Stewart Sec 3.2

SECTION 3.2 THE PRODUCT AND QUOTIENT RULES
L¡
183
segments L 1 and L 2 to be tangent to the parabola at the transition points P and Q. (See the figure.) To simplify the equations you decide to place the origin at P.
f
P
||||
1. (a) Suppose the horizontal distance between P and Q is 100 ft. Write equations in a, b, and c
Q
L™
;
that will ensure that the track is smooth at the transition points.
(b) Solve the equations in part (a) for a, b, and c to find a formula for f !x".
(c) Plot L 1, f , and L 2 to verify graphically that the transitions are smooth.
(d) Find the difference in elevation between P and Q.
2. The solution in Problem 1 might look smooth, but it might not feel smooth because the piece-
wise defined function [consisting of L 1!x" for x & 0, f !x" for 0 # x # 100, and L 2!x" for
x $ 100] doesn’t have a continuous second derivative. So you decide to improve the design by
using a quadratic function q!x" ! ax 2 " bx " c only on the interval 10 # x # 90 and connecting it to the linear functions by means of two cubic functions:
t!x" ! k x 3 " lx 2 " m x " n
3
2
90 & x # 100
h!x" ! px " qx " rx " s
CAS
3.2
0 # x & 10
(a) Write a system of equations in 11 unknowns that ensure that the functions and their first
two derivatives agree at the transition points.
(b) Solve the equations in part (a) with a computer algebra system to find formulas for
q!x", t!x", and h!x".
(c) Plot L 1, t, q, h, and L 2, and compare with the plot in Problem 1(c).
THE PRODUCT AND QUOTIENT RULES
The formulas of this section enable us to differentiate new functions formed from old functions by multiplication or division.
THE PRODUCT RULE
| By analogy with the Sum and Difference Rules, one might be tempted to guess, as Leibniz
Î√
√
u Î√
Îu Î√
u√
√ Îu
did three centuries ago, that the derivative of a product is the product of the derivatives. We
can see, however, that this guess is wrong by looking at a particular example. Let f !x" ! x
and t!x" ! x 2. Then the Power Rule gives f %!x" ! 1 and t%!x" ! 2x. But ! ft"!x" ! x 3, so
! ft"%!x" ! 3x 2. Thus ! ft"% " f %t%. The correct formula was discovered by Leibniz (soon
after his false start) and is called the Product Rule.
Before stating the Product Rule, let’s see how we might discover it. We start by assuming that u ! f !x" and v ! t!x" are both positive differentiable functions. Then we can
interpret the product uv as an area of a rectangle (see Figure 1). If x changes by an amount
(x, then the corresponding changes in u and v are
(u ! f !x " (x" ! f !x"
u
Îu
FIGURE 1
The geometry of the Product Rule
(v ! t!x " (x" ! t!x"
and the new value of the product, !u " (u"!v " (v", can be interpreted as the area of the
large rectangle in Figure 1 (provided that (u and (v happen to be positive).
The change in the area of the rectangle is
1
(!uv" ! !u " (u"!v " (v" ! uv ! u (v " v (u " (u (v
! the sum of the three shaded areas
184
||||
CHAPTER 3 DIFFERENTIATION RULES
If we divide by (x, we get
(!uv"
(v
(u
(v
!u
"v
" (u
(x
(x
(x
(x
Recall that in Leibniz notation the definition of
a derivative can be written as
N
If we now let (x l 0, we get the derivative of uv :
(
(y
dy
! lim
( x l 0 (x
dx
d
(!uv"
(v
(u
(v
!uv" ! lim
! lim u
"v
" (u
(x l 0
(x l 0
dx
(x
(x
(x
(x
)
(v
(u
" v lim
"
(x l 0 (x
(x
(v
(x
! u lim
(x l 0
!u
2
(
)(
lim (u
(x l 0
lim
(x l 0
)
dv
du
dv
"v
"0!
dx
dx
dx
d
dv
du
!uv" ! u
"v
dx
dx
dx
(Notice that (u l 0 as (x l 0 since f is differentiable and therefore continuous.)
Although we started by assuming (for the geometric interpretation) that all the quantities are positive, we notice that Equation 1 is always true. (The algebra is valid whether u,
v, (u, and (v are positive or negative.) So we have proved Equation 2, known as the
Product Rule, for all differentiable functions u and v.
