Capitulo 2

CHAPTER 2 The Derivative
2.1 Concepts Review 4.

1. tangent line

2. secant line

f (c + h ) − f ( c )
3.
h

4. average velocity

Problem Set 2.1

5–3
1. Slope = =4
2– 3
2 Slope ≈ 1.5

6–4 5.
2. Slope = = –2
4–6

3.

5
Slope ≈
2
Slope ≈ −2 6.

3
Slope ≈ –
2

94 Section 2.1 Instructor’s Resource Manual
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7. y = x 2 + 1 [(2.01)3 − 1.0] − 7
d. msec =
2.01 − 2
a., b.
0.120601
=
0.01
= 12.0601

f (2 + h) – f (2)
e. mtan = lim
h →0 h
[(2 + h)3 – 1] – (23 − 1)
= lim
h →0 h
12h + 6h 2 + h3
= lim
h→0 h
h(12 + 6h + h 2 )
c. m tan = 2 = lim
h→0 h
(1.01)2 + 1.0 − 2 = 12
d. msec =
1.01 − 1 9. f (x) = x 2 – 1
0.0201
= f (c + h ) – f (c )
.01 mtan = lim
h→0 h
= 2.01
[(c + h)2 – 1] – (c 2 – 1)
= lim
f (1 + h) – f (1) h→0 h
e. mtan = lim
h →0 h c 2 + 2ch + h 2 – 1 – c 2 + 1
= lim
[(1 + h)2 + 1] – (12 + 1) h→0 h
= lim
h →0 h h(2c + h)
= lim = 2c
2 + 2h + h 2 − 2 h→0 h
= lim At x = –2, m tan = –4
h →0 h
h(2 + h) x = –1, m tan = –2
= lim x = 1, m tan = 2
h →0 h
= lim (2 + h) = 2 x = 2, m tan = 4
h →0
10. f (x) = x 3 – 3x
3
8. y = x – 1 f (c + h ) – f (c )
mtan = lim
h→0 h
a., b.
[(c + h)3 – 3(c + h)] – (c3 – 3c)
= lim
h→0 h
c3 + 3c 2 h + 3ch 2 + h3 – 3c – 3h – c3 + 3c
= lim
h→0 h
h(3c 2 + 3ch + h 2 − 3)
= lim = 3c 2 – 3
h→0 h
At x = –2, m tan = 9
x = –1, m tan = 0
x = 0, m tan = –3
x = 1, m tan = 0
c. m tan = 12 x = 2, m tan = 9

Instructor’s Resource Manual Section 2.1 95

11. 13. a. 16(12 ) –16(02 ) = 16 ft

b. 16(22 ) –16(12 ) = 48 ft

144 – 64
c. Vave = = 80 ft/sec
3–2

16(3.01) 2 − 16(3)2
d. Vave =
3.01 − 3
0.9616
=
0.01
1 = 96.16 ft/s
f ( x) =
x +1
f (1 + h) – f (1) e. f (t ) = 16t 2 ; v = 32c
mtan = lim v = 32(3) = 96 ft/s
h→0 h
1− 1
= lim 2+ h 2 (32 + 1) – (22 + 1)
h →0 h 14. a. Vave = = 5 m/sec
3– 2
− 2(2h h)
+
= lim [(2.003)2 + 1] − (22 + 1)
h →0 h b. Vave =
1 2.003 − 2
= lim − 0.012009
h→0 2(2 + h) =
1 0.003
=– = 4.003 m/sec
4
1 1
y – = – ( x –1) [(2 + h) 2 + 1] – (22 + 1)
2 4 Vave =
2+h–2
1 4h + h 2
12. f (x) = c. =
x –1 h
f (0 + h) − f (0) = 4 +h
mtan = lim
h →0 h
1 +1 d. f (t ) = t2 + 1
= lim h −1 f (2 + h) – f (2)
h →0 h v = lim
h →0 h
h
h −1 [(2 + h)2 + 1] – (22 + 1)
= lim
h →0
h = lim
h →0 h
1
= lim 4h + h 2
h →0 h − 1 = lim
h →0 h
= −1
y + 1 = –1(x – 0); y = –x – 1 = lim (4 + h)
h →0
=4


f (α + h) – f (α )
15. a. v = lim
h →0 h
2(α + h) + 1 – 2α + 1
= lim
h →0 h
2α + 2h + 1 – 2α + 1
= lim
h →0 h
( 2α + 2h + 1 – 2α + 1)( 2α + 2h + 1 + 2α + 1)
= lim
h →0 h( 2α + 2h + 1 + 2α + 1)
2h
= lim
h →0 h( 2α + 2h + 1 + 2α + 1)
2 1
= = ft/s
2α + 1 + 2α + 1 2α + 1

1 1
b. =
2α + 1 2
2α + 1 = 2
3
2 α + 1= 4; α =
2
The object reaches a velocity of 1 ft/s when t = 3 .
2 2

16. f (t ) = – t2 + 4 t 18. a. 1000(3)2 – 1000(2)2 = 5000
[–(c + h)2 + 4(c + h)] – (– c 2 + 4c)
v = lim
h →0 h 1000(2.5)2 – 1000(2)2 2250
b. = = 4500
– c 2 – 2ch – h 2 + 4c + 4h + c 2 – 4c 2.5 – 2 0.5
= lim
h →0 h
c. f (t ) = 1000t 2
h(–2c – h + 4)
= lim = –2c + 4 1000(2 + h)2 − 1000(2) 2
h →0 h r = lim
–2c + 4 = 0 when c = 2 h→0 h
The particle comes to a momentary stop at 4000 + 4000h + 1000h 2 – 4000
t = 2. = lim
h→0 h
h(4000 + 1000h)
⎡1 ⎤ ⎡1 2 ⎤ = lim = 4000
⎢ 2 (2.01) + 1⎥ – ⎢ 2 (2) + 1⎥ = 0.02005 g
2
17. a. h→0 h
⎣ ⎦ ⎣ ⎦
53 – 33 98
b. rave =
0.02005
= 2.005 g/hr 19. a. dave = = = 49 g/cm
2.01 – 2 5–3 2

