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Calculus - Chapter 6 Exponentials Logarithms and other...

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Chapter 6 Exponentials, Logarithms and other Transcendental Functions 184 Chapter 6 Section 6.1 5. 4 1 1 ln 4 dt t = 7. 8.2 1 1 ln8.2 dt t = 9. 1 1 (ln 4 ) 4 4 x x x ′ = = 11. [ ] 1 ln(cos ) ( sin ) tan cos x x x x = = − 13. [ ] 1 ln ln 1 ln x x x x x x = + = + 15. 2 2 2 1 2 ln( 2) 2 2 2 x x x x x + = = + + 17. 1/ 2 ln 2 3ln 2 ln 2 3ln 2 1 ln 2 3ln 2 2 7 ln 2 2 + = + = + = 19. 2 1/ 2 2ln3 ln9 ln 3 2ln3 ln3 ln3 1 2ln3 2ln3 ln3 2 1 ln3 2 + = + = + = 21. Use 2 1, u x = + then 2 du x dx = . 2 2 2 1 1 ln ln 1 x dx du u x u c x c = + = + = + + 23. Use u = cos x , then du = –sin x dx . sin sin cos cos 1 ln ln cos x x dx dx x x du u u c x c = − = − = − + = − + 25. Use 2 2 1 u x x = + , then (2 2) 2( 1) du x dx x dx = + = + . 2 2 2 2( 1) 1 1 2 2 1 2 1 1 1 2 1 ln 2 1 ln 2 1 2 x x dx dx x x x x du u u c x x c + + = + + = = + = + + 27. Use u = ln x , then 1 du dx x = . 1 1 1 ln ln 1 ln ln ln dx dx x x x x du u u c x c = = = + = + 29. Use ln 1 u x = + , then 1 du dx x = . 2 2 2 3 3 (ln 1) 1 (ln 1) 1 3 1 (ln 1) 3 x dx x dx x x u du u c x c + = + = = + = + + 31. 2 2 0 0 1 ln 1 ln3 1 dx x x =  + = + 33. 2 1 1 3 3 0 0 1 ln 4 3 4 1 (ln3 ln 4) 3 1 3 ln 3 4 x dx x x = = = 35. 2 2 1/ 2 2 2 2 ln 1 ln( 1) 1 ln( 1) 2 1 1 2 2 1 1 d d x x dx dx d x dx x x x x + = + = + = + = +
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Chapter 6 Exponentials, Logarithms and other Transcendental Functions 185 37. 4 4 5 5 4 5 4 5 ln ln ln( 1) 1 ln ln( 1) 4 5 1 d x d x x dx dx x d d x x dx dx x x x = + + = + = + 39. 2 1 1 1 1 1 2 2ln ln ln dx dx dx x x x x x x = = Use ln u x = , then 1 du dx x = . 1 1 1 1 1 2 2 ln 1 ln 2 1 ln ln 2 dx du x u x u c x c = = + = + 41. y 1 x 0 1 x = 2 y = ln( x – 2) Increasing and concave down on (2, ). 43. y 1 x 0 1 y = ln( x 2 + 1) Increasing on (0, ), decreasing on ( , 0). Concave up on ( 1,1); concave down on ( , 1) (1, ). − ∞ − ∞ − 45. y 1 x 0 1 y = x ln x 1 1 Decreasing on 0, , increasing on , . Concave up on (0, ). e e 47. We will make use of the Reciprocal Property, 1 ln ln b b = − , in our proof. To prove this property, notice that 1 1 ln ln ln ln b b b b b b = = + And since ln ln1 0 b b = = we must have that 1 ln ln b b = − (since the additive inverse of a real number is unique and since y = ln x is one-to-one). Therefore, 1 ln ln 1 ln ln ln ln a a b b a b a b = = + = as desired. 49. 2 2 1 ( ) ln ln s x x x x x = = − 2 1 ( ) ln 2 2 ln (1 2ln ) s x x x x x x x x x x = − = − = − + The critical values are 1/ 2 0 and 1 2ln 0 1 ln 2 x x x x e = + = = − = The function is undefined for x = 0 and the derivative changes from + to – at 1/ 2 x e = . Therefore, there is a maximum at 1/ 2 x e = .
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Chapter 6 Exponentials, Logarithms and other Transcendental Functions 186 51. 2 2 3 ( ) ln ln 1 ( ) , positive for . (ln ) 2 ln ( ) , negative for .
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