Summer Problem Set
Noah Rousseau
23 August 2021
1
Summer Problem Set
Rousseau 2
Chapter 2
Problem 0
2
Summer Problem Set
1
Rousseau 3
2.2, #25: lim sin(2x)
x
x→0
Solution:
Start with a graph:
4
3
2
1
0
−1
−6
−4
−2
0
2
4
6
The point marked should be where (0, f (0)) lies on the graph, such that f (x) = sin(2x)
. To find the limit at
x
that point, one needs to plug in values ranging from any number to 0, I am going to use π2 ( τ4 ) for simplicity:
f ( π2 ) =
f ( π3 ) =
f ( π4 )
f ( π6 )
=
=
f ( π8 ) =
π
f ( 12
)
=
sin(2( π
2 ))
π
2
sin(2( π
))
3
π
3
sin(2( π
4 ))
π
2
sin(2( π
6 ))
π
6
sin(2( π
8 ))
π
8
π
sin(2( 12
))
π
12
=
=
=
=
sin(π)
π
2
sin( 2π
)
3
π
3
sin( π
2)
π
4
sin( pi
3 )
π
6
pi
sin( 4 )
π
8
sin( pi
6 )
π
12
=
√
3
2
=
=
=
×
6
π
=
=
×
8
π
=
×
12
π
√
3
2
√
2
2
=
1
2
≈0
=0
3
π
4
π
×
=1×
=
=
0
π
2
=
√
3 3
2π
4
π
√
3 3
π
√
4 2
π
=
6
π
≈ 0.8269...
≈ 1.2732...
≈ 1.6539...
≈ 1.8006...
≈ 1.9098...
With the graph and the approximated values above, the value that f (x) approaches, as x approaches, is 2.
2
#26: lim sin(5x)
x
x→0
Solution:
Graph:
6
4
2
0
−2
Problem 2
−6
−4
−2
0
3
2
4
6
Summer Problem Set
Rousseau 4
Looking at the graph, I’m noticing a pattern of lim
x→0
3
sin(nx)
x
= n, using this theorem, the solution is 5.
#27: lim sin(2x)
x2
x→0
Solution:
Graph:
4
2
0
−2
−4
−6
−4
−2
0
2
4
6
There is no solution to this expression due to the function being non-continuous, that is the one sided limits are
not equal:
lim
x→0+
sin(2x)
=∞
x2
sin(2x)
= −∞
x→0−
x2
lim
4
#34: lim |x|x
x→0
Solution:
Graph:
4
3
2
1
0
−3
Problem 4
−2
−1
0
4
1
2
3
Summer Problem Set
Rousseau 5
The graph is continuous at x=0, therefor the limit is f(0):
f (0) = |0|0 :Plug it in
= 00 :Simplify
= 1 :Anything to the 0th power = 1
5
#38: Determine the one sided limits at c = 1, 2, 3, 4 of the
function g(t) shown in figure 12 and state wether the limit exists
at these points.
4
3
2
1
0
0
1
2
3
4
5
Solution:
This graph is non-continuous, a pice-wise function that took me a while to approximate the different
functions.
lim g(t) = 2.5
lim g(t) = .5 No two sided limit
t→1−
lim g(t) = 2
t→2−
lim g(t) = 2.25
t→3+
lim g(t) = 2
t→4+
Problem 6
x→1+
lim g(t) = 3
x→2+
lim g(t) = 2.25
x→3−
lim g(t) = 2
x→4−
5
No two sided limit
lim g(t) = 2.25
x→3
lim g(t) = 2
x→4
Summer Problem Set
6
#41: lim
x→0±
Rousseau 6
sin(x)
|x|
Solution:
Graph:
2
1
0
−1
−2
−6
−4
−2
0
2
sin(x)
= −1
|x|
sin(x)
lim
=1
x→0+
|x|
lim
x→0−
Problem 6
6
4
6
Summer Problem Set
7
Rousseau 7
2.3, #5: lim (3x + 4)
x→−3
Solution:
The limit of a liner function is equal to that function:
lim (3x + 4) :Given
x→−3
= 3(−3) + 4 :Plug it in
= −5 :Simplify
8
#8: lim y(y + 14)
y→−3
Solution:
The limit of this function tending toward a value is the function of that value.
lim y(y + 14) :Given
y→−3
= −3(14 − 3) :Plug it in
= −33 :Simplify
9
−1
+z
#22: lim( z z+1
)
z→1
Solution:
What is equivalent are the values of the limit and the function of the value that the limit tends towards.
