?1 ? (?1)

=

1 2;

so

f 0(1)

=

1 2:

(c) The domain of f is [?1; 1) because the xcoordinates of the points of f start at x = ?1 and continue in...nitely through the positive real numbers.

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

49. f(x) = 1 ? f(x) kk

Use the rule for the derivative of a constant times

a function.

??

?

?

d f(x) = d 1 ? f(x)

dx k

dx k

= 1 f 0(x) k

= f0(x) k

50. Graph the numerical derivative of f (x) = 1:25x3 + 0:01x2 ? 2:9x + 1

for x ranging from ?5 to 5. (a) When x = 4, the derivative equals 57.18.

(b) The derivative crosses the x-axis at approximately ?0:88 and 0:88.

244

51. The demand is given by q = 5000 ? 100p:

Solve for p:

p

=

5000 ? 100

q

?

R(q) = q

5000 ? q 100

= 5000q ? q2 100

R0(q)

=

5000 ? 100

2q

(a)

R0(1000) =

5000 ? 2(1000) 100

= 30

(b)

R0(2500) =

5000 ? 2(2500) 100

=0

(c)

R0(3000) =

5000 ? 2(3000) 100

= ?10

52. C(q) = 3000 ? 20q + 0:03q2

?

R(q) = qp = q

5000 ? q 100

=

q

? 50

?

q 100

?

(from Exercise 51)

The pro...t equation is found as follows.

P =R?C

=

q

? 50

?

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 ? 3000

Therefore, 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

Chapter 4 CALCULATING THE DERIVATIVE

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

53. S(t) = 100 ? 100t?1 S0(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

1000 q2

+

1000

R(q) = 1000q?1 + 1000q R0(q) = ?1000q?2 + 1000

R0(q)

=

1000

?

1000 q2

?

R0(q) = 1000

1

?

1 q2

?

R0(10) = 1000

1

?

1 102

?

R0(10) = 1000

99 100

= 990

The marginal revenue is $990.

55. Pro...t = Revenue ? Cost

P (q) = qp(q) ? C(q)

?

P (q) = q

1000 q2 + 1000

? (0:2q2 + 6q + 50)

= 1000 + 1000q ? 0:2q2 ? 6q ? 50 q

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

=

994

?

0:4q

?

1000 q2

P 0(10)

=

994

?

0:4(10)

?

1000 (10)2

= 994 ? 4 ? 10 = 980

The marginal pro...t is $980.

Section 4.1 Techniques for Finding Derivatives

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)

=

6x

?

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 is 0 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 ? 17:9 cents

2002 when t = 70:

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

(b) C0(t) = 0:00875(2t) ? 0:108(1) = 0:0175t ? 0:108

1982 when t = 50:

C0(50) = 0:0175(50) ? 0:108 = 0:767 cents/year

2002 when t = 70: C0(70) = 0:0175(70) ? 0:108

? 1:12 cents/year

(c) Using a graphing calculator, a cubic function that models the data is C(t) = (?1:790 ? 10?4)t3 + 0:02947t2

? 0:7105t + 3:291:

245

Using the values from the calculator, the rate of change in 1982 is C0(50) ? 0:894 cents/year. Using the values from the calculator, the rate of change in 2002 is C0(70) ? 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) ? 2743

(b) 1960 ? 1900 = 60 M0(60) ? 1598

(c) 1980 ? 1900 = 80 M 0(80) ? 11,983

(d) 2000 ? 1900 = 100 M0(100) ? 29,675

(e) The amount of money in circulation was increasing at the rate of $2743 million/year in 1920 but then began to decrease. The decrease of money continued through the early 1950's. Since then, the amount has again been increasing.

59. N(t) = 0:00437t3:2 N 0(t) = 0:013984t2:2 (a) N 0(5) ? 0:4824 (b) N 0(10) ? 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) = ?4

After 10 units of insulin are injected, the blood sugar level is decreasing at a rate of 4 points per unit of insulin.

(d) G0(25) = ?0:4(25) = ?10

After 25 units of insulin are injected, the blood sugar level is decreasing at a rate of 10 points per unit of insulin.

246

Chapter 4 CALCULATING THE DERIVATIVE

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 ? 121:64t V 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

+

1500 c2

When

c

=

30;

dw dc

=

28

2 3

g/cm.

When the cir-

cumference of the brain is 30 cm, it is increasing

by

28

2 3

g

with

every

centimeter

the

circumference

increases.

63. v = 2:69l1:86 dv = (1:86)2:69l1:86?1 ? 5:00l0:86 dl

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

(a) The problem states that a fetus this formula concerns is at least 18 weeks old. So, the minimum x value should be 18. Considering the gestation time of a cow in general, a meaninful range for this 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 ? 238:1 seconds, or 3 minutes, 58.1 seconds.

(b) dt = 0:0588(1:125s1:125?1) ds = 0:06615s0:125

When

s

=

100;

dt ds

? 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, the world record in the mile stood at 3:43.13. (Ref: )

66. V = C(R0 ? R)R2 = CR0R2 ? CR3

dV dR

=

2CR0R ? 3CR2 CR(2R0 ? 3R)

= =

0 0

CR = 0 or 2R0 ? 3R = 0

R=0

R

=

2 3 R0

Discard R = 0, since a closed windpipe produces

no

airow.

Velocity

is

maximized

when

R

=

2 3

R0

.

67.

BMI =

703w h2

(a) 60200 = 74 in.

703(220) BMI = 742 ? 28

(b)

BMI

=

703w 742

=

24:9

implies

w

=

24:9(74)2 703

?

194:

A 220-lb person needs to lose 26 pounds to get

down 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)

=

?

175,750 653

?

?0:64

For a 125-lb female with a height of 65 in. (50500),

the BMI decreases by 0.64 for each additional inch

of height.

(e) Sample Chart

ht=wt 140 160 180 200 60 27 31 35 39 65 23 27 30 33 70 20 23 26 29 75 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

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