1 2 b. f (x) = x 3
c. f (t ) = t +1
2 (3 + h)3 – 33
d = lim
⎡ 1 (2 + h)2 + 1⎤ – ⎡ 1 22 + 1⎤ h →0 h
r = lim ⎣ ⎦ ⎣2 ⎦
2
27 + 27h + 9h 2 + h3 – 27
h →0 h = lim
h→0 h
2 + 2h + 1 h 2 + 1 − 2 − 1
= lim 2
h(27 + 9h + h 2 )
h→0 h = lim = 27 g/cm
h→0 h
= lim
(
h 2+ 1 h
2 )=2
h→0 h
At t = 2, r = 2


R (c + h ) – R (c )
20. MR = lim
h→0 h
[0.4(c + h) – 0.001(c + h)2 ] – (0.4c – 0.001c 2 )
= lim
h→0 h
0.4c + 0.4h – 0.001c 2 – 0.002ch – 0.001h 2 – 0.4c + 0.001c 2
= lim
h→0 h
h(0.4 – 0.002c – 0.001h)
= lim = 0.4 – 0.002c
h→0 h
When n = 10, MR = 0.38; when n = 100, MR = 0.2

2(1 + h)2 – 2(1) 2 c. The building averages 84/7=12 feet from
21. a = lim floor to floor. Since the velocity is zero for
h →0 h
two intervals between time 0 and time 85, the
2 + 4h + 2h 2 – 2 elevator stopped twice. The heights are
= lim
h→0 h approximately 12 and 60. Thus, the elevator
h(4 + 2h) stopped at floors 1 and 5.
= lim =4
h→0 h 25. a. A tangent line at t = 91 has slope
approximately (63 − 48) /(91 − 61) = 0.5 . The
p (c + h ) – p (c )
22. r = lim normal high temperature increases at the rate
h→0 h
of 0.5 degree F per day.
[120(c + h)2 – 2(c + h)3 ] – (120c 2 – 2c3 )
= lim b. A tangent line at t = 191 has approximate
h →0 h
slope (90 − 88) / 30 ≈ 0.067 . The normal
h(240c – 6c 2 + 120h – 6ch – 2h 2 )
= lim high temperature increases at the rate of
h →0 h 0.067 degree per day.
= 240c – 6c 2 c. There is a time in January, about January 15,
When t = 10, r = 240(10) – 6(10) 2 = 1800 when the rate of change is zero. There is also
a time in July, about July 15, when the rate of
t = 20, r = 240(20) – 6(20)2 = 2400
change is zero.
t = 40, r = 240(40) – 6(40)2 = 0
d. The greatest rate of increase occurs around
day 61, that is, some time in March. The
100 – 800 175
23. rave = =– ≈ –29.167 greatest rate of decrease occurs between day
24 – 0 6 301 and 331, that is, sometime in November.
29,167 gal/hr
700 – 400 26. The slope of the tangent line at t = 1930 is
At 8 o’clock, r ≈ ≈ −75 approximately (8 − 6) /(1945 − 1930) ≈ 0.13 . The
6 − 10
75,000 gal/hr rate of growth in 1930 is approximately 0.13
million, or 130,000, persons per year. In 1990,
24. a. The elevator reached the seventh floor at time the tangent line has approximate slope
t = 80 . The average velocity is (24 − 16) /(20000 − 1980) ≈ 0.4 . Thus, the rate of
v avg = (84 − 0) / 80 = 1.05 feet per second growth in 1990 is 0.4 million, or 400,000,
persons per year. The approximate percentage
b. The slope of the line is approximately growth in 1930 is 0.107 / 6 ≈ 0.018 and in 1990 it
60 − 12 is approximately 0.4 / 20 ≈ 0.02 .
= 1.2 . The velocity is
55 − 15
27. In both (a) and (b), the tangent line is always
approximately 1.2 feet per second.
positive. In (a) the tangent line becomes steeper
and steeper as t increases; thus, the velocity is
increasing. In (b) the tangent line becomes flatter
and flatter as t increases; thus, the velocity is
decreasing.


1 2.2 Concepts Review
28. f (t ) = t 3 + t
3
f (c + h) – f (c) f (t ) – f (c)
f ( c + h ) – f (c ) 1. ;
current = lim h t –c
h →0 h
⎡
= lim ⎣
(
1 ( c + h )3 + ( c + h ) ⎤ – 1 c 3 + c
3 ⎦ 3 ) 2. f ′(c )
h→0 h

= lim
(
h c 2 + ch + 1 h 2 + 1
3 ) = c2 + 1 3. continuous; f ( x) = x

h→0 h dy
When t = 3, the current =10 4. f '( x);
dx
c 2 + 1 = 20
2
c = 19
Problem Set 2.2
c = 19 ≈ 4.4
A 20-amp fuse will blow at t = 4.4 s.
f (1 + h) – f (1)
1. f ′(1) = lim
h →0 h
29. A = πr 2 , r = 2t
(1 + h)2 – 12 2h + h 2
A = 4πt2 = lim = lim
h→0 h h →0 h
4π(3 + h)2 – 4π(3)2
rate = lim = lim (2 + h ) = 2
h →0 h h→0
h(24π + 4πh)
= lim = 24π km2/day f (2 + h) – f (2)
h→0 h 2. f ′(2) = lim
h →0 h
4 1 [2(2 + h)]2 – [2(2)]2
30. V = π r 3 , r = t = lim
3 4 h→0 h
1
V = π t3 16h + 4h 2
48 = lim = lim (16 + 4h) = 16
h→0 h h →0
1 (3 + h)3 − 33 27
rate = π lim = π f (3 + h) – f (3)
48 h→0 h 48 3. f ′(3) = lim
9 h →0 h
= π inch 3 / sec
16 [(3 + h)2 – (3 + h)] – (32 – 3)
= lim
h→0 h
31. y = f ( x) = x 3 – 2 x 2 + 1 5h + h 2
= lim = lim (5 + h) = 5
h→0 h h →0
a. m tan = 7 b. m tan = 0
f (4 + h) – f (4)
c. m tan = –1 d. m tan = 17. 92 4. f ′(4) = lim
h →0 h
3–(3+ h )
32. y = f ( x) = sin x sin 2 x 2 1
3+ h
1
– 4–1 3(3+ h ) –1
= lim = lim = lim
h →0 h h →0 h h →0 3(3 + h)
a. m tan = –1.125 b. m tan ≈ –1.0315
1
=–
c. m tan = 0 d. m tan ≈ 1.1891
9

