206 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS

6.

d

dx2y2 -

d

dx3x3 -

d

dx5 =

d

dx(0)

4yy' - 9x2 - 0 = 0

y' =

9x2

4y

dy

dx (1,2) =

9 ! 12

4 ! 2=9

8

(4-5)

7. y = 3x2 - 5

dy

dt =

d(3x2)

dt!d(5)

dt

dy

dt = 6x

dx

dt

x = 12;

dx

dt = 3

dy

dt = 6·12·3 = 216 (4-6)

8. 25p + x = 1,000 (A) x = 1,000 - 25p (B) x = f(p) = 1,000 - 25p

f'(p) = -25

E(p) = -

pf'(p)

f(p) =

25p

1,000 ! 25p=

p

40 ! p

(C) E(15) =

15

40 ! 15=

15

25=

3

5 = 0.6

Demand is inelastic and insensitive to small changes in price.

(D) Revenue: R(p) = pf(p) = 1,000p - 25p2

(E) From (B), E(25) =

25

40 ! 25=25

15=5

3 = 1.6

Demand is elastic; a price cut will increase revenue. (4-7)

9. y = 100e-0.1x

y’ = 100(-0.1)e-0.1x; y’(0) = 100(-0.1) = -10 (4-2)

10. n 1000 100,000 10,000,000 100,000,000

1 +

2

n

! "

# $

n

7.374312 7.388908 7.389055 7.389056

limn!"

1 +2

n

! "

# $

n ≈ 7.38906 (5 decimal places);

limn!"

1 +2

n

! "

# $

n = e2 (4-1)

11.

d

dz[(ln z)7 + ln z7] =

d

dz[ln z]7 +

d

dz7 ln z

= 7[ln z]6

d

dzln z + 7

d

dzln z

= 7[ln z]6

1

z+7

z

=

!

7(ln z)6 + 7

z =

!

7[(ln z)6 + 1]

z (4-4)

CHAPTER 4 REVIEW 207

12.

d

dxx6 ln x = x6

d

dxln x + (ln x)

d

dxx6

= x6

!

1

x

"

# $

%

& ' + (ln x)6x5 = x5(1 + 6 ln x) (4-3)

13.

d

dx

ex

x6

!

" # $

% & =

x6 d

dxex! e

x d

dxx6

(x6)2 =

x6ex ! 6x5ex

x12 =

xex ! 6ex

x7 =

ex(x ! 6)

x7

(4-3)

14. y = ln(2x3 - 3x)

y' =

1

2x3 ! 3x(6x2 - 3) =

6x2 ! 3

2x3 ! 3x

(4-4)

15. f(x) = ex3-x2

f'(x) = ex3-x2(3x2 - 2x)

= (3x2 - 2x)ex3-x2 (4-4)

16. y = e-2x ln 5x

dy

dx = e-2x

1

5x

! "

# $ (5) + (ln 5x)(e

-2x)(-2)

= e-2x

1

x! 2 ln 5x

" #

$ % =

1 ! 2x ln 5x

xe2x (4-4)

17. f(x) = 1 + e-x f'(x) = e-x(-1) = -e-x

An equation for the tangent line to the graph of f at x = 0 is:

y - y1 = m(x - x1),

where x1 = 0, y1 = f(0) = 1 + e0 = 2, and m = f'(0) = -e0 = -1.

Thus, y - 2 = -1(x - 0) or y = -x + 2. An equation for the tangent line to the graph of f at x = -1 is

y - y1 = m(x - x1), where x1 = -1, y1 = f(-1) = 1 + e, and m = f'(-1) = -e. Thus, y - (1 + e) = -e[x - (-1)] or y - 1 - e = -ex - e and y = -ex + 1. (4-4)

18. x2 - 3xy + 4y2 = 23 Differentiate implicitly: 2x - 3(xy' + y·1) + 8yy' = 0 2x - 3xy' - 3y + 8yy' = 0 8yy' - 3xy' = 3y - 2x (8y - 3x)y' = 3y - 2x

y' =

3y ! 2x

8y ! 3x

y' (!1,2)

=

3 ! 2 " 2("1)

8 ! 2 " 3("1)=

8

19 [Slope at (-1, 2)] (4-5)

208 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS

19. x3 - 2t2x + 8 = 0 3x2x' - (2t2x' + x·4t) + 0 = 0 3x2x' - 2t2x' - 4xt = 0 (3x2 - 2t2)x' = 4xt x' =

4xt

3x2 ! 2t2

x' (!2,2)

=

4 ! 2 ! ("2)

3(22) " 2("2)2=

"16

12 " 8=

"16

4 = -4 (4-5)

20. x - y2 = ey Differentiate implicitly: 1 - 2yy' = eyy' 1 = eyy' + 2yy' 1 = y'(e y + 2y) y' =

1

ey + 2y

y' (1,0)

=

1

e0 + 2 ! 0 = 1

(4-5)

21. ln y = x2 - y2 Differentiate implicitly:

y'

y = 2x - 2yy'

y'

!

