SalasSV_07_05 - 406 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS...

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Unformatted text preview: 406 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS PROJECT 7.4 Estimating the Number e Since e is an irrational number, we cannot hope to express e as a repeating or terminating decimal. In this project, we derive a numerical estimate for e from the numerical representation of the natural logarithm function: x by showing: a. ln b. ln 1+ 1 n 1 1+ n ≤ ≥ 1 n (HINT : 1 ≤ 1 on[1, 1 + 1/n]. ) t ln x = 1 1 dt , t x > 0. Let n be a positive integer. Then Problem 2. 1 ln 1 + n = 1 1+1/n 1 n+1 1 1 (HINT : ≥ on [1, 1 + 1/n]. ) t 1 + 1/n n n+1 Show that 1+ 1 n ≤e ≤ 1+ 1 n . 1 dt t (1) y The result in Problem 2 is an elegant characterization of e but not a very efficient tool for calculating e. For example, 1+ 1 100 100 ≈ 2. 7048138 and 1+ 1 101 100 ≈ 2. 7318619 gives e rounded off to one decimal place: e ≈ 2. 7. Problem 3. Evaluate (1) for n = 1, 000, n = 10, 000, n = 100, 000, and n = 1, 000, 000. What is the accuracy of these estimates for e? y= 1 1+ 1 n 1 t x Based on these results of Problem 2, we conjecture that (2) 1+ 1 x x → e as x → ∞. Problem 1. Show that 1 1 ≤ ln 1 + n+1 n ≤ 1 n Problem 4. Prove that (2) holds. ( HINT: At x = 1, the logarithm function has derivative h→ 0 lim ln (1 + h) − ln 1 =1 h . 7.5 ARBITRARY POWERS; OTHER BASES Arbitrary Powers: The Function f (x) = xr The elementary notion of exponent applies only to rational numbers. Expressions such as 105 , 21/3 , 7−4/5 , π −1/2 make sense, but so far we have attached no meaning to expressions such as 10 2 , √ 2π , 7− 3 , √ π e. The extension of our sense of exponent to allow for irrational exponents is conveniently done by making use of the logarithm function and the exponential function. The heart of the matter is to observe that for x > 0 and p/q rational xp/q = e ( p/q) ln x . 7.5 ARBITRARY POWERS; OTHER BASES 407 (To verify this, take the natural log of both sides.) We define xz for irrational z by setting x z = e z ln x . We then have the following result: if x > 0, then x r = e r ln x (7.5.1) for all real numbers r . In particular √ 10 2 =e √ 2 ln 10 , 2π = eπ ln 2 , 7− √ 3 = e− √ 3 ln 7 , π e = ee ln π . With this extended sense of exponent, the usual laws of exponents xr+s = xr xs , xr−s = xr , xs (7.5.2) (xr )s = xrs still hold: xr+s = e(r+s) ln x = er ln x · es ln x = xr xs , xr−s = e(r−s) ln x = er ln x · e−s ln x = (xr )s = es ln x = ers ln x = xrs . In Chapter 3 we proved that d (xp )/dx = pxp−1 for any rational number p. We can now extend this result to arbitrary powers. For any real number r: dr (x ) = r xr−1 dx r er ln x xr = s, es ln x x (7.5.3) for all x > 0. PROOF d dr d r (x ) = (er ln x ) = er ln x (r ln x) = xr = r xr−1 . dx dx dx x You can also write f (x) = xr and use logarithmic differentiation: ln f (x) = r ln x r f (x ) = f (x ) x r xr r f (x ) f (x ) = = = r xr−1 . x x Thus √√ d √2 (x ) = 2 x 2−1 , dx dπ ( x ) = π x π −1 . dx 408 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS If u is a positive differentiable function of x and r is any real number, then, by the chain rule, dr du (u ) = r ur−1 . dx dx (7.5.4) PROOF dr d r du du (u ) = (u ) = r ur−1 . dx du dx dx For example, √ √ √ √ √ d [(x2 + 5) 3 ] = 3(x2 + 5) 