N
THE PRODUCT RULE If f and t are both differentiable, then
In prime notation:
d
d
d
& f !x"t!x"' ! f !x"
&t!x"' " t!x"
& f !x"'
dx
dx
dx
! ft"% ! ft% " t f %
In words, the Product Rule says that the derivative of a product of two functions is the
first function times the derivative of the second function plus the second function times the
derivative of the first function.
EXAMPLE 1
Figure 2 shows the graphs of the function f
of Example 1 and its derivative f %. Notice that
f %!x" is positive when f is increasing and negative when f is decreasing.
N
(a) If f !x" ! xe x, find f %!x".
(b) Find the nth derivative, f !n"!x".
SOLUTION
(a) By the Product Rule, we have
3
f %!x" !
d
d x
d
!xe x " ! x
!e " " e x
!x"
dx
dx
dx
! xe x " e x ) 1 ! !x " 1"e x
_3
FIGURE 2
fª
f
1.5
_1
(b) Using the Product Rule a second time, we get
f '!x" !
d
d x
d
&!x " 1"e x ' ! !x " 1"
!e " " e x
!x " 1"
dx
dx
dx
! !x " 1"e x " e x ! 1 ! !x " 2"e x
SECTION 3.2 THE PRODUCT AND QUOTIENT RULES
||||
185
Further applications of the Product Rule give
f *!x" ! !x " 3"e x
f !4"!x" ! !x " 4"e x
In fact, each successive differentiation adds another term e x, so
f !n"!x" ! !x " n"e x
In Example 2, a and b are constants. It is
customary in mathematics to use letters near the
beginning of the alphabet to represent constants
and letters near the end of the alphabet to represent variables.
N
M
EXAMPLE 2 Differentiate the function f !t" ! st !a " bt".
SOLUTION 1 Using the Product Rule, we have
f %!t" ! st
d
d
!a " bt" " !a " bt"
(st )
dt
dt
! st ! b " !a " bt" ! 12 t !1%2
! bst "
a " bt
a " 3bt
!
2st
2st
SOLUTION 2 If we first use the laws of exponents to rewrite f !t", then we can proceed
directly without using the Product Rule.
f !t" ! ast " btst ! at 1%2 " bt 3%2
f %!t" ! 12 at!1%2 " 32 bt 1%2
which is equivalent to the answer given in Solution 1.
M
Example 2 shows that it is sometimes easier to simplify a product of functions than
to use the Product Rule. In Example 1, however, the Product Rule is the only possible
method.
EXAMPLE 3 If f !x" ! sx t!x", where t!4" ! 2 and t%!4" ! 3, find f %!4".
SOLUTION Applying the Product Rule, we get
f %!x" !
d
d
d
[
&t!x"' " t!x"
[sx ]
sx t!x"] ! sx
dx
dx
dx
! sx t%!x" " t!x" ) 12 x !1%2 ! sx t%!x" "
f %!4" ! s4 t%!4" "
So
t!x"
2sx
t!4"
2
!2)3"
! 6.5
2s4
2)2
M
THE QUOTIENT RULE
We find a rule for differentiating the quotient of two differentiable functions u ! f !x" and
v ! t!x" in much the same way that we found the Product Rule. If x, u, and v change by
amounts (x, (u, and (v, then the corresponding change in the quotient u%v is
()
(
u
v
!
u " (u
u
!u " (u"v ! u!v " (v"
v (u ! u(v
!
! !
v " (v
v
v!v " (v"
v!v " (v"
186
||||
CHAPTER 3 DIFFERENTIATION RULES
so
d
dx
()
u
v
! lim
(x l 0
(!u%v"
! lim
(x l 0
(x
v
(u
(v
!u
(x
(x
v!v " (v"
As (x l 0, (v l 0 also, because v ! t!x" is differentiable and therefore continuous. Thus,
using the Limit Laws, we get
d
dx
N
()
u
v
(u
(v
du
dv
! u lim
v
!u
(x
l
0
(x
(x
dx
dx
!
v lim !v " (v"
v2
v lim
(x l 0
!
(x l 0
THE QUOTIENT RULE If f and t are differentiable, then
In prime notation:
()
f %
t f % ! ft%
!
t
t2
d
dx
* +
f !x"
t!x"
t!x"
!
d
d
& f !x"' ! f !x"
&t!x"'
dx
dx
&t!x"' 2
In words, the Quotient Rule says that the derivative of a quotient is the denominator
times the derivative of the numerator minus the numerator times the derivative of the
denominator, all divided by the square of the denominator.
