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Capitulo 2
1.
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.
2.
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 © 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.
3.
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 96 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.
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 Instructor’s Resource Manual Section 2.1 97 © 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.
5.
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. 98 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.
6.
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 Instructor’s Resource Manual Section 2.2 99 © 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.
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) 2 100 Section 2.2 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.
8.
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 Instructor’s Resource Manual Section 2.2 101 © 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.
9.
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→x t–x t→x t–x t 2 – x 2 – (3t – 3x) t 3 – x3 + 5t – 5 x = lim = lim t→x t–x t→x t–x (t – x)(t + x) – 3(t – x) (t – x)(t 2 + tx + x 2 ) + 5(t – x) = lim = lim t→x t–x t→x t–x (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 102 Section 2.2 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.
10.
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 . Instructor’s Resource Manual Section 2.2 103 © 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.
11.
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 104 Section 2.2 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.
12.
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. Instructor’s Resource Manual Section 2.2 105 © 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.
13.
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. 106 Section 2.2 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.
14.
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 Instructor’s Resource Manual Section 2.3 107 © 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.
15.
⎛ 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)2 108 Section 2.3 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.
16.
⎛
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 Instructor’s Resource Manual Section 2.3 109 © 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.
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