3−1 (2x) = 2 3 x(x2 + 5) 3−1 . dx Example 1 SOLUTION Find d [(x2 + 1)3x ]. dx 2 +1) One way to find this derivative is to observe that (x2 + 1)3x = e3x ln (x and then differentiate: d d 2x 2 2 [(x2 + 1)3x ] = [e3x ln (x +1) ] = e3x ln (x +1) 3x · 2 + 3 ln (x2 + 1) dx dx x +1 = (x2 + 1)3x 6x2 + 3 ln (x2 + 1) . x2 + 1 Another way to find the derivative of (x2 + 1)3x is to set f (x) = (x2 + 1)3x and use logarithmic differentiation: ln f (x) = 3x · ln (x2 + 1) 2x f (x ) 6x2 = 3x · 2 + [ ln (x2 + 1)](3) = 2 + 3 ln (x2 + 1) f (x ) x +1 x +1 f (x ) = f (x ) 6x 2 + 3 ln (x2 + 1) x2 + 1 6x2 + 3 ln (x2 + 1) . x2 + 1 = (x2 + 1)3x Each derivative formula gives rise to a companion integral formula. The integral version of (7.5.3) takes the form xr+1 + C, r+1 (7.5.5) xr dx = for r = −1. Note the exclusion of r = −1. What is the integral if r = −1? Example 2 Find x3 dx. (2x4 + 1)π 7.5 ARBITRARY POWERS; OTHER BASES SOLUTION 409 Set u = 2x4 + 1, du = 8x3 dx. 1 8 u1−π 1−π +C = (2x4 + 1)1−π + C. 8(1 − π ) 1 x3 dx = 4 + 1)π (2x 8 u−π du = Base p : The Function f (x) = p x To form the function f (x) = xr we take a positive variable x and raise it to a constant power r . To form the function f (x) = p x we take a positive constant p and raise it to a variable power x. Since 1x = 1 for all x, the function is of interest only if p = 1. Functions of the form f (x) = p x are called exponential functions with base p. The high status enjoyed by Euler’s number e comes from the fact that dx (e ) = e x . dx For other bases the derivative has an extra factor: (7.5.6) d (p x ) = p x ln p. dx PROOF dx d (p ) = (e x ln p ) = e x ln p ln p = p x ln p. dx dx For example, dx (2 ) = 2x ln 2 dx and d (10x ) = 10x ln 10. dx If u is a differentiable function of x, then, by the chain rule, du du (p ) = pu ln p . dx dx (7.5.7) PROOF d u du du du (p ) = (p ) = pu ln p . dx du dx dx For example, d 3x 2 2 2 (2 ) = 23x (ln 2)(6x) = 6x 23x ln 2. dx The integral version of (7.5.6) reads p x dx = 1 p x + C, ln p p > 0, p = 1. (7.5.8) For example, 2x dx = 1x 2 + C. ln 2 410 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS Example 3 SOLUTION Find x5−x dx. 2 Set u = −x2 , 2 du = −2x dx. 5u du = − 1 2 1 5u + C ln 5 x5−x dx = − 1 2 = −1 −x2 5 + C. 2 ln 5 2 Example 4 SOLUTION Evaluate 1 32x−1 dx. du = 2 dx. Set u = 2x − 1, At x = 1, u = 1; at x = 2, u = 3. Thus 2 1 32x−1 dx = 3 1 2 1 3u du = 1 2 1 · 3u ln 3 3 = 1 12 ∼ = 10. 923. ln 3 Base p : The Function f (x) = logp x If p > 0, then ln pt = t ln p for all t . If p is also different from 1, then ln p = 0, and we have ln pt = t. ln p This indicates that the function f (x ) = satisfies the relation f ( pt ) = t In view of this we call ln x ln p the logarithm of x to the base p and write: ln x . ln p for all real t . ln x , ln p x > 0, (7.5.9) logp x = The relation holds for all x > 0 and assumes that p is a positive number different from 1. 7.5 ARBITRARY POWERS; OTHER BASES 411 For example, log2 32 = ln 32 ln 25 5 ln 2 = = =5 ln 2 ln 2 ln 2 and log100 1 10 = 1 ln( 10 ) ln 10−1 1 −ln 10 = =− . = ln 100 ln 102 2 ln 10 2 We can obtain these same results more directly from the relation (7.5.10) logp pt = t . Accordingly log2 32 = log2 25 = 5 and log100 1 10 = log100 (100−1/2 ) = − 1 . 