−1
+z
lim ( z z+1
:Given
z→1
=
=
10
1−1 +1
1+1
2
2 =1
:Plug it in
:Simplify
#29: Can the quotient law be applied to evaluate lim ( sin(x)
x )?
x→0
Explain.
Solution:
Let’s try it:
lim ( sin(x)
x )
x→0
lim (sin(x))
x→0
lim (x)
:Given
:Apply the Rule
x→0
lim (sin(x))
x→0
0
:Most simple though this rout
Because of the simple rule that lim (x) = c then the result of the application of this rule, in this case, is a
x→c
”divide by zero” which, at this point, cannot be evaluated. Therefor, the quotient rule cannot be applied to this
expression.
11
#31: Give an example where lim (f (x) + g(x)) exists but neither
x→0
lim f (x) nor lim g(x) exists.
x→0
x→0
Solution:
I started with two functions, |x| and sin(x)
x . In one function, the limit exist, in the other the limit does not,
and the limit of the sum does not exist. Now, I do algebra to it. So we want the limit sum of these two
functions to exist, meaning that f (x) + g(x) = |x|. Now, let f (x) = sin(x)
|x| who’s limit does not exist. To find
Problem 11
7
Summer Problem Set
Rousseau 8
g(x) its simple Algieba:
f (x) + g(x) = |x| :Given
sin(x)
|x| + g(x) = |x|
sin(x)
|x| − |x| = −g(x)
− sin(x)
|x| + |x| = g(x)
:Plug in Def.
:Subtract |x| and g(x) from both sides
:Multiply by -1
Now we have our functions:
f (x) =
g(x) = −
Problem 11
sin(x)
|x|
sin(x)
+ |x|
|x|
8
Summer Problem Set
12
Rousseau 9
2.4, #10: Use the Laws of continuity to show the functions’
continuity:
f (x) = 3x3 + 8x2 − 20x
1
x3 is continuous
: xn is always continuous for n > 0
2
3x3 is continuous
: nf (x)is always continuous
2
3
x is continuous
: See line 1 reasoning
2
4
8x is continuous
: See line 2 reasoning
5
x is continuous
: See line 1 reasoning
−20x is continuous
: See line 2 reasoning
6
3
2
7 3x + 8x − 20x is continuous : Given f and g are continuous so is f + g see lines 2, 4, 6
13
#12: f (x) =
x2 −cos(x)
3+cos(x)
Solution:
1
3 + cos(x) > 0
2
cos(x) is continuous
: There is no x value that will make this less than 0
: This function is continuous
3 3 + cos(x) is continuous : n + f (x) is continuous, see line 2
x2 is continuous
4
5
− cos(x) is continuous
: xn is continuous for n > 0
: nf (x) is continuous, see line 2
2
6 x − cos(x) is continuous : f + g is continuous given f and g are continuous see lines 4,5
7
14
x2 −cos(x)
3+cos(x)
is continuous
:
f
g
is continuous given f and g is continuous, and that g > 0. See lines 1,3,6
#15: f (x) = ex cos(3x)
Solution:
1
ex is continuous
: This function is continuous
2
x1 is continuous
: xn is continuous such that n > 0
3
3x is continuous
: nf (x) is continuous, see line 2
4
cos(x) is continuous
: This function is continuous
cos(3x) is continuous
: f (g(x)) is continuous given f and g are continuous see line 3,4
5
x
6 e cos(3x) is continuous : f × g is continuous given f and g are continuous, see lines 1,5
15
#27: Determine the points at which the function is
discontinuous and the type of discontinuity
h(z) =
1 − 2z
z2 − z − 6
Solution:
This type of rational function has infinite discontinuities at the x values that make the denominator equal to
zero:
0 = z 2 − z − 6 :Setup
0 = (z − 3)(z + 2) :Factor
z = {3, −2} :Solve for z
Problem 16
9
Summer Problem Set
16
#28: h(z) =
Rousseau 10
1−2z
z 2 +9
Solution:
This rational function can only have a discontinuity when the denominator is equal to zero, however, this
denominator will never be equal to zero, making the function continuous at all points.