The Quotient Rule and the other differentiation formulas enable us to compute the
derivative of any rational function, as the next example illustrates.
We can use a graphing device to check that
the answer to Example 4 is plausible. Figure 3
shows the graphs of the function of Example 4
and its derivative. Notice that when y grows
rapidly (near !2), y% is large. And when y
grows slowly, y% is near 0.
N
V EXAMPLE 4
Let y !
!x 3 " 6"
y% !
1.5
4
y
_1.5
FIGURE 3
V EXAMPLE 5
point (1, e).
1
2
d
d
!x 2 " x ! 2" ! !x 2 " x ! 2"
!x 3 " 6"
dx
dx
!x 3 " 6"2
!
!x 3 " 6"!2x " 1" ! !x 2 " x ! 2"!3x 2 "
!x 3 " 6"2
!
!2x 4 " x 3 " 12x " 6" ! !3x 4 " 3x 3 ! 6x 2 "
!x 3 " 6"2
!
!x 4 ! 2x 3 " 6x 2 " 12x " 6
!x 3 " 6"2
yª
_4
x2 " x ! 2
. Then
x3 " 6
Find an equation of the tangent line to the curve y ! e x%!1 " x 2 " at the
SOLUTION According to the Quotient Rule, we have
dy
!
dx
!
!1 " x 2 "
d
d
!e x " ! e x
!1 " x 2 "
dx
dx
!1 " x 2 "2
!1 " x 2 "e x ! e x !2x"
e x !1 ! x"2
!
!1 " x 2 "2
!1 " x 2 "2
M
SECTION 3.2 THE PRODUCT AND QUOTIENT RULES
2.5
y=
´
1+≈
dy
dx
3.5
0
187
So the slope of the tangent line at (1, 12 e) is
1
y=2 e
_2
||||
FIGURE 4
,
x!1
!0
This means that the tangent line at (1, 12 e) is horizontal and its equation is y ! 12 e. [See
Figure 4. Notice that the function is increasing and crosses its tangent line at (1, 12 e).]
M
NOTE Don’t use the Quotient Rule every time you see a quotient. Sometimes it’s easier to rewrite a quotient first to put it in a form that is simpler for the purpose of differentiation. For instance, although it is possible to differentiate the function
F!x" !
3x 2 " 2sx
x
using the Quotient Rule, it is much easier to perform the division first and write the function as
F!x" ! 3x " 2x !1%2
before differentiating.
We summarize the differentiation formulas we have learned so far as follows.
TABLE OF DIFFERENTIATION FORMULAS
d
!c" ! 0
dx
d
!x n " ! nx n!1
dx
d
!e x " ! e x
dx
!cf "% ! cf %
! f " t"% ! f % " t%
! f ! t"% ! f % ! t%
! ft"% ! ft% " tf %
3.2
f %
tf % ! ft%
!
t
t2
EXERCISES
1. Find the derivative of y ! !x 2 " 1"!x 3 " 1" in two ways: by
using the Product Rule and by performing the multiplication
first. Do your answers agree?
2. Find the derivative of the function
F!x" !
x ! 3x sx
sx
ex
x2
4. t!x" ! sx e x
6. y !
3x ! 1
2x " 1
8. f !t" !
2t
4 " t2
9. V!x" ! !2x 3 " 3"!x 4 ! 2x"
11. F! y" !
3–26 Differentiate.
3. f !x" ! !x 3 " 2x"e x
7. t!x" !
10. Y!u" ! !u!2 " u!3 "!u 5 ! 2u 2 "
in two ways: by using the Quotient Rule and by simplifying
first. Show that your answers are equivalent. Which method do
you prefer?
5. y !
()
ex
1"x
(
)
1
3
! 4 ! y " 5y 3 "
y2
y
12. R!t" ! !t " e t "(3 ! st )
13. y !
x3
1 ! x2
14. y !
x"1
x3 " x ! 2
15. y !
t2 " 2
t ! 3t 2 " 1
16. y !
t
!t ! 1"2
18. y !
1
s " ke s
4
17. y ! !r 2 ! 2r"e r
188
||||
19. y !
CHAPTER 3 DIFFERENTIATION RULES
v 3 ! 2v sv
20. z ! w 3%2!w " ce w "
v
21. f !t" !
