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57317-0136147054_04 - Section 4.2 4.2.1 y = y(x = 3x2 4x2...

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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 , so dy = 6 x + 8 x 3 dx . 4.2.2: y = y ( x ) = 2 x 1 / 2 3 x 1 / 3 : dy dx = x 1 / 2 + x 4 / 3 , so dy = 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 = 1 + 3 x 2 2 4 x 3 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 ) = 3 x 2 ( x 3) 3 / 2 , so dy = 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 , so 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 , so dy = ( x 2 + 25) 1 / 4 + 1 4 x ( x 2 + 25) 3 / 4 · 2 x dx = 3 x 2 + 50 2( x 2 + 25) 3 / 4 dx. 4.2.8: y = y ( x ) = ( x 2 1) 4 / 3 , so dy = 8 x 3( x 2 1) 7 / 3 dx . 4.2.9: y = y ( x ) = cos x , so dy = sin x 2 x dx . 4.2.10: y = y ( x ) = x 2 sin x , so dy = ( x 2 cos x + 2 x sin x ) dx . 4.2.11: y = y ( x ) = sin 2 x cos 2 x , so dy = (2 cos 2 2 x 2 sin 2 2 x ) dx . 4.2.12: y = y ( x ) = (cos 3 x ) 3 , so dy = 9 cos 2 3 x sin 3 x dx . 4.2.13: y = y ( x ) = sin 2 x 3 x , so dy = 2 x cos 2 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 , so 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 ( x ) = 1 (1 x ) 2 , so f (0) = 1. Therefore f ( x ) = 1 1 x f (0) + f (0)( x 0) = 1 + 1 · x = 1 + x. 4.2.18: f ( x ) = 1 2(1 + x ) 3 / 2 , so f (0) = 1 2 . Therefore f ( x ) = 1 1 + x f (0) + f (0)( x 0) = 1 1 2 x. 4.2.19: f ( x ) = 2(1 + x ), so f (0) = 2. Therefore f ( x ) = (1 + x ) 2 f (0) + f (0)( x 0) = 1 + 2 x . 4.2.20: f ( x ) = 3(1 x ) 2 , so f (0) = 3. Therefore f ( x ) = (1 x ) 3 f (0) + f (0)( x 0) = 1 3 x . 4.2.21: f ( x ) = 3 1 2 x , so f (0) = 3; f ( x ) = (1 2 x ) 3 / 2 f (0) + f (0)( x 0) = 1 3 x . 4.2.22: f (0) = 1, so L ( x ) = 1 x . 4.2.23: If f ( x ) = sin x , then f ( x ) = cos x , so that f (0) = 1. Therefore f ( x ) = sin x f (0) + f (0)( x 0) = 0 + 1 · x = x. 4.2.24: f (0) = 1, so L ( x ) = x . 4.2.25: Choose f ( x ) = x 1 / 3 and a = 27. Then f ( x ) = 1 3 x 2 / 3 , so that f ( a ) = 1 27 . So the linear approximation to f ( x ) near a = 27 is 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 ( x ) = 1 2 x , so that f ( a ) = 1 20 . So the linear approximation to f ( x ) near 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 ( x ) = 1 4 x 3 / 4 , so that f ( a ) = 1 32 . So the linear approximation to f ( x ) near a = 16 is 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 ( x ) = 1 2 x , so that f ( a ) = 1 18 . So the linear approximation to f ( x ) near a = 81 is L ( x ) = 9 2 + 1 18 x . Hence 80 = f (80) L (80) = 161 18 8 . 9444 .
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