2 Since logp x and ln x differ only by a constant factor, there is no reason to introduce new differentiation and integration formulas. For the record , we simply point out that d d (logp x) = dx dx ln x ln p = 1 x ln p and d d (logp u) = dx dx ln u ln p = du 1 · u ln p dx if u is a positive, differentiable function of x. If p is e, the factor ln p is 1 and we have 1 d (loge x) = . dx x The logarithm to the base e, ln = loge , is called the natural logarithm because it is the logarithm with the simplest derivative. Example 5 Find (b) d [log2 (3x2 + 1)] dx (c) 1 dx x ln 10 (a) d (log5 |x|) dx SOLUTION (a) d d ln |x| 1 (log5 |x|) = . = dx dx ln 5 x ln 5 d d ln (3x2 + 1) 6x 1 [log2 (3x2 + 1)] = = · 6x = . 2 + 1) ln 2 2 + 1) ln 2 dx dx ln 2 (3x (3x 1 1 dx = x ln 10 ln 10 1 ln |x| dx = + C = log10 |x| + C . x ln 10 (b) (c) 412 CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS EXERCISES 7.5 Evaluate. 1. log2 64. 3. log64 1 . 2 5. log5 1. 7. log5 125. 2. log2 1 . 64 In Exercises 41 and 42, derive the formula for f (x) by logarithmic differentiation. 41. f (x) = p x . d [(x + 1)x ]. dx d [(ln x)ln x ]. dx 42. f (x) = p g (x) . d [(ln x)x ]. dx d dx 1 x x 4. log10 0. 01. 6. log5 0. 2. 8. log2 43 . Find the derivative by logarithmic differentiation. 43. 45. 47. 44. 46. 48. Show that the given identity holds. 9. logp xy = logp x + logp y. 1 10. logp = −logp x. 11. logp xy = y logp x. x x 12. logp = logp x − logp y. y Find the numbers x, if any, which satisfy the given equation. 13. 10x = e x . 15. logx 10 = log4 100. b . 14. log5 x = 0. 04. 16. logx 2 = log3 x. d sin x [x ]. dx d 49. [(sin x)cos x ]. dx d (2x ) 51. [x ]. dx d 2 [(cos x)(x +1) ]. dx d (x 2 ) 50. [x ]. dx d 52. [(tan x)sec x ]. dx c Sketch figures in which you compare the following pairs of graphs. 53. f (x) = e x 54. f (x) = e x 55. f (x) = ln x and and and and and and g ( x ) = 2x . g ( x ) = 3x . g (x) = log3 x. g (x) = 2−x . g (x) = log2 x. g (x) = log2 x. 1 17. Estimate ln a given that et1 < a < et2 . 18. Estimate e given that ln x1 < b < ln x2 . Differentiate the function. 19. f (x) = 3 . 2x 20. g (x) = 4 3x2 . 2 +x 2 56. f (x) = 2 . 57. f (x) = 2 x x 21. f (x) = 25x 3ln x . 23. g (x) = log3 x. 22. F (x) = 5−2x 24. h(x) = 7sin x . log10 x . x2 28. h(x) = a−x cos bx. 26. g (x) = 58. f (x) = ln x 2 Evaluate the integral. 59. 1 4 25. f (x) = tan(log5 x). 27. F (x) = cos(2x + 2−x ). Calculate the integral. 29. 31. 33. 35. 3x dx. (x + 3 ) dx. 3 −x 2−x dx. dx . x ln 2 x10 1+x2 60. 0 2 4x dx. px/2 dx. 5p x+1 dx. √ x+1 √ 61. 30. 32. 34. 2−x dx. 63. x10 −x2 0 1 1 62. 0 1 dx. 64. 0 dx. 65. 0 1 dx . x ln 5 log2 x3 dx. x log5 x dx. x (2x + x2 ) dx. c Evaluate numerically and then give the exact value. 66. 71/ln 7 . 68. (16)1/ln 2 . 67. 5(ln 17)/(ln 5) . 36. Show that, if a, b, c are positive, then loga c = loga b logb c provided that a and b are both different from 1. Find f (e) in Exercises 37–40. 37. f (x) = log3 x. 39. f (x) = ln (ln x). 38. f (x) = x log3 x. 40. f (x) = log3 (log2 x). c 69. (a) Use a graphing utility to graph f (x) = 2x and g (x) = x2 − 1 together. (b) Use a CAS to find the points of intersection of the two graphs. (c) Use a CAS to find the area of the region bounded by the two graphs. c 70. Repeat Exercise 69 with the functions f (x) = 2−x and g (x) = 1/x2 , x > 0. ...
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.

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