5128_Ch03_pp098-184

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Section 3.9 Derivatives of Exponential and Logarithmic Functions 177 The domain of f appears to be all x 3. However, since f is not defined for x 3, neither is f . Thus, 1 f x , x 3. x 3 That is, the domain of f is 3, . Now try Exercise 37. Sometimes the properties of logarithms can be used to simplify the differentiation process, even if we must introduce the logarithms ourselves as a step in the process. Example 7 shows a clever way to differentiate y x x for x 0. EXAMPLE 7 Find dydx for y x x , x 0. SOLUTION Logarithmic Differentiation ln y ln x x y x x ln y x ln x d d ln y x ln x d x d x 1 y d y 1 • ln x x • 1 dx x d y yln x 1 dx Logs of both sides Property of logs Differentiate implicitly. d y x dx x ln x 1 Now try Exercise 43. [–5, 10] by [–25, 120] Figure 3.58 The graph of 100 Pt , 1 e3t modeling the spread of a flu. (Example 8) EXAMPLE 8 How Fast does a Flu Spread? The spread of a flu in a certain school is modeled by the equation 100 Pt , 1 e3t where Pt is the total number of students infected t days after the flu was first noticed. Many of them may already be well again at time t. (a) Estimate the initial number of students infected with the flu. (b) How fast is the flu spreading after 3 days? (c) When will the flu spread at its maximum rate? What is this rate? SOLUTION The graph of P as a function of t is shown in Figure 3.58. (a) P0 1001 e 3 5 students (to the nearest whole number). continued
178 Chapter 3 Derivatives [–5, 10] by [–10, 30] Figure 3.59 The graph of dPdt, the rate of spread of the flu in Example 8. The graph of P is shown in Figure 3.58. (b) To find the rate at which the flu spreads, we find dPdt. To find dPdt, we need to invoke the Chain Rule twice: d P d 1001 e dt d t 3t 1 100 • 11 e 3t 2 d • 1 e d t 3t 1001 e 3t 2 • 0 e 3t d • 3 t d t 1001 e 3t 2 e 3t • 1 100e 1 e 3t 3t 2 At t 3, then, dPdt 1004 25. The flu is spreading to 25 students per day. (c) We could estimate when the flu is spreading the fastest by seeing where the graph of y Pt has the steepest upward slope, but we can answer both the “when” and the “what” parts of this question most easily by finding the maximum point on the graph of the derivative (Figure 3.59). We see by tracing on the curve that the maximum rate occurs at about 3 days, when (as we have just calculated) the flu is spreading at a rate of 25 students per day. Now try Exercise 51. Quick Review 3.9 (For help, go to Sections 1.3 and 1.5.) 1. Write log 5 8 in terms of natural logarithms. l n 8 ln 5 2. Write 7 x as a power of e. e x ln 7 In Exercises 3–7, simplify the expression using properties of exponents and logarithms. 3. ln e tan x tan x 4. ln x 2 4 ln x 2 ln (x 2) 5. log 2 8 x5 3x 15 6. log 4 x 15 log 4 x 12 54 7. 3lnx ln 3x ln 12x 2 ln (4x 4 ) 13. 3 csc x (ln 3)(csc x cot x) Section 3.9 Exercises 1 1 23. y log 2 1x, x 0 24. y 1log 2 x x l n2 x(ln 2)( log 2 x) 2 25. y ln 2 • log 2 x 26. y log 3 1 x ln 3 1 x , x 0 1 1 26., x 1 x ln 3 ln 3 In Exercises 1–28, find dydx. Remember that you can use NDER to support your computations. 1. y 2e x 2e x 2. y e 2x 2e 2x 3. y e x e x 4. y e 5x 5e 5x 5. y e 2x3 2 3 e2x3 6. y e x4 1 4 ex/4 7. y xe 2 e x e 2 e x 8. y x 2 e x xe x x 2 e x xe x e x 9. y e x ex/2x 10. y e x2 2xe (x2 ) 11. y 8 x 8 x ln 8 12. y 9 x 9 x ln 9 13. y 3 csc x 14. y 3 cot x 3 cot x (ln 3)(csc 2 x) 15. y ln x 2 2 x 16. y ln x2 2l n x x 17. y ln 1x See page 180. 18. y ln 10x See page 180. 1 19. y ln ln x x l 20. y x ln x x ln x n x 21. y log 4 x 2 See page 180. 22. y log 5 x See page 180. In Exercises 8–10, solve the equation algebraically using logarithms. Give an exact answer, such as ln 23, and also an approximate answer to the nearest hundredth. 8. 3 x 19 x l n 19 ln 3 2.68 9. 5 t ln 5 18 x ln 18 ln (ln 5) 1.50 ln 5 10. 3 x1 2 x ln 3 x 2.71 ln 2 ln 3 27. y log 10 e x 1 28. y ln 10 x ln 10 ln 10 29. At what point on the graph of y 3 x 1 is the tangent line parallel to the line y 5x 1? (1.379, 5.551) 30. At what point on the graph of y 2e x 1 is the tangent line perpendicular to the line y 3x 2? (1.792, 0.667) 31. A line with slope m passes through the origin and is tangent to y ln (2x). What is the value of m? 2e 1 32. A line with slope m passes through the origin and is tangent to y ln (x3). What is the value of m? 1 3 e1 In Exercises 33–36, find dydx. 33. y x p x 1 34. y x 12 (1 2)x 2 35. y x 2 2x 21 36. y x 1e (1 e)x e In Exercises 37–42, find f(x) and state the domain of f . 1 37. f (x) ln (x 2) , x 2 x 2 38. f (x) ln (2x 2) 1 , x 1 x 1
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