2t
2 " st
22. t!t" !
1 ! xe x
24. f !x" !
x " ex
A
23. f !x" !
B " Ce x
25. f !x" !
t ! st
t 1%3
x
26. f !x" !
c
x"
x
ax " b
cx " d
;
;
39. (a) If f !x" ! !x ! 1"e x, find f %!x" and f '!x".
(b) Check to see that your answers to part (a) are reasonable
by comparing the graphs of f , f %, and f '.
40. (a) If f !x" ! x%!x 2 " 1", find f %!x" and f '!x".
(b) Check to see that your answers to part (a) are reasonable
by comparing the graphs of f , f %, and f '.
41. If f !x" ! x 2%!1 " x", find f '!1".
42. If t!x" ! x%e x, find t !n"!x".
43. Suppose that f !5" ! 1, f %!5" ! 6, t!5" ! !3, and t%!5" ! 2.
Find the following values.
(a) ! ft"%!5"
(c) ! t%f "%!5"
27–30 Find f %!x" and f '!x".
27. f !x" ! x 4e x
28. f !x" ! x 5%2e x
x2
29. f !x" !
1 " 2x
x
30. f !x" !
3 " ex
44. Suppose that f !2" ! !3, t!2" ! 4, f %!2" ! !2, and
t%!2" ! 7. Find h%!2".
(a) h!x" ! 5f !x" ! 4 t!x"
(c) h!x" !
31–32 Find an equation of the tangent line to the given curve at
the specified point.
31. y !
2x
,
x"1
!1, 1"
32. y !
ex
,
x
!1, e"
(b) ! f%t"%!5"
(b) h!x" ! f !x" t!x"
f !x"
t!x"
(d) h!x" !
t!x"
1 " f !x"
45. If f !x" ! e x t!x", where t!0" ! 2 and t%!0" ! 5, find f %!0".
46. If h!2" ! 4 and h%!2" ! !3, find
d
dx
( ),
h!x"
x
x!2
33–34 Find equations of the tangent line and normal line to the
given curve at the specified point.
33. y ! 2xe x,
!0, 0"
34. y !
sx
, !4, 0.4"
x"1
47. If f and t are the functions whose graphs are shown, let
u!x" ! f !x"t!x" and v!x" ! f !x"%t!x".
(a) Find u%!1".
(b) Find v%!5".
y
35. (a) The curve y ! 1%!1 " x 2 " is called a witch of Maria
;
36. (a) The curve y ! x%!1 " x 2 " is called a serpentine. Find
;
an equation of the tangent line to this curve at the point
!3, 0.3".
(b) Illustrate part (a) by graphing the curve and the tangent
line on the same screen.
;
0
x
are the functions whose graphs are shown.
(a) Find P%!2".
(b) Find Q%!7".
y
F
(b) Check to see that your answer to part (a) is reasonable by
comparing the graphs of f and f %.
(b) Check to see that your answer to part (a) is reasonable by
comparing the graphs of f and f %.
1
48. Let P!x" ! F!x"G!x" and Q!x" ! F!x"%G!x", where F and G
3
38. (a) If f !x" ! x%!x 2 ! 1", find f %!x".
g
1
37. (a) If f !x" ! e %x , find f %!x".
x
;
f
Agnesi. Find an equation of the tangent line to this curve
at the point (!1, 12 ).
(b) Illustrate part (a) by graphing the curve and the tangent
line on the same screen.
G
1
0
1
x
SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
49. If t is a differentiable function, find an expression for the deriv-
ative of each of the following functions.
x
(a) y ! xt!x"
(b) y !
t!x"
(c) y !
t!x"
x
50. If f is a differentiable function, find an expression for the
derivative of each of the following functions.
f !x"
(b) y !
x2
(a) y ! x f !x"
2
(c) y !
x2
f !x"
(d) y !
1 " x f !x"
sx
51. How many tangent lines to the curve y ! x%!x " 1) pass
through the point !1, 2"? At which points do these tangent lines
touch the curve?
52. Find equations of the tangent lines to the curve
y!
x!1
x"1
53. In this exercise we estimate the rate at which the total personal
income is rising in the Richmond-Petersburg, Virginia, metropolitan area. In 1999, the population of this area was 961,400,
and the population was increasing at roughly 9200 people per
year. The average annual income was $30,593 per capita, and
this average was increasing at about $1400 per year (a little
above the national average of about $1225 yearly). Use the
Product Rule and these figures to estimate the rate at which
total personal income was rising in the Richmond-Petersburg
area in 1999. Explain the meaning of each term in the Product
Rule.