17
#35: f (x) =
1
ex −e−x
Solution:
This function is another rational function, with an infinite discontinuity when the denominator is equal to
zero, which will only happen when x is equal to zero:
e0 − e−0 = 0
18
#71: Evaluate using the the substitutional method
3
lim (1 − 8x3 ) 2
x→−1
Solution:
3
f (x) = (1 − 8x3 ) 2 :Given
3
f (−1) = (1 − 8(−1)3 ) 2 :Substitute
3
= (1 + 8) 2 :Simplify
=9
3
2
= (±3)3
= ±27
= 27 :Make a decision
19
#78: lim (tan−1 (ex ))
x→0
Solution:
f (x) = tan−1 (ex ) :Given
f (0) = tan−1 (e0 ) :Substitute
= tan−1 (1) :Simplify
= { π4 , τ8 }
Problem 19
10
Summer Problem Set
20
Rousseau 11
2
−64
)
2.5, #6 Evaluate: lim ( xx−9
x→8
Solution:
f (x) =
f (8) =
=
x2 −64
x−9
82 −64
8−9
64−64
1
:Given
:Substitute
:Simplify
=0
21
#7: lim ( x
2
x→2
−3x+2
x−2 )
Solution:
x2 −3x+2
x−2
(x−1)(x−2)
x−2
f (x) =
=
:Given
:Factor
= x − 1 :Simplify
= 2 − 1 :Substitute
=1
√
22
#24: lim (
x→4
5−x−1
√ )
2− x
Solution:
pain.
√
f (4.1) =
5−4.1−1
√
2− 4.1
Evaluate for 4.1
≈ 2.065418058
√
f (3.9) =
5−3.9−1
√
2− 3.9
Evaluate for 3.9
≈ 1.940074482
√
lim (
x→4
23
5−x−1
√
)
2− x
= 2 Using approaching values ± .1
#43: Evaluate with respect to constant: lim (2a + x)
x→0
Solution:
f (0) = 2a + 0 :Substitute 0 With Given Function
lim (2a + x) = 2a :Respectfully to the Constant
x→0
24
2
2
#46: lim ( (3a+h)h −9a )
h→0
Solution:
(3a+h)2 −9a2
h
(3a+h)(3a+h)−9a2
h
9a2 +6ah+h2 −9a2
h
6ah+h2
h
:Given
:Split Square
:Foil
:Simplify
f (x) = 6a + h :Simplest Function
f (0) = 6a + 0 :Substitute for h = 0
2
2
lim ( (3a+h)h −9a )
h→0
Problem 24
= 6a :With respect to the constant
11
Summer Problem Set
25
Rousseau 12
2
2.6, #11: Evaluate using theorem 2: lim( sin(t)
t )
t→0
Solution:
sin(t)2
t
2
= t sin(t)
2
t
2
= t( sin(t)
t )
sin(t) 2
lim (t( t ) )
t→0
lim (t(1)2 )
t→0
f (t) =
:Given Function
:Multiply top and bottom by t
:Re-arrange
:Take the Limit
:Using Theorem 2
= 0 :Substitute 0 for t
26
2
)
#16: lim( cos(t)−cos(t)
t
t→0
Solution:
f (t) =
f (0) =
cos(t)−cos(t)2
t
cos(0)−cos(0)2
0
2
= 1−1
0
= 1−1
0
:Given Function
:Substitute for f (0)
:Simplify
=0
27
x
)
#24: limπ ( sin(11x)
x→ 4
Solution:
f (x) =
f ( π4 )
=
x
sin(11x)
π
4
sin( 11π
4√ )
π
4
π
4
28
2
2
× √22
√
2π
4
÷
:Given
:Substitute for f ( π4 )
:Simplify
sin(2x)
#33: lim ( sin(5x)
sin(3x) sin(5x) )
x→0
Solution:
sin(5x) sin(2x)
sin(3x) sin(5x)
= sin(2x)
sin(3x)
sin(2×0)
f (0) = sin(3×0)
f (x) =
:Given Function
:Cancel out sin(5x)
:Evaluate f (0)
= 0 :Evaluated using substitutional method
Problem 28
12
Summer Problem Set
29
Rousseau 13
2.7, #17: Carry out three steps of the Bisection Method for
f (x) = 2x − x3
as follows:
29.1
#17a: show that f (x) has a zero in [1, 1.5]
Solution:
f (1) = 2(1) − (1)2 :Evaluate function for x = 1
= −1
f (1.5) = 2(1.5) − (1.5)3 : Evaluate for x = 1.5
= −0.546572875
There is a zero in that interval due to the difference in signs for f (x): one is positive, above the x axis, and the
other is negative, below the x axis; therefor the function crosses the axis in that interval.
