57317-0136147054_04

# 57317-0136147054_04 - Section 4.2 4.2.1: y = y (x) = 3x2...

This preview shows pages 1–4. Sign up to view the full content.

Section 4.2 4.2.1: y = y ( x )=3 x 2 4 x 2 : dy dx =6 x +8 x 3 ,s o dy = ± 6 x + 8 x 3 ² dx . 4.2.2: y = y ( x )=2 x 1 / 2 3 x 1 / 3 : dy dx = x 1 / 2 + x 4 / 3 o dy = ³ x 1 / 2 + x 4 / 3 ´ dx = 1+ x 5 / 6 x 4 / 3 dx. 4.2.3: y = y ( x )= x (4 x 3 ) 1 / 2 : dy dx =1 1 2 (4 x 3 ) 1 / 2 · ( 3 x 2 ), so dy = ± 3 x 2 2 4 x 3 ² dx = 3 x 2 +2 4 x 3 2 4 x 3 dx. 4.2.4: y = y ( x 1 x x : dy = 1 1 2 x 1 / 2 ( x x ) 2 dx = 1 2 x 2 x ( x x ) 2 dx . 4.2.5: y = y ( x x 2 ( x 3) 3 / 2 o dy = µ 6 x ( x 3) 3 / 2 + 9 2 x 2 ( x 3) 1 / 2 dx = 3 2 (7 x 2 12 x ) x 3 dx. 4.2.6: y = y ( x x x 2 4 o dy = ( x 2 4) 2 x 2 ( x 2 4) 2 dx = x 2 +4 ( x 2 4) 2 dx . 4.2.7: y = y ( x x ( x 2 + 25) 1 / 4 o dy =( x 2 + 25) 1 / 4 + 1 4 x ( x 2 + 25) 3 / 4 · 2 xdx = 3 x 2 +50 2( x 2 + 25) 3 / 4 dx. 4.2.8: y = y ( x )=( x 2 1) 4 / 3 o dy = 8 x 3( x 2 1) 7 / 3 dx . 4.2.9: y = y ( x )=cos x o dy = sin x 2 x dx . 4.2.10: y = y ( x x 2 sin x o dy x 2 cos x x sin x ) dx . 4.2.11: y = y ( x )=s in2 x cos2 x o dy =(2cos 2 2 x 2sin 2 2 x ) dx . 4.2.12: y = y ( x )=(cos3 x ) 3 o dy = 9cos 2 3 x sin 3 . 4.2.13: y = y ( x sin 2 x 3 x o dy = 2 x x sin 2 x 3 x 2 dx . 4.2.14: dy =(3 x 2 2 x 3 ) e 2 x dx . 260

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
4.2.15: y = y ( x )= 1 1 x sin x ,s o dy = x cos x +sin x (1 x sin x ) 2 dx . 4.2.16: dy = 1 ln x x 2 dx . 4.2.17: f 0 ( x 1 (1 x ) 2 o f 0 (0) = 1. Therefore f ( x 1 1 x f (0) + f 0 (0)( x 0) = 1 + 1 · x =1+ x. 4.2.18: f 0 ( x 1 2(1 + x ) 3 / 2 o f 0 (0) = 1 2 . Therefore f ( x 1 1+ x f (0) + f 0 (0)( x 0) = 1 1 2 x. 4.2.19: f 0 ( x )=2(1+ x ), so f 0 (0) = 2. Therefore f ( x )=(1+ x ) 2 f (0) + f 0 (0)( x 0) = 1 + 2 x . 4.2.20: f 0 ( x 3(1 x ) 2 o f 0 (0) = 3. Therefore f ( x )=(1 x ) 3 f (0) + f 0 (0)( x 0) = 1 3 x . 4.2.21: f 0 ( x 3 1 2 x o f 0 (0) = 3; f ( x 2 x ) 3 / 2 f (0) + f 0 (0)( x 0) = 1 3 x . 4.2.22: f 0 (0) = 1, so L ( x )=1 x . 4.2.23: If f ( x )=s in x ,t h e n f 0 ( x )=cos x ot h a t f 0 (0) = 1. Therefore f ( x x f (0) + f 0 (0)( x 0) = 0 + 1 · x = x. 4.2.24: f 0 (0) = 1, so L ( x x . 4.2.25: Choose f ( x x 1 / 3 and a = 27. Then f 0 ( x 1 3 x 2 / 3 oth a t f 0 ( a 1 27 . So the linear approximation to f ( x )near a =27is L ( x )=2+ 1 27 x . Hence 3 25 = f (25) L (25) = 79 27 2 . 9259 . A calculator reports that f (25) is actually closer to 2 . 9240, but the linear approximation is fairly accurate, with an error of only about 0 . 0019. 4.2.26: Choose f ( x x and a = 100. Then f 0 ( x 1 2 x a t f 0 ( a 1 20 . So the linear approximation to f ( x a = 100 is L ( x )=5+ 1 20 x . Hence 102 = f (102) L (102) = 101 10 =10 . 1000 . 261
A calculator reports that f (25) is actually closer to 10 . 0995, but the linear approximation is quite accurate, with an error of only about 0 . 0005. 4.2.27: Choose f ( x )= x 1 / 4 and a = 16. Then f 0 ( x 1 4 x 3 / 4 ,s oth a t f 0 ( a 1 32 . So the linear approximation to f ( x )near a =16is L ( x 3 2 + 1 32 x . Hence 4 15 = f (15) L (15) = 63 32 =1 . 96875 . A calculator reports that f (15) is actually closer to 1 . 96799. 4.2.28: Choose f ( x x and a = 81. Then f 0 ( x 1 2 x a t f 0 ( a 1 18 . So the linear approximation to f ( x a =81is L ( x 9 2 + 1 18 x . Hence 80 = f (80) L (80) = 161 18 8 . 9444 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This document was uploaded on 11/28/2010.

### Page1 / 169

57317-0136147054_04 - Section 4.2 4.2.1: y = y (x) = 3x2...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online