54. A manufacturer produces bolts of a fabric with a fixed width.
The quantity q of this fabric (measured in yards) that is sold is
a function of the selling price p (in dollars per yard), so we can
A review of the trigonometric functions is
given in Appendix D.
N
189
write q ! f ! p". Then the total revenue earned with selling price
p is R! p" ! pf ! p".
(a) What does it mean to say that f !20" ! 10,000 and
f %!20" ! !350?
(b) Assuming the values in part (a), find R%!20" and interpret
your answer.
55. (a) Use the Product Rule twice to prove that if f , t, and h are
differentiable, then ! fth"% ! f %th " ft%h " fth%.
(b) Taking f ! t ! h in part (a), show that
d
& f !x"' 3 ! 3& f !x"' 2 f %!x"
dx
(c) Use part (b) to differentiate y ! e 3x.
56. (a) If F!x" ! f !x"t!x", where f and t have derivatives of all
orders, show that F ' ! f 't " 2 f %t% " f t '.
(b) Find similar formulas for F * and F !4".
(c) Guess a formula for F !n".
57. Find expressions for the first five derivatives of f !x" ! x 2e x.
that are parallel to the line x ! 2y ! 2.
3.3
||||
Do you see a pattern in these expressions? Guess a formula for
f !n"!x" and prove it using mathematical induction.
58. (a) If t is differentiable, the Reciprocal Rule says that
d
dx
* +
1
t!x"
!!
t%!x"
& t!x"' 2
Use the Quotient Rule to prove the Reciprocal Rule.
(b) Use the Reciprocal Rule to differentiate the function in
Exercise 18.
(c) Use the Reciprocal Rule to verify that the Power Rule is
valid for negative integers, that is,
d
!x !n " ! !nx!n!1
dx
for all positive integers n.
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
Before starting this section, you might need to review the trigonometric functions. In particular, it is important to remember that when we talk about the function f defined for all
real numbers x by
f !x" ! sin x
it is understood that sin x means the sine of the angle whose radian measure is x. A similar convention holds for the other trigonometric functions cos, tan, csc, sec, and cot. Recall
from Section 2.5 that all of the trigonometric functions are continuous at every number in
their domains.
If we sketch the graph of the function f !x" ! sin x and use the interpretation of f %!x"
as the slope of the tangent to the sine curve in order to sketch the graph of f % (see Exer-
A74
||||
APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES
(b) At 0 and 3
(c)
9. f 共t兲苷 t 3
11. y 苷 5 x7兾5
13. V共r兲苷 4
r 2
15. A共s兲苷 60兾s 6
17. G共x兲苷 1兾(2sx ) 2e x
5
19. F共x兲苷 32 x 4
21. y 苷 2ax b
2
y
3
0
x
3
31. ⺢
35. (a) 8
(b) y 苷 8x 17
37. (a) (i) 3 m兾s (ii) 2.75 m兾s (iii) 2.625 m兾s
(iv) 2.525 m兾s
39. (a) 10
(b) 2.5 m兾s
(b) y 苷 10 x 16
(c)
12
23.
25.
29.
33.
35.
37.
43.
y 苷 2 sx (2兾sx ) 3兾(2x sx )
27. H共x兲苷 3x 2 3 3x 2 3x 4
y 苷 0
31. z 苷 10A兾y 11 Be y
u 苷 15 t 4兾5 10t 3兾2
1
3
y 苷 4x 4
1
Tangent: y 苷 2x 2; normal: y 苷 2 x 2
x
y 苷 3x 1
39. e 5
41. 45x 14 15x 2
(a)
(c) 4x 3 9x 2 12x 7
3
50
–4
4
3
–12
41. (a) The rate at which the cost changes with respect to the
interest rate; dollars兾(percent per year)
(b) As the interest rate increases past 10%, the cost is increasing
at a rate of $1200兾(percent per year).
(c) Always positive
43.
100
y
0
x
5
10
40
45. f 共x兲苷 4x 9x 16, f 共x兲苷 12x 18x
3
49.