s ( x + h) – s ( x )
33. s = f (t ) = t + t cos 2 t 5. s ′( x) = lim
h →0 h
At t = 3, v ≈ 2.818 [2( x + h) + 1] – (2 x + 1)
= lim
h →0 h
(t + 1)3
34. s = f (t ) = 2h
t+2 = lim =2
h →0 h
At t = 1.6, v ≈ 4.277


f ( x + h) – f ( x ) g ( x + h) – g ( x )
6. f ′( x) = lim 12. g ′( x) = lim
h →0 h h →0 h
[α ( x + h) + β ] – (α x + β ) [( x + h)4 + ( x + h) 2 ] – ( x 4 + x 2 )
= lim = lim
h →0 h h →0 h
αh 4hx3 + 6h 2 x 2 + 4h3 x + h 4 + 2hx + h 2
= lim =α = lim
h →0 h
h →0 h
r ( x + h) – r ( x ) = lim (4 x3 + 6hx 2 + 4h 2 x + h3 + 2 x + h)
7. r ′( x) = lim h →0
h →0 h 3
= 4x + 2x
[3( x + h)2 + 4] – (3 x 2 + 4)
= lim h( x + h) – h( x )
h →0 h 13. h′( x) = lim
h →0 h
6 xh + 3h 2
= lim = lim (6 x + 3h) = 6 x ⎡⎛ 2 2 ⎞ 1⎤
h →0 h h →0 = lim ⎢⎜ – ⎟⋅ ⎥
h → 0 ⎣⎝ x + h x ⎠ h ⎦
f ( x + h) – f ( x ) ⎡ –2h 1 ⎤
8. f ′( x) = lim = lim ⎢ ⋅ ⎥ = lim
–2
=–
2
h →0 h h → 0 ⎣ x ( x + h ) h ⎦ h →0 x ( x + h ) x2
[( x + h)2 + ( x + h) + 1] – ( x 2 + x + 1)
= lim
h →0 h S ( x + h) – S ( x )
14. S ′( x) = lim
2 xh + h + h
2 h →0 h
= lim = lim (2 x + h + 1) = 2 x + 1
h →0 h h →0 ⎡⎛ 1 1 ⎞ 1⎤
= lim ⎢⎜ – ⎟⋅ ⎥
h →0 ⎣⎝ x + h + 1 x + 1 ⎠ h ⎦
f ( x + h) – f ( x ) ⎡
9. f ′( x) = lim –h 1⎤
h →0 h = lim ⎢ ⋅ ⎥
h→0 ⎣ ( x + 1)( x + h + 1) h ⎦
[a( x + h) 2 + b( x + h) + c] – (ax 2 + bx + c)
= lim = lim
–1
=−
1
h →0 h
h→0 ( x + 1)( x + h + 1) ( x + 1) 2
2axh + ah 2 + bh
= lim = lim (2ax + ah + b)
h →0 h h →0 F ( x + h) – F ( x )
= 2ax + b 15. F ′( x) = lim
h →0 h
f ( x + h) – f ( x ) ⎡⎛ 6 6 ⎞ 1⎤
10. f ′( x) = lim = lim ⎢⎜ – ⎟⋅ ⎥
h →0 h →0 ⎢⎜ ( x + h) 2 + 1 x 2 + 1 ⎟ h ⎥
h ⎣⎝ ⎠ ⎦
( x + h) 4 – x 4 ⎡ 6( x + 1) – 6( x + 2hx + h 2 + 1) 1 ⎤
2 2
= lim = lim ⎢ ⋅ ⎥
h →0 h h→0 ⎢ ( x 2 + 1)( x 2 + 2hx + h 2 + 1)
⎣ h⎥
⎦
4hx3 + 6h 2 x 2 + 4h3 x + h 4
= lim ⎡ –12hx – 6h 2
1 ⎤
h →0 h = lim ⎢ ⋅ ⎥
h→0 ⎢ ( x 2 + 1)( x 2 + 2hx + h 2 + 1) h ⎥
= lim (4 x3 + 6hx 2 + 4h 2 x + h3 ) = 4 x3 ⎣ ⎦
h →0 –12 x – 6h 12 x
= lim =−
f ( x + h) – f ( x )
h →0 ( x 2 + 1)( x + 2hx + h + 1)
2 2
( x + 1)2
2
11. f ′( x) = lim
h →0 h
F ( x + h) – F ( x )
[( x + h)3 + 2( x + h)2 + 1] – ( x3 + 2 x 2 + 1) 16. F ′( x) = lim
= lim h →0 h
h →0 h
⎡⎛ x + h –1 x –1 ⎞ 1 ⎤
3hx 2 + 3h 2 x + h3 + 4hx + 2h 2 = lim ⎢⎜ – ⎟⋅ ⎥
= lim h →0 ⎣ ⎝ x + h + 1 x + 1 ⎠ h ⎦
h →0 h
⎡ x 2 + hx + h –1 – ( x 2 + hx – h –1) 1 ⎤
= lim (3x + 3hx + h + 4 x + 2h) = 3 x + 4 x
2 2 2 = lim ⎢ ⋅ ⎥
h →0 h →0 ⎢
⎣ ( x + h + 1)( x + 1) h⎥⎦
⎡ 2h 1⎤ 2
= lim ⎢ ⋅ ⎥=
h→0 ⎣ ( x + h + 1)( x + 1) h ⎦ ( x + 1) 2