29.2
#17b: and at [1.25, 1.5]
Solution:
f (1.25) = 2(1.25) − (1.25)3 :Plug in Values
= 0.42528923
f (1.25) = + whereas f (1.5) = − therefor the function must cross in that interval.
29.3
#17c: Determine weather [1.25, 1.375] or [1.375, 1.5] contains a zero.
Solution:
f (1.375) = 2(1.375) − (1.375)3 :Plug it in
= −0.005930265698
The interval of [1.25, 1.375] contains a zero because it maintains the positive negative values of f (x) where the
function will cross the x axis and have a zero.
Problem 29
13
Summer Problem Set
30
Rousseau 14
#21: Write and Draw f (x) on [0, 4] such that there is a Jump to
discontinuity at x = 2 and does not satisfy the conclusion of the
IVT
f (x) =
sin(x − 2)
|x − 2|
Graph:
4
2
0
−2
−4
−1
31
0
1
2
3
4
5
#22 Jump to discontinuity at x = 2 and does satisfy the
conclusion of the IVT on [0, 4]
Solution:
⌈x⌉ + 0.8 sin(16x)
Graph:
6
4
2
0
−1
32
0
1
2
3
4
5
#23 Infinite one sided limits at x = 2 and does not satisfy the
conclusion of the IVT
f (x) =
Problem 32
1
x−2
14
Summer Problem Set
Rousseau 15
Graph:
4
2
0
−2
−4
−1
33
0
1
2
3
4
5
#24 Infinite one sided limits at x = 2 yet satisfies the conclusion
of the IVT on [0, 4]
Solution:
f (x) =
3(x − 2)2
1
−
4(x − 2)
9(x − 2)
Graph:
2
1
0
−1
−2
−1
Problem 33
0
1
2
15
3
4
5
Summer Problem Set
Rousseau 16
Chapter 3
Problem 33
16
Summer Problem Set
34
Rousseau 17
d
2
dx [x ]
Solution:
2x
35
d
3
dx [x ]
Solution:
3x2
36
d
17
dx [x ]
Solution:
17x16
37
d
18
dx [x ]
Solution:
18x17
38
d
82
dx [x ]
Solution:
82x81
39
d
49
dx [x ]
Solution:
49x48
40
d
37
dx [x ]
Solution:
37x36
41
d
93
dx [x ]
Solution:
93x92
42
d
1
dx [x ]
Solution:
1
43
d
53
dx [x ]
Solution:
53x52
44
d
2
dx [6x ]
Solution:
12x
Problem 45
17
Summer Problem Set
45
Rousseau 18
d
5
dx [4x ]
Solution:
20x4
46
d
8
dx [6x ]
Solution:
48x7
47
d
dx [10x]
Solution:
10
48
d
4
dx [9x ]
Solution:
36x3
49
d
3
dx [4x ]
Solution:
13x2
50
d
2
dx [x ]
Solution:
x
51
d
dx [sin(x)]
Solution:
cos(x)
52
d
dx [cos(x)]
Solution:
− sin(x)
53
d
dx [− sin(x)]
Solution:
− cos(x)
54
d
dx [− cos(x)]
Solution:
cos(x)
55
d
dx [100, 000]
Solution:
0
Problem 56
18
Summer Problem Set
56
Rousseau 19
d
dx [−18]
Solution:
0
57
d
dx [0]
Solution:
0
58
d
−2
dx [x ]
Solution:
−2
x3
59
d
−22
]
dx [x
Solution:
−22
x23
60