1

y+ 2y

"

# $

%

& ' = 2x

y'

!

1 + 2y2

y

"

# $ $

%

& ' ' = 2x

y' =

2xy

1 + 2y2

y' (1,1)

=

2 ! 1 ! 1

1 + 2(1)2=

2

3 (4-5)

22. y2 - 4x2 = 12 Differentiate with respect to t: 2y

dy

dt - 8x

dx

dt = 0

Given:

dx

dt = -2 when x = 1 and y = 4. Therefore,

2·4

dy

dt - 8·1·(-2) = 0

8

dy

dt + 16 = 0

dy

dt = -2.

The y coordinate is decreasing at 2 units per second. (4-6)

23. From the figure, x2 + y2 = 172. Differentiate with respect to t:

2x

dx

dt + 2y

dy

dt = 0 or x

dx

dt + y

dy

dt =0

We are given

dx

dt = -0.5 feet per second. Therefore,

x(-0.5) + y

dy

dt = 0 or

dy

dt =

0.5x

y =

x

2y

x

y 17

CHAPTER 4 REVIEW 209

Now, when x = 8, we have: 82 + y2 = 172 y2 = 289 - 64 = 225 y = 15

Therefore,

dy

dt (8,15) =

8

2(15)=

4

15 ≈ 0.27 ft/sec. (5-5)

(4-6)

24. A = πR2. Given:

dA

dt = 24 square inches per minute.

Differentiate with respect to t:

dA

dt = 2πR

dR

dt

24 = 2πR

dR

dt

Therefore,

dR

dt =

24

2!R=

12

!R.

dR

dt R =12 =

12

! " 12=

1

! ≈ 0.318 inches per minute (4-6)

25. x = f(p) = 20(p - 15)2 0 ≤ p ≤ 15 f'(p) = 40(p - 15)

E(p) = -

pf'(p)

f(p) =

!

"40p(p " 15)

20(p " 15)2=

"2p

p " 15

Elastic: E(p) =

!2p

p ! 15 > 1

-2p < p - 15 (p - 15 < 0 reverses inequality) -3p < -15 p > 5; 5 < p < 15

Inelastic: E(p) =

!2p

p ! 15 < 1

-2p > p - 15 (p - 15 < 0 reverses inequality) -3p > -15 p < 5; 0 < p < 5 (4-7)

26. x = f(p) = 5(20 - p) 0 ≤ p ≤ 20 R(p) = pf(p) = 5p(20 - p) = 100p - 5p2 R'(p) = 100 - 10p = 10(10 - p)

Critical values: p = 10 Sign chart for R'(p):

R'(p)

R(p)

x

Increasing Decreasing

Demand: Inelastic Elastic

0 10 20

+ + + 0 - - -

Test Numbers

p R'(p)

5 50(+)

15 !50(!)

R(p)

p10 20

100

200

300

400

500

Inelastic Elastic

(4-7)

210 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS

27. y = w3, w = ln u, u = 4 - ex

(A) y = [ln(4 - ex)]3

(B)

dy

dx =

dy

dw!dw

du!du

dx

= 3w2 ·

1

u · (-ex) = 3[ln(4 - ex)]2

!

1

4 " ex

#

$ %

&

' ( ("e

x)

=

!

"3ex[ln(4 " ex)]2

4 " ex (4-4)

28. y = 5x2-1

y' = 5x2-1(ln 5)(2x) = 2x5x

2-1(ln 5) (4-4)

29.

d

dxlog5(x

2 - x) =

1

x2 ! x"

1

ln 5"d

dx(x2 - x) =

1

ln 5!2x " 1

x2 " x (4-4)

30.

d

dx

!

ln(x2 + x) =

d

dx[ln(x2 + x)]1/2 =

1

2[ln(x2 + x)]-1/2

d

dxln(x2 + x)

=

1

2[ln(x2 + x)]-1/2

1

x2 + x

d

dx(x

2+ x)

=

1

2[ln(x2 + x)]-1/2 ·

2x + 1

x2 + x =

!

2x + 1

2(x2 + x)[ln(x2 + x)]1 2

(4-4)

31. exy = x2 + y + 1 Differentiate implicitly:

d

dxexy =

d

dxx2 +

d

dxy +

d

dx1

exy(xy' + y) = 2x + y' xexyy' - y' = 2x - yexy

y' =

2x ! yexy

xexy ! 1

y' (0,0)

=

2 ! 0 " 0 ! e0

0 ! e0 " 1 = 0 (4-5)

32. A = πr2, r ≥ 0 Differentiate with respect to t:

dA

dt = 2πr

dr

dt = 6πr since

dr

dt = 3

The area increases at the rate 6πr. This is smallest when r = 0; there is no largest value. (4-6)

CHAPTER 4 REVIEW 211

33. y = x3 Differentiate with respect to t:

dy

dt = 3x2

dx

dt

Solving for

dx

dt, we get

dx

dt =

1

3x2 ·

dy

dt =

5

3x2 since

dy

dt = 5

To find where

dx

dt >

dy

dt, solve the inequality

5

3x2 > 5

1

3x2 > 1

3x2 < 1

-

!