G ( x + h) – G ( x )
17. G ′( x ) = lim
h →0 h
⎡⎛ 2( x + h) –1 2 x –1 ⎞ 1 ⎤
= lim ⎢⎜ ⎟⋅
x – 4 ⎠ h⎥
–
h→0 ⎣⎝ x + h – 4 ⎦
⎡ 2 x 2 + 2hx − 9 x − 8h + 4 − (2 x 2 + 2hx − 9 x − h + 4) 1 ⎤ ⎡ –7h 1⎤
= lim ⎢ ⋅ ⎥ = lim ⎢ ⋅ ⎥
h →0 ⎢
⎣ ( x + h − 4)( x − 4) h⎥
⎦ h→0 ⎣ ( x + h – 4)( x – 4) h ⎦
–7 7
= lim =–
h→0 ( x + h – 4)( x – 4) ( x – 4)2

G ( x + h) – G ( x )
18. G ′( x ) = lim
h →0 h
⎡⎛ 2( x + h) 2x ⎞ 1 ⎤ ⎡ (2 x + 2h)( x 2 – x ) – 2 x( x 2 + 2 xh + h 2 – x – h) 1 ⎤
= lim ⎢⎜ – ⎟ ⋅ ⎥ = lim ⎢ ⋅ ⎥
⎜ ⎟
h →0 ⎢⎝ ( x + h) 2 – ( x + h) x 2 – x ⎠ h ⎥ ( x 2 + 2hx + h 2 – x – h)( x 2 – x)
⎣ ⎦ h→0 ⎢⎣ h⎥⎦
⎡ 2
–2h x – 2hx 2
1⎤
= lim ⎢ ⋅ ⎥
h→0 ⎢ ( x + 2hx + h – x – h)( x – x ) h ⎥
⎣
2 2 2
⎦
–2hx – 2 x 2
= lim
h→0 ( x 2 + 2hx + h 2 – x – h)( x 2 – x)
–2 x 2 2
= =–
2 2
( x – x) ( x – 1) 2

g ( x + h) – g ( x )
19. g ′( x) = lim
h →0 h
3( x + h) – 3x
= lim
h→0 h
( 3x + 3h – 3x )( 3x + 3h + 3 x )
= lim
h→0 h( 3 x + 3h + 3x )
3h 3 3
= lim = lim =
h→0 h( 3 x + 3h + 3 x ) h →0 3x + 3h + 3x 2 3x

g ( x + h) – g ( x )
20. g ′( x) = lim
h →0 h
⎡⎛ 1 1 ⎞ 1⎤
= lim ⎢⎜ – ⎟⋅ ⎥
h→0 ⎢⎜ 3( x + h) 3x ⎟ h ⎥
⎣⎝ ⎠ ⎦
⎡ 3x – 3x + 3h 1 ⎤
= lim ⎢ ⋅ ⎥
h→0 ⎢
⎣ 9 x ( x + h) h⎥
⎦
⎡ ( 3 x – 3 x + 3h )( 3 x + 3 x + 3h ) 1 ⎤
= lim ⎢ ⋅ ⎥
h→0 ⎢
⎣ 9 x( x + h)( 3x + 3x + 3h ) h⎥⎦
–3h –3 1
= lim = =–
h→0 h 9 x( x + h)( 3x + 3x + 3h ) 3x ⋅ 2 3x 2 x 3x


H ( x + h) – H ( x )
21. H ′( x ) = lim
h →0 h
⎡⎛ 3 3 ⎞ 1⎤
= lim ⎢⎜ – ⎟⋅ ⎥
h→0 ⎣⎝ x + h – 2 x – 2 ⎠ h⎦
⎡3 x – 2 – 3 x + h – 2 1 ⎤
= lim ⎢ ⋅ ⎥
⎣ ( x + h – 2)( x – 2) h ⎥
h→0 ⎢ ⎦
3( x – 2 – x + h – 2)( x – 2 + x + h – 2)
= lim
h→0 h ( x + h – 2)( x – 2)( x – 2 + x + h – 2)
−3h
= lim
h→0 h[( x – 2) x + h – 2 + ( x + h – 2) x – 2]
–3
= lim
h→0 ( x – 2) x + h – 2 + ( x + h – 2) x – 2
3 3
=– =−
2( x – 2) x – 2 2( x − 2)3 2

H ( x + h) – H ( x )
22. H ′( x) = lim
h →0 h
( x + h) 2 + 4 – x 2 + 4
= lim
h→0 h
⎛ x 2 + 2hx + h 2 + 4 – x 2 + 4 ⎞ ⎛ x 2 + 2hx + h 2 + 4 + x 2 + 4 ⎞
⎜ ⎟⎜ ⎟
= lim ⎝ ⎠⎝ ⎠
h →0
h⎜⎛ x 2 + 2hx + h 2 + 4 + x 2 + 4 ⎞
⎟
⎝ ⎠
2hx + h 2
= lim
h→0
h ⎛ x 2 + 2hx + h 2 + 4 + x 2 + 4 ⎞
⎜ ⎟
⎝ ⎠
2x + h
= lim
h →0 x 2 + 2hx + h 2 + 4 + x 2 + 4
2x x
= =
2 x +4
2
x +4
2

f (t ) – f ( x) f (t ) – f ( x)
23. f ′( x) = lim 24. f ′( x) = lim
t→x t–x t→x t–x
(t − 3t ) – ( x – 3 x)
2 2
(t + 5t ) – ( x3 + 5 x)
3
= lim = lim
t 2 – x 2 – (3t – 3x) t 3 – x3 + 5t – 5 x
= lim = lim
(t – x)(t + x) – 3(t – x) (t – x)(t 2 + tx + x 2 ) + 5(t – x)
= lim = lim
(t – x)(t + x – 3) (t – x)(t 2 + tx + x 2 + 5)
= lim = lim (t + x – 3) = lim
t→x t–x t→x
t→x t–x
=2x–3
= lim (t 2 + tx + x 2 + 5) = 3x 2 + 5
t→x


f (t ) – f ( x) 38. The slope of the tangent line is always −1 .
25. f ′( x) = lim
t→x t–x
⎡⎛ t x ⎞ ⎛ 1 ⎞⎤
= lim ⎢⎜ – ⎟⎜ ⎟⎥
t → x ⎣⎝ t – 5 x – 5 ⎠ ⎝ t – x ⎠ ⎦
tx – 5t – tx + 5 x
= lim
t → x (t – 5)( x – 5)(t – x)
–5(t – x) –5
= lim = lim
t → x (t – 5)( x – 5)(t – x) t → x (t – 5)( x – 5)
5 39. The derivative is positive until x = 0 , then
=− becomes negative.
( x − 5)
2

f (t ) – f ( x)
26. f ′( x) = lim
t→x t–x
⎡⎛ t + 3 x + 3 ⎞ ⎛ 1 ⎞ ⎤
= lim ⎢⎜ ⎟⎜ ⎟
x ⎠ ⎝ t – x ⎠⎥
–
t → x ⎣⎝ t ⎦
3x – 3t –3 3
= lim = lim =–
t → x xt (t – x ) t → x xt x2
40. The derivative is negative until x = 1 , then
27. f (x) = 2 x 3 at x = 5 becomes positive.
28. f (x) = x 2 + 2 x at x = 3