(c)
2
2
5兾4
x1兾4, f 共x兲苷 15
16 x
(a) v共t兲苷 3t 3, a共t兲苷 6t (b) 12 m兾s2
a共1兲苷 6 m兾s2
51. 共2, 21兲, 共1, 6兲
1
1
y 苷 12x 15, y 苷 12x 17
57. y 苷 3 x 3
共2, 4兲
63. P共x兲苷 x 2 x 3
3 3
y 苷 16 x 94 x 3
No
y
y
47. f 共x兲苷 2 55.
59.
65.
67.
fª
3
5
15
4
2
ƒ
45. (a) f 共x兲苷 2 共3 5x兲1兾2
5
(c)
]
(b) (, 35 , (, 35 )
0
(1, 1)
6
fª
f 共x兲苷
_6
47. 4 (discontinuity), 1 (corner), 2 (discontinuity),
再
(b)
ⱍ ⱍ
ⱍ ⱍ
2x
if x 3
2x if x 3
y
y
ƒ
9
ƒ
5 (vertical tangent)
49. The rate at which the total value of US currency in circulation
is changing in billions of dollars per year; $22.2 billion兾year
51. 0
N
0
3
3
PAGE 170
0
3
1
1
71. y 苷 2x 2 x
73. a 苷 2 , b 苷 2
77. 1000
79. 3; 1
EXERCISES 3.2
EXERCISES 3.1
N
PAGE 180
1. (a) See Definition of the Number e (page 179).
(b) 0.99, 1.03; 2.7 e 2.8
2
5. f 共t兲苷 3
3. f 共x兲苷 0
7. f 共x兲苷 3x 2 4
N
x
75. m 苷 4, b 苷 4
PAGE 187
1. y 苷 5x 3x 2 2x
3. f 共x兲苷 e x共x 3 3x 2 2x 2兲
5. y 苷共x 2兲e x兾x 3
7. t共x兲苷 5兾共2x 1兲2
6
3
9. V共x兲苷 14x 4x 6
11. F共 y兲苷 5 14兾y 2 9兾y 4
2t共t 4 4t 2 7兲
x 2共3 x 2兲
13. y 苷
15. y 苷
2 2
共1 x 兲
共t 4 3t 2 1兲 2
4
CHAPTER 3
3
x
1
1. 3
3. 4
5. 1
7. a 苷 2 2 s5
3
(c) Yes; no
9. 4
11. (b) Yes
13. (a) 0
(b) 1
(c) f 共x兲苷 x 2 1
2
x
69. (a) Not differentiable at 3 or 3
1
PROBLEMS PLUS
1
x
0
f
_3
ƒ
APPENDIX I ANSWERS TO ODD-NUMBERED EXERCISES
17. y 苷共r 2 2兲e r
21. f 共t兲苷
4t
23. f 共x兲苷 ACe x兾共B Ce x 兲 2
25. f 共x兲苷 2cx兾共x 2 c兲2
27. 共x 4 4x 3兲e x; 共x 4 8x 3 12x 2 兲e x
35. (a) y 苷 x 1
(b)
1.5
13.
17.
19.
4
0.5
39. xe x, 共x 1兲e x
20
(b) 9
(c) 20
2
45. 7
47. (a) 0
(b) 3
49. (a) y 苷 xt共x兲 t共x兲 (b) y 苷关 t共x兲 xt共x兲兴兾关t共x兲兴 2
(c) y 苷关xt共x兲 t共x兲兴兾x 2
51. Two, (2 s3, (1 s3 )兾2)
53. $1.627 billion兾year
55. (c) 3e 3x
2
x
57. f 共x兲苷共x 2x兲e , f 共x兲苷共x 2 4x 2兲e x,
f 共x兲苷共x 2 6x 6兲e x, f 共4兲共x兲苷共x 2 8x 12兲e x,
f 共5兲共x兲苷共x 2 10x 20兲e x; f (n)共x兲苷关x 2 2nx n共n 1兲兴e x
1
4
43. (a) 16
EXERCISES 3.3
N
21.
25.
27.
31.
35.
37.
39.
41.
43.
45.
47.
PAGE 195
1. f 共x兲苷 6x 2 sin x
PAGE 203
2 3x 2
12t 3
t共t兲
苷
11.
