57317-0136147054_04

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

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

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