29. f (x) = x 2 at x = 2

30. f (x) = x 3 + x at x = 3

31. f (x) = x 2 at x

32. f (x) = x 3 at x 41. The derivative is −1 until x = 1 . To the right of
x = 1 , the derivative is 1. The derivative is
2 undefined at x = 1 .
33. f (t ) = at t
t

34. f(y) = sin y at y

35. f(x) = cos x at x

36. f(t) = tan t at t

37. The slope of the tangent line is always 2.
42. The derivative is −2 to the left of x = −1 ; from
−1 to 1, the derivative is 2, etc. The derivative is
not defined at x = −1, 1, 3 .


43. The derivative is 0 on ( −3, −2 ) , 2 on ( −2, −1) , 0 1
Δy x +Δx +1 – x +1
1
53. =
on ( −1, 0 ) , −2 on ( 0,1) , 0 on (1, 2 ) , 2 on ( 2,3) Δx Δx
and 0 on ( 3, 4 ) . The derivative is undefined at ⎛ x + 1 – ( x + Δx + 1) ⎞ ⎛ 1 ⎞
=⎜ ⎟⎜ ⎟
x = −2, − 1, 0, 1, 2, 3 . ⎝ ( x + Δx + 1)( x + 1) ⎠ ⎝ Δx ⎠
– Δx
=
( x + Δx + 1)( x + 1)Δx
1
=–
( x + Δx + 1)( x + 1)
dy ⎡ 1 ⎤ 1
= lim − =−
dx Δx →0 ⎢ ( x + Δx + 1)( x + 1) ⎥
⎣ ⎦ ( x + 1) 2

1 ⎛ 1⎞
1+ − ⎜1 + ⎟
44. The derivative is 1 except at x = −2, 0, 2 where Δy x + Δx ⎝ x ⎠
54. =
it is undefined. Δx Δx
1 1 −Δx
− x ( x + Δx )
= x + Δx x =
1
=−
Δx Δx x ( x + Δx )
dy 1 1
= lim − =− 2
dx Δx →0 x ( x + Δx ) x

55.
x + Δx − 1 x − 1
−
45. Δy = [3(1.5) + 2] – [3(1) + 2] = 1.5 Δy x + Δx + 1 x + 1
=
Δx Δx
46. Δy = [3(0.1) 2 + 2(0.1) + 1] – [3(0.0) 2 + 2(0.0) + 1] ( x + 1)( x + Δx − 1) − ( x − 1)( x + Δx + 1) 1
= ×
= 0.23 ( x + Δx + 1)( x + 1) Δx

47. Δy = 1/1.2 – 1/1 = – 0.1667 x 2 + xΔx − x + x + Δx − 1 − ⎡ x 2 + xΔx − x + x − Δx − 1⎤ 1
= ⎣ ⎦×
x + x Δx + x + x + Δ x + 1
2 Δx
48. Δy = 2/(0.1+1) – 2/(0+1) = – 0.1818
2Δx 1 2
= 2 × =
3 3 x + xΔx + x + x + Δx + 1 Δx x 2 + xΔx + x + x + Δx + 1
49. Δy = – ≈ 0.0081
2.31 + 1 2.34 + 1 dy
= lim
2
=
2
=
2
dx Δx →0 x 2 + xΔx + x + x + Δx + 1 x 2 + 2 x + 1 ( x + 1)2
50. Δy = cos[2(0.573)] – cos[2(0.571)] ≈ –0.0036

Δy ( x + Δx) 2 – x 2 2 xΔx + (Δx) 2
51. = = = 2 x + Δx ( x + Δx ) 2 − 1 − x 2 − 1
Δx Δx Δx Δy
56. = x + Δx x
dy Δx Δx
= lim (2 x + Δx) = 2 x
dx Δx →0
(
⎡ x ( x + Δx )2 − x − ( x + Δx ) x 2 − 1 ⎤
=⎢
)
⎥× 1
⎢ x ( x + Δx ) ⎥ Δx
Δy [( x + Δx)3 – 3( x + Δx) 2 ] – ( x3 – 3 x 2 ) ⎣ ⎦
52. =
Δx Δx
( ( )) (
⎡ x x + 2 xΔx + Δx − x − x + x 2 Δx − x − Δx
=⎢
2 2 3
)⎤× 1
⎥
3 x 2 Δx + 3x(Δx)2 – 6 xΔx – 3(Δx) 2 + Δx3 ⎢ x 2 + x Δx ⎥ Δx
= ⎢
⎣ ⎥
⎦
Δx
x 2 Δx + x ( Δx ) + Δx
2
= 3x 2 + 3xΔx – 6 x – 3Δx + (Δx)2 1 x 2 + x Δx + 1
= × = 2
x + x Δx
2 Δx x + x Δx
dy
= lim (3 x 2 + 3 xΔx – 6 x – 3Δx + (Δx)2 ) dy x 2 + xΔx + 1 x 2 + 1
dx Δx→0 = lim = 2
dx Δx → 0 x 2 + xΔx x
= 3x2 – 6 x


1 63. The derivative is 0 at approximately t = 15 and
57. f ′(0) ≈ – ; f ′(2) ≈ 1 t = 201 . The greatest rate of increase occurs at
2
2 about t = 61 and it is about 0.5 degree F per day.
f ′(5) ≈ ; f ′(7) ≈ –3 The greatest rate of decrease occurs at about
3
t = 320 and it is about 0.5 degree F per day. The
58. g ′(–1) ≈ 2; g ′(1) ≈ 0 derivative is positive on (15,201) and negative on
(0,15) and (201,365).
1
g ′(4) ≈ –2; g ′(6) ≈ –
3

59.