d
−14
]
dx [x
Solution:
−14
x15
61
d
−99
]
dx [x
Solution:
−99
x100
62
d
−20
]
dx [25x
Solution:
−500
x21
Problem 62
19
Summer Problem Set
63
Use the product rule:
Rousseau 20
d
2
dx [(x )(cos(x))
Solution:
(x2 )(− sin(x)) + (cos(x))(2x)
64
d
3
dx [x sin(x)]
Solution:
x3 cos(x) + 3x2 sin(x)
65
d
7
dx [(− sin(x))x )]
Solution:
x7 cos(x) − 7x6 sin(x)
66
d
58
dx [x cos(x)]
Solution:
58x57 cos(x) − x58 sin(x)
67
d
25
dx [4x sin(x)]
Solution:
100x24 sin(x) + 4x25 sin(x)
68
d
89
dx [(− sin(x))2x ]
Solution:
−178x88 sin(x) − 2x89 cos(x)
69
d
37
dx [− cos(x)(3x )]
Solution:
−111x36 cos(x) + 3x37 sin(x)
70
d
9
dx [x cos(x)]
Solution:
9x8 cos(x) − x9 sin(x)
71
d
1
dx [15x sin(x)]
Solution:
15 sin(x) + 15x cos(x)
72
d
52
dx [82x (− cos(x)]
Solution:
−4264x51 cos(x) + 82x52 sin(x)
73
d
2
dx [(− cos(x))6x ]
Solution:
−12x cos(x) + 6x2 sin(x)
Problem 74
20
Summer Problem Set
74
Rousseau 21
d
6
dx [4x cos(x)]
Solution:
24x5 cos(x) − 4x6 sin(x)
75
d
9
dx [6x sin(x)]
Solution:
54x8 sin(x) + 6x9 cos(x)
76
d
dx [(19)(− cos(x))]
Solution:
19 sin(x)
77
d
73
dx [9x (− sin(x))]
Solution:
−657x72 sin(x) − 9x73 sin(x)
78
d
dx [4x cos(x)]
Solution:
4 cos(x) − 4x sin(x)
79
d
2
dx [x sin(x)]
Solution:
2x sin(x) + x2 cos(x)
80
d
dx [cos(x) sin(x)]
Solution:
− sin(x)2 + cos(x)2
81
d
dx [sin(x) cos(x)]
Solution:
cos(x)2 − sin(x)2
82
d
dx [− sin(x) cos(x)]
Solution:
− cos(x)2 + sin(x)2
Problem 82
21
Summer Problem Set
83
Rousseau 22
Use the Quotient Rule:
d cos(x)
dx [ x2 ]
Solution:
−x2 sin(x)−2x cos(x)
x4
−x sin(x)−2 cos(x)
x3
84
:Plug it in
:Simplify
d
x3
[
dx sin(x) ]
Solution:
2
3x sin(x)−x3 cos(x)
sin(x)2
85
d − sin(x)
dx [ x10 ]
Solution:
−x10 cos(x)+10x9 sin(x)
x20
−x cos(x)+10x sin(x)
x11
86
:Plug it in
:Simplify
d
x58
[
dx cos(x) ]
Solution:
57
58x
87
cos(x)+x58 sin(x)
cos(x)2
d 10x81
dx [ sin(x) ]
Solution:
80
sin(x)−10x81 cos(x)
cos(x)2
810x
88
d − sin(x)
dx [ 7x40 ]
Solution:
−7x40 cos(x)+280x39 sin(x)
49x80
−x cos(x)+40 sin(x)
x41
89
d 15x
dx [ sin(x) ]
Solution:
15 sin(x)−15x cos(x)
sin(x)2
90
d
11x60
[
dx − cos(x) ]
Solution:
59
−660x
91
cos(x)−11x60 sin(x)
− cos(x)2
d − cos(x)
dx [ 6x54 ]
Solution:
53
−324x
cos(x)−6x54 sin(x)
− cos(x)2
Problem 92
22
:Plug it in
Simplify
Summer Problem Set
92
Rousseau 23
4x6
d
[
dx cos(x) ]