4共1 2x x 3兲3兾4
共t 4 1兲4
y 苷 3x 2 sin共a 3 x 3 兲
15. y 苷 ekx 共k x 1兲
t共x兲苷 4共1 4x兲4共3 x x 2 兲7共17 9x 21x 2 兲
y 苷 8共2x 5兲 3 共8x 2 5兲4 共4x 2 30x 5兲
12 x 共x 2 1兲 2
y 苷
23. y 苷共cos x x sin x兲e x cos x
共x 2 1兲 4
F共z兲苷 1兾关共z 1兲1兾2共z 1兲3兾2 兴
y 苷共r 2 1兲3兾2
29. y 苷 2 cos共tan 2x兲 sec 2 共2x兲
sin x
y 苷 2
共
ln 2兲 cos x
33. y 苷 4 sec 2x tan x
2x
2x
4e
1e
y 苷
sin
共1 e 2x 兲 2
1 e 2x
y 苷 2 cos cot共sin 兲 csc 2 共sin 兲
f 共t兲苷 sec 2 共e t 兲e t e tan t sec 2t
f 共t兲苷 4 sin 共e sin t 兲 cos 共e sin t 兲 e sin t sin t cos t
t共x兲苷 2r 2 p共ln a兲共2ra rx n兲 p1 a rx
cos共tan x兲 sec 2 共
x兲 sinssin 共tan x兲
y 苷
2ssin 共tan x兲
h共x兲苷 x兾sx 2 1, h 共x兲苷 1兾共x 2 1兲3兾2
9. F共x兲苷
4
41.
N
1. 4 cos 4 x
3. 20x共1 x 2 兲9
5. e sx兾(2 sx )
4
2
4
2
7. F共x兲苷 10 x共x 3x 2兲共2x 3兲
(_1, 0.5)
37. (a) e x 共x 3兲兾x 4
(b) sec x tan x 苷共sin x兲兾cos 2x
(c) cos x sin x 苷共cot x 1兲兾csc x
51. 1
EXERCISES 3.4
2x 2 2x
2
;
29.
共1 2x兲 2 共1 2x兲3
1
1
1
31. y 苷 2 x 2
33. y 苷 2x ; y 苷 2 x
1
2
3. f 共x兲苷 cos x 2 csc2x
1
5. t共t兲苷 3t 2 cos t t 3 sin t
7. h共兲苷 csc cot e 共cot csc 2 兲
2 tan x x sec 2 x
sec tan 11. f 共兲苷
共2 tan x兲 2
共1 sec 兲2
y 苷共x cos x 2 sin x兲兾x 3
f 共x兲苷 e x csc x 共x cot x x 1兲
2
y 苷 2s3x 3 s3 2
23. y 苷 x 1
(a) y 苷 2 x (b) 3π
2
2
2
49. e x共 cos x sin x兲; e x 关共 2 2兲 sin x 2 cos x兴
51. y 苷 20x 1
53. y 苷 x 1
55. (a) y 苷 2 x 1
(b)
3
9. y 苷
13.
15.
21.
25.
2
π
” 2 , π’
π
0
A75
49. (a) sec 2x 苷 1兾cos 2x
19. y 苷 2v 1兾sv
1兾2
(2 st ) 2
||||
(0, 1)
3
_3
_1.5
57.
59.
61.
65.
67.
69.
77.
(a) f 共x兲苷共2 2x 兲兾s2 x
共共
兾2兲 2n
, 3兲, 共共3
兾2兲 2n
, 1兲, n an integer
24
63. (a) 30
(b) 36
(a) 34
(b) Does not exist
(c) 2
(a) F共x兲苷 e x f 共e x 兲 (b) G共x兲苷 e f 共x兲 f 共x兲
120
71. 96
75. 250 cos 2x
5
v共t兲苷 2 cos共10
t兲 cm兾s
2
2
dB
7
2
t
苷
cos
(b) 0.16
dt
54
5.4
81. v 共t兲苷 2e1.5t共2
cos 2
t 1.5 sin 2
t兲
79. (a)
27. (a) sec x tan x 1
29. cos sin ; 2 cos sin 31. (a) f 共x兲苷共1 tan x兲兾sec x
(b) f 共x兲苷 cos x sin x
2
15
33. 共2n 1兲
3 , n an integer
1
35. (a) v共t兲苷 8 cos t, a共t兲苷 8 sin t
(b) 4 s3, 4, 4s3; to the left
37. 5 ft兾rad
39. 3
41. 3
1
45. 2
47. s2
√
s
0
2
0
43. sin 1
1
7
2

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