64. The slope of a tangent line for the dashed
function is zero when x is approximately 0.3 or
1.9. The solid function is zero at both of these
points. The graph indicates that the solid
function is negative when the dashed function
60. has a tangent line with a negative slope and
positive when the dashed function has a tangent
line with a positive slope. Thus, the solid
function is the derivative of the dashed function.

65. The short-dash function has a tangent line with
zero slope at about x = 2.1 , where the solid
function is zero. The solid function has a tangent
line with zero slope at about x = 0.4, 1.2 and 3.5.
The long-dash function is zero at these points.
5 3 The graph shows that the solid function is
61. a. f (2) ≈ ; f ′(2) ≈ positive (negative) when the slope of the tangent
2 2 line of the short-dash function is positive
f (0.5) ≈ 1.8; f ′(0.5) ≈ –0.6 (negative). Also, the long-dash function is
2.9 − 1.9 positive (negative) when the slope of the tangent
b. = 0.5 line of the solid function is positive (negative).
2.5 − 0.5 Thus, the short-dash function is f, the solid
c. x=5 function is f ' = g , and the dash function is g ' .
d. x = 3, 5
66. Note that since x = 0 + x, f(x) = f(0 + x) = f(0)f(x),
e. x = 1, 3, 5 hence f(0) = 1.
f. x=0 f ( a + h) – f ( a )
f ′(a ) = lim
h →0 h
3
g. x ≈ −0.7, and 5 < x < 7 f ( a ) f ( h) – f ( a )
2 = lim
h→0 h
62. The derivative fails to exist at the corners of the f ( h) – 1 f (h) – f (0)
graph; that is, at t = 10, 15, 55, 60, 80 . The = f (a ) lim = f (a) lim
h →0 h h →0 h
derivative exists at all other points on the interval = f (a ) f ′(0)
(0,85) . f ′ ( a) exists since f ′ (0 ) exists.


67. If f is differentiable everywhere, then it is b. If f is an even function,
continuous everywhere, so f (t ) − f ( x0 )
lim − f ( x ) = lim− ( mx + b ) = 2 m + b = f (2) = 4 f ′(– x0 ) = lim . Let u = –t, as
x→2 x→ 2 t →− x0 t + x0
and b = 4 – 2m. f (−u ) − f ( x0 )
For f to be differentiable everywhere, above, then f ′(− x0 ) = lim
u → x0 −u + x0
f ( x) − f (2)
f ′(2) = lim must exist. f (u ) − f ( x0 ) f (u ) − f ( x0 )
x→2 x−2 = lim = − lim
u → x0 −(u − x0 ) u → x0 u − x0
f ( x) − f (2) x2 − 4
lim = lim = lim ( x + 2) = 4 = − f ′ (x 0 ) = −m.
x → 2+ x−2 x → 2+ x − 2 x → 2+
f ( x) − f (2) mx + b − 4 70. Say f(–x) = –f(x). Then
lim = lim
− x−2 − x−2 f (– x + h) – f (– x)
x→2 x→2 f ′(– x) = lim
mx + 4 − 2m − 4 m( x − 2) h →0 h
= lim = lim =m – f ( x – h) + f ( x ) f ( x – h) – f ( x )
x → 2− x−2 x →2 − x−2 = lim = – lim
h→0 h h →0 h
Thus m = 4 and b = 4 – 2(4) = –4
f [ x + (– h)] − f ( x)
= lim = f ′( x) so f ′ ( x ) is
f ( x + h) – f ( x ) + f ( x ) – f ( x – h ) – h →0 –h
68. f s ( x) = lim
h →0 2h an even function if f(x) is an odd function.
⎡ f ( x + h ) – f ( x ) f ( x – h) – f ( x ) ⎤ Say f(–x) = f(x). Then
= lim ⎢ + ⎥ f (– x + h) – f (– x)
h→0 ⎣ 2h –2h ⎦ f ′(– x) = lim
h →0 h
1 f ( x + h) – f ( x ) 1 f [ x + (– h)] – f ( x)
= lim + lim f ( x – h) – f ( x )
2 h →0 h 2 – h →0 –h = lim
h→0 h
1 1
= f ′( x) + f ′( x ) = f ′( x). f [ x + (– h)] – f ( x)
2 2 = – lim = – f ′( x) so f ′ (x)
For the converse, let f (x) = x . Then – h→0 –h
is an odd function if f(x) is an even function.
h – –h h–h
f s (0) = lim = lim =0
h→0 2h h →0 2h 71.
but f ′ (0) does not exist.

f (t ) − f ( x0 )
69. f ′( x0 ) = lim , so
t→x
0
t − x0
f (t ) − f (− x0 )
f ′(− x0 ) = lim
t →− x t − (− x0 )
0 8 ⎛ 8⎞
a. 0< x< ; ⎜ 0, ⎟
f (t ) − f (− x0 ) 3 ⎝ 3⎠
= lim
t →− x 0 t + x0
8 ⎡ 8⎤
b. 0≤ x≤ ; 0,
a. If f is an odd function, 3 ⎢ 3⎥
⎣ ⎦
f (t ) − [− f (− x0 )]
f ′(− x0 ) = lim
t →− x0 t + x0 c. A function f(x) decreases as x increases when
f ′ ( x ) < 0.
f (t ) + f (− x0 )
= lim .
t →− x0 t + x0 72.
Let u = –t. As t → − x0 , u → x 0 and so
f (−u ) + f ( x0 )
f ′(− x0 ) = lim
u → x0 −u + x0
− f (u ) + f ( x0 ) −[ f (u ) − f ( x0 )]
= lim = lim
u → x0 −(u − x0 ) u → x0 −(u − x0 )
f (u ) − f ( x0 ) a. π < x < 6.8 b. π < x < 6.8
= lim = f ′( x0 ) = m.
u → x0 u − x0
c. A function f(x) increases as x increases when
f ′ ( x ) > 0.