1

3 < x <

!

1

3 or

!

" 3

3 < x <

!

3

3 (4-6)

34. (A) The compound interest formula is: A = P(1 + r)t. Thus, the time for P to double when r = 0.05 and interest is compounded annually can be found by solving

2P = P(1 + 0.05)t or 2 = (1.05)t for t. ln(1.05)t = ln 2 t ln(1.05) = ln 2

t =

ln 2

ln(1.05) ≈ 14.2 or 15 years

(B) The continuous compound interest formula is: A = Pert. Proceeding as above, we have

2P = Pe0.05t or e0.05t = 2. Therefore, 0.05t = ln 2 and

t =

ln 2

.05 ≈ 13.9 years (4-1)

35. A(t) = 100e0.1t A'(t) = 100(0.1)e0.1t = 10e0.1t A'(1) = 11.05 or $11.05 per year A'(10) = 27.18 or $27.18 per year (4-1)

36. R(x) = xp(x) = 1000xe-0.02x R'(x) = 1000[xDxe

-0.02x + e-0.02xDxx]

= 1000[x(-0.02)e-0.02x + e-0.02x] = (1000 - 20x)e-0.02x (4-4)

212 CHAPTER 4 ADDITIONAL DERIVATIVE TOPICS

37. x =

!

5000 " 2p3 = (5000 - 2p3)1/2 Differentiate implicitly with respect to x: 1 =

1

2(5000 - 2p3)-1/2(-6p2)

dp

dx

1 =

!3p2

(5000 ! 2p3)1 2dp

dx

dp

dx =

!(5000 ! 2p3)1 2

3p2 (4-5)

38. Given: R(x) = 36x -

x2

20 and

dx

dt = 10 when x = 250.

Differentiate with respect to t:

dR

dt = 36

dx

dt -

1

20(2x)

dx

dt = 36

dx

dt -

x

10

dx

dt

Thus,

dR

dt x =250 and

dx

dt=10

= 36(10) -

250

10(10)

= $110 per day (4-6)

39. p = 16.8 - 0.002x

x = f(p) =

16.8

0.002!

1

0.002p = 8,400 - 500p

f'(p) = -500

Elasticity of demand: E(p) =

!pf'(p)

f(p) =

500p

8,400 ! 500p =

5p

84 ! 5p

E(8) =

40

84 ! 40=40

44=10

11 < 1

Demand is inelastic, a (small) price increase will increase revenue. (4-7)

40. f(t) = 1,700t + 20,500 f'(t) = 1,700

Relative rate of change:

f'(t)

f(t) =

1,700

1,700t + 20,500

Relative rate of change at t = 30:

1,700

1,700(30) + 20,500 ≈ 0.02378 (4-7)

41. C(t) = 5e-0.3t C'(t) = 5e-0.3t(-0.3) = -1.5e-0.3t After one hour, the rate of change of concentration is C'(1) = -1.5e-0.3(1) = -1.5e-0.3 ≈ -1.111 mg/ml per hour. After five hours, the rate of change of concentration is C'(5) = -1.5e-0.3(5) = -1.5e-1.5 ≈ -0.335 mg/ml per hour. (4-4)

CHAPTER 4 REVIEW 213

42. Given: A = πR2 and

dA

dt = -45 mm2 per day (negative because the area is

decreasing). Differentiate with respect to t:

dA

dt = π2R

dR

dt

-45 = 2πR

dR

dt

dR

dt = -

45

2!R

dR

dt R =15 =

!45

2" # 15=

!3

2" ≈ -0.477 mm per day (4-6)

43. N(t) = 10(1 - e-0.4t) (A) N'(t) = -10e-0.4t(-0.4) = 4e-0.4t N'(1) = 4e-0.4(1) = 4e-0.4 ≈ 2.68. Thus, learning is increasing at the rate of 2.68 units per day after 1 day. N'(5) = 4e-0.4(5) = 4e-2 = 0.54 Thus, learning is increasing at the rate of 0.54 units per day after 5 days.

(B) We solve N’(t) = 0.25 = 4e-0.4t for t

e-0.4t =

!

0.25

4 = 0.625

-0.4t = ln(0.625)

t =

!

ln(0.625)

"0.4 ≈ 6.93

The rate of learning is less than 0.25 after 7 days. (4-4)

44. Given: T = 2

!

1 +1

x3 2

"

# $

%

& ' = 2 + 2x-3/2, and

dx

dt = 3 when x = 9.

Differentiate with respect to t:

dT

dt = 0 + 2

!

"3

2x"5 2

#

$ %

&

' ( dx

dt = -3x-5/2

dxdt

dT

dt x =9 and

dx

dt=3 = -3(9)-5/2(3) = -3 · 3-5 · 3 = -3-3 =

!1

27

≈ -0.037 minutes per operation per hour (4-6)