5128_Ch03_pp098-184
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178 Chapter 3 Derivatives<br />
[–5, 10] by [–10, 30]<br />
Figure 3.59 The graph of dPdt, the rate<br />
of spread of the flu in Example 8. The<br />
graph of P is shown in Figure 3.58.<br />
(b) To find the rate at which the flu spreads, we find dPdt. To find dPdt, we need to<br />
invoke the Chain Rule twice:<br />
d P d<br />
1001 e<br />
dt<br />
d t<br />
3t 1 100 • 11 e 3t 2 d<br />
• 1 e d t<br />
3t <br />
1001 e 3t 2 • 0 e 3t d<br />
• 3 t d t<br />
1001 e 3t 2 e 3t • 1<br />
100e<br />
1 e<br />
3t<br />
3t<br />
2<br />
At t 3, then, dPdt 1004 25. The flu is spreading to 25 students per day.<br />
(c) We could estimate when the flu is spreading the fastest by seeing where the graph of<br />
y Pt has the steepest upward slope, but we can answer both the “when” and the<br />
“what” parts of this question most easily by finding the maximum point on the graph of<br />
the derivative (Figure 3.59).<br />
We see by tracing on the curve that the maximum rate occurs at about 3 days, when<br />
(as we have just calculated) the flu is spreading at a rate of 25 students per day.<br />
Now try Exercise 51.<br />
Quick Review 3.9 (For help, go to Sections 1.3 and 1.5.)<br />
1. Write log 5 8 in terms of natural logarithms. l n<br />
<br />
8<br />
ln<br />
5<br />
2. Write 7 x as a power of e. e x ln 7<br />
In Exercises 3–7, simplify the expression using properties of exponents<br />
and logarithms.<br />
3. ln e tan x tan x 4. ln x 2 4 ln x 2 ln (x 2)<br />
5. log 2 8 x5 3x 15 6. log 4 x 15 log 4 x 12 54<br />
7. 3lnx ln 3x ln 12x 2 ln (4x 4 )<br />
13. 3 csc x (ln 3)(csc x cot x)<br />
Section 3.9 Exercises<br />
1<br />
1<br />
23. y log 2 1x, x 0 24. y 1log 2 x <br />
x l n2<br />
x(ln 2)( log 2 x) 2<br />
25. y ln 2 • log 2 x 26. y log 3 1 x ln 3<br />
1 x , x 0 1<br />
1<br />
26., x <br />
1 x ln 3 ln 3<br />
In Exercises 1–28, find dydx. Remember that you can use NDER<br />
to support your computations.<br />
1. y 2e x 2e x 2. y e 2x 2e 2x<br />
3. y e x e x 4. y e 5x 5e 5x<br />
5. y e 2x3 2 3 e2x3 6. y e x4 1 4 ex/4<br />
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<br />
9. y e x ex/2x 10. y e x2 2xe (x2 )<br />
11. y 8 x 8 x ln 8 12. y 9 x 9 x ln 9<br />
13. y 3 csc x 14. y 3 cot x 3 cot x (ln 3)(csc 2 x)<br />
15. y ln x 2 2 x 16. y ln x2 2l n x<br />
<br />
x<br />
17. y ln 1x See page 180. 18. y ln 10x See page 180.<br />
1<br />
19. y ln ln x x l<br />
20. y x ln x x ln x<br />
n x<br />
21. y log 4 x 2 See page 180. 22. y log 5 x See page 180.<br />
In Exercises 8–10, solve the equation algebraically using logarithms.<br />
Give an exact answer, such as ln 23, and also an approximate answer<br />
to the nearest hundredth.<br />
8. 3 x 19 x l n 19<br />
<br />
ln<br />
3<br />
2.68<br />
9. 5 t ln 5 18 x ln 18 ln (ln 5)<br />
1.50<br />
ln 5<br />
10. 3 x1 2 x ln 3<br />
x 2.71<br />
ln 2 ln 3<br />
27. y log 10 e x 1<br />
<br />
28. y ln 10 x ln 10<br />
ln 10<br />
29. At what point on the graph of y 3 x 1 is the tangent line<br />
parallel to the line y 5x 1? (1.379, 5.551)<br />
30. At what point on the graph of y 2e x 1 is the tangent line<br />
perpendicular to the line y 3x 2? (1.792, 0.667)<br />
31. A line with slope m passes through the origin and is tangent to<br />
y ln (2x). What is the value of m? 2e 1<br />
32. A line with slope m passes through the origin and is tangent to<br />
y ln (x3). What is the value of m? 1 3 e1<br />
In Exercises 33–36, find dydx.<br />
33. y x p x 1 34. y x 12 (1 2)x 2 <br />
35. y x 2 2x 21 36. y x 1e (1 e)x e<br />
In Exercises 37–42, find f(x) and state the domain of f .<br />
1<br />
37. f (x) ln (x 2) , x 2<br />
x 2<br />
38. f (x) ln (2x 2)<br />
1<br />
, x 1<br />
x 1