2.3 Concepts Review 13. Dx ( x 4 + x3 + x 2 + x + 1)
1. the derivative of the second; second; = Dx ( x 4 ) + Dx ( x3 ) + Dx ( x 2 ) + Dx ( x) + Dx (1)
f (x) g′ (x ) + g(x) f ′( x)
= 4 x3 + 3 x 2 + 2 x + 1
2. denominator; denominator; square of the
g ( x) f ′( x) – f ( x) g ′( x) 14. Dx (3x 4 – 2 x3 – 5 x 2 + πx + π2 )
denominator;
g 2 ( x) = 3Dx ( x 4 ) – 2 Dx ( x3 ) – 5Dx ( x 2 )

n– 1 n –1
+ πDx ( x) + Dx (π2 )
3. nx h ; nx
= 3(4 x3 ) – 2(3 x 2 ) – 5(2 x) + π(1) + 0
4. kL(f); L(f) + L(g); Dx
= 12 x3 – 6 x 2 –10 x + π

Problem Set 2.3 15. Dx (πx 7 – 2 x5 – 5 x –2 )
= πDx ( x7 ) – 2 Dx ( x5 ) – 5 Dx ( x –2 )
1. Dx (2 x ) = 2 Dx ( x ) = 2 ⋅ 2 x = 4 x
2 2
= π(7 x6 ) – 2(5 x 4 ) – 5(–2 x –3 )

2. Dx (3x3 ) = 3Dx ( x3 ) = 3 ⋅ 3x 2 = 9 x 2 = 7 πx 6 –10 x 4 + 10 x –3

3. Dx (πx ) = πDx ( x) = π ⋅1 = π 16. Dx ( x12 + 5 x −2 − πx −10 )
= Dx ( x12 ) + 5Dx ( x −2 ) − πDx ( x −10 )
4. Dx (πx ) = πDx ( x ) = π ⋅ 3 x = 3πx
3 3 2 2
= 12 x11 + 5(−2 x −3 ) − π(−10 x −11 )

5. Dx (2 x –2 ) = 2 Dx ( x –2 ) = 2(–2 x –3 ) = –4 x –3 = 12 x11 − 10 x −3 + 10πx −11

⎛ 3 ⎞
6. Dx (–3 x –4 ) = –3Dx ( x –4 ) = –3(–4 x –5 ) = 12 x –5 17. Dx ⎜ + x –4 ⎟ = 3Dx ( x –3 ) + Dx ( x –4 )
3
⎝x ⎠
⎛π⎞ = 3(–3 x –4 ) + (–4 x –5 ) = –
9
– 4 x –5
7. Dx ⎜ ⎟ = πDx ( x –1 ) = π(–1x –2 ) = – πx –2
⎝x⎠ x4
π
=– 2
x 18. Dx (2 x –6 + x –1 ) = 2 Dx ( x –6 ) + Dx ( x –1 )
= 2(–6 x –7 ) + (–1x –2 ) = –12 x –7 – x –2
⎛α ⎞
8. Dx ⎜ ⎟ = α Dx ( x –3 ) = α (–3x –4 ) = –3α x –4
⎝ x3 ⎠ ⎛2 1 ⎞
⎟ = 2 Dx ( x ) – Dx ( x )
–1 –2
19. Dx ⎜ –
3α ⎝ x x2 ⎠
=–
x4 2 2
= 2(–1x –2 ) – (–2 x –3 ) = – +
2
x x3
⎛ 100 ⎞
9. Dx ⎜ = 100 Dx ( x –5 ) = 100(–5 x –6 )
5 ⎟
⎝ x ⎠ ⎛ 3 1 ⎞
⎟ = 3 Dx ( x ) – Dx ( x )
–3 –4
20. Dx ⎜ –
500 ⎝x 3
x4 ⎠
= –500 x –6 = –
x6 9 4
= 3(–3 x –4 ) – (–4 x –5 ) = – +
4
x x5
⎛ 3α ⎞ 3α 3α
10. Dx ⎜ = Dx ( x –5 ) = (–5 x –6 )
5⎟
⎝ 4x ⎠ 4 4 ⎛ 1 ⎞ 1
21. Dx ⎜ + 2 x ⎟ = Dx ( x –1 ) + 2 Dx ( x)
15α –6 15α ⎝ 2x ⎠ 2
=– x =–
4 4 x6 1 1
= (–1x –2 ) + 2(1) = – +2
2 2 x2
11. Dx ( x 2 + 2 x) = Dx ( x 2 ) + 2 Dx ( x ) = 2 x + 2

12. Dx (3x 4 + x3 ) = 3Dx ( x 4 ) + Dx ( x3 )
= 3(4 x3 ) + 3x 2 = 12 x3 + 3 x 2


⎛ 2 2⎞ 2 ⎛2⎞ 26. Dx [(–3 x + 2)2 ]
22. Dx ⎜ – ⎟ = Dx ( x –1 ) – Dx ⎜ ⎟
⎝ 3x 3 ⎠ 3 ⎝3⎠ = (–3 x + 2) Dx (–3 x + 2) + (–3 x + 2) Dx (–3x + 2)
2 2 = (–3x + 2)(–3) + (–3x + 2)(–3) = 18x – 12
= (–1x –2 ) – 0 = –
3 3x 2
27. Dx [( x 2 + 2)( x3 + 1)]
23. Dx [ x( x 2 + 1)] = x Dx ( x 2 + 1) + ( x 2 + 1) Dx ( x) = ( x 2 + 2) Dx ( x3 + 1) + ( x3 + 1) Dx ( x 2 + 2)
= x(2 x) + ( x + 1)(1) = 3 x + 1
2 2
= ( x 2 + 2)(3x 2 ) + ( x3 + 1)(2 x)
= 3x 4 + 6 x 2 + 2 x 4 + 2 x
24. Dx [3 x( x3 –1)] = 3 x Dx ( x3 –1) + ( x3 –1) Dx (3 x)
= 5x4 + 6 x2 + 2 x
= 3x(3 x 2 ) + ( x3 –1)(3) = 12 x3 – 3
28. Dx [( x 4 –1)( x 2 + 1)]
25. Dx [(2 x + 1) ] 2
= ( x 4 –1) Dx ( x 2 + 1) + ( x 2 + 1) Dx ( x 4 –1)
= (2 x + 1) Dx (2 x + 1) + (2 x + 1) Dx (2 x + 1)
= (2 x + 1)(2) + (2 x + 1)(2) = 8 x + 4 = ( x 4 –1)(2 x) + ( x 2 + 1)(4 x3 )
= 2 x5 – 2 x + 4 x5 + 4 x3 = 6 x 5 + 4 x3 – 2 x

29. Dx [( x 2 + 17)( x3 – 3 x + 1)]
= ( x 2 + 17) Dx ( x3 – 3 x + 1) + ( x3 – 3x + 1) Dx ( x 2 + 17)
= ( x 2 + 17)(3 x 2 – 3) + ( x3 – 3x + 1)(2 x)
= 3x 4 + 48 x 2 – 51 + 2 x 4 – 6 x 2 + 2 x
= 5 x 4 + 42 x 2 + 2 x – 51

30. Dx [( x 4 + 2 x)( x3 + 2 x 2 + 1)] = ( x 4 + 2 x) Dx ( x3 + 2 x 2 + 1) + ( x3 + 2 x 2 + 1) Dx ( x 4 + 2 x)
= ( x 4 + 2 x)(3 x 2 + 4 x) + ( x3 + 2 x 2 + 1)(4 x3 + 2)
= 7 x6 + 12 x5 + 12 x3 + 12 x 2 + 2

31. Dx [(5 x 2 – 7)(3x 2 – 2 x + 1)] = (5 x 2 – 7) Dx (3x 2 – 2 x + 1) + (3x 2 – 2 x + 1) Dx (5 x 2 – 7)
= (5 x 2 – 7)(6 x – 2) + (3 x 2 – 2 x + 1)(10 x)
= 60 x3 – 30 x 2 – 32 x + 14

32. Dx [(3 x 2 + 2 x)( x 4 – 3 x + 1)] = (3 x 2 + 2 x) Dx ( x 4 – 3 x + 1) + ( x 4 – 3 x + 1) Dx (3x 2 + 2 x)
= (3 x 2 + 2 x)(4 x3 – 3) + ( x 4 – 3 x + 1)(6 x + 2)
= 18 x5 + 10 x 4 – 27 x 2 – 6 x + 2

⎛ 1 ⎞ (3x 2 + 1) Dx (1) – (1) Dx (3 x 2 + 1)
33. Dx ⎜ ⎟=
⎝ 3x2 + 1 ⎠ (3 x 2 + 1)2
(3 x 2 + 1)(0) – (6 x) 6x
= =–
(3 x + 1)
2 2
(3x + 1) 2
2

⎛ 2 ⎞ (5 x 2 –1) Dx (2) – (2) Dx (5 x 2 –1)
34. Dx ⎜ ⎟=
⎝ 5 x 2 –1 ⎠ (5 x 2 –1) 2
(5 x 2 –1)(0) – 2(10 x) 20 x
= =–
2 2
(5 x –1) (5 x 2 –1)2


⎛ 1 ⎞ (4 x 2 – 3x + 9) Dx (1) – (1) Dx (4 x 2 – 3 x + 9)
35. Dx ⎜ ⎟=
⎝ 4 x 2 – 3x + 9 ⎠ (4 x 2 – 3x + 9)2
(4 x 2 – 3x + 9)(0) – (8 x – 3) 8x − 3
= =–
(4 x – 3 x + 9)
2 2
(4 x – 3x + 9)2
2

−8 x + 3
=
(4 x 2 – 3x + 9)2

⎛ 4 ⎞ (2 x3 – 3 x) Dx (4) – (4) Dx (2 x3 – 3 x)
36. Dx ⎜ ⎟ =
⎝ 2 x3 – 3x ⎠ (2 x3 – 3 x)2
(2 x3 – 3 x)(0) – 4(6 x 2 – 3) –24 x 2 + 12
= =
(2 x3 – 3 x)2 (2 x3 – 3x) 2

⎛ x –1 ⎞ ( x + 1) Dx ( x –1) – ( x –1) Dx ( x + 1)
37. Dx ⎜ ⎟=
⎝ x +1⎠ ( x + 1)2
( x + 1)(1) – ( x –1)(1) 2
= =
( x + 1) 2
( x + 1)2

⎛ 2 x –1 ⎞ ( x –1) Dx (2 x –1) – (2 x –1) Dx ( x –1)
38. Dx ⎜ ⎟=
⎝ x –1 ⎠ ( x –1) 2
( x –1)(2) – (2 x –1)(1) 1
= =–
2
( x –1) ( x –1) 2

⎛ 2 x 2 – 1 ⎞ (3 x + 5) Dx (2 x 2 –1) – (2 x 2 –1) Dx (3 x + 5)
39. Dx ⎜ ⎟ =
⎜ 3x + 5 ⎟ (3 x + 5)2
⎝ ⎠
(3 x + 5)(4 x) – (2 x 2 – 1)(3)
=
(3x + 5) 2
6 x 2 + 20 x + 3
=
(3x + 5)2

⎛ 5 x – 4 ⎞ (3 x 2 + 1) Dx (5 x – 4) – (5 x – 4) Dx (3x 2 + 1)
40. Dx ⎜ ⎟=
⎝ 3x2 + 1 ⎠ (3 x 2 + 1) 2
(3 x 2 + 1)(5) – (5 x – 4)(6 x)
=
(3x 2 + 1)2
−15 x 2 + 24 x + 5
=
(3x 2 + 1)2

⎛ 2 x 2 – 3 x + 1 ⎞ (2 x + 1) Dx (2 x 2 – 3x + 1) – (2 x 2 – 3x + 1) Dx (2 x + 1)
41. Dx ⎜ ⎟ =
⎜ 2x +1 ⎟ (2 x + 1)2
⎝ ⎠
(2 x + 1)(4 x – 3) – (2 x 2 – 3 x + 1)(2)
=
(2 x + 1)2
4 x2 + 4 x – 5
=
(2 x + 1) 2


Capitulo 2

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