Solution:
5
24x cos(x)+4x6 sin(x)
cos(x)2
93
d 6x19
dx [ sin(x) ]
Solution:
18
sin(x)−6x19 cos(x)
sin(x)2
114x
94
d
19
dx [ − cos(x) ]
Solution:
−19 sin(x)
sin(x)2
−18
sin(x)
95
:Plug it in
:Simplify
9x73
d
dx [ − sin(x) ]
Solution:
72
−657x
96
sin(x)+9x73 cos(x)
− sin(x)2
d
4x
dx [ cos(x) ]
Solution:
4 cos(x)+4x sin(x)
cos(x)2
97
x2
d
[
dx sin(x) ]
Solution:
2x sin(x)−x2 cos(x)
sin(x)2
98
d cos(x)
dx [ sin(x) ]
Solution:
− sin(x) sin(x)−cos(x) cos(x)
sin(x)2
− sin(x)2 −cos(x)2
sin(x)2
2
sin(x)2
− sin(x)2 − cos(x)
sin(x)2
2
−1 − cot(x)
:Plug it in
:Simplify
:Split Fraction
:Simplify
2
−(1 + cot(x) ) :Factor out Negative
− csc(x)2 Pythagorean Trig Identity
Problem 99
23
Summer Problem Set
99
Rousseau 24
d sin(x)
dx [ cos(x) ]
Solution:
cos(x) cos(x)+sin(x) sin(x)
cos(x)2
cos(x)2 +sin(x)2
cos(x)2
cos(x)2
sin(x)2
cos(x)2 + cos(x)2
2
:Plug it in
:Simplify
:Split Fraction
tan(x) + 1 :Simplify
sec(x)2 :Pythagorean trig Identity
Problem 99
24
Summer Problem Set
100
Rousseau 25
Use the chain rule:
d
2
dx [cos(x )]
Solution:
−2x sin(x2 )
101
d
2
dx [(sin(x)) ]
Solution:
2 sin(x) cos(x)
102
d
42
dx [(− sin(x)) ]
Solution:
42 sin(x)41 cos(x)
103
d
8
dx [(− cos(x)) ]
Solution:
−8 cos(x)7 sin(x)
104
d
5
dx [cos(10x )]
Solution:
−50x4 sin(10x5 )
105
d
3
dx [− sin(6x )]
Solution:
−18x2 cos(6x3 )
106
d
83
dx [− cos(8x )]
Solution:
664x82 sin(8x83 )
107
d
61
dx [sin(13x ]
Solution:
793x60 cos(13x61 )
108
d
11
dx [(6 sin(x)) ]
Solution:
11(6 sin(x))10 6 cos(x) :Plug it in
11 × 6 × 610 sin(x)10 cos(x) :Re-arrange
399076716 sin(x)10 cos(x) :Simplify
Problem 109
25
Summer Problem Set
109
Rousseau 26
d
17
dx [(4 sin(x)) ]
Solution:
16(4 sin(x))16 4 cos(x) :Plug it in
16 × 416 × 4 sin(x)16 cos(x) :Re-arrange
42 × 416 × 4 sin(x)16 cos(x)
419 sin(x)16 cos(x) :Simplify
274877906900 sin(x)16 cos(x)
110
d
dx [− cos(3x)]
Solution:
3 sin(3x)
111
d
dx [− sin(4x)]
Solution:
−4 sin(4x)
112
d
65
dx [10(sin(x)) ]
Solution:
650 cos(x)64 cos(x)
113
d
65
dx [5(cos(x)) ]
Solution:
−325 cos(x)64 sin(x)
114
d
89
dx [cos(x )]
Solution:
−89x88 sin(x89 )
115
d
83
dx [− sin(−8x )]
Solution:
664x82 cos(−8x83 )
116
d
23
dx [(sin(x)) ]
Solution:
23 sin(x)22 cos(x)
117
d
50
dx [(cos(x)) ]
Solution:
50 cos(x)49 sin(x)
118
d
9
dx [sin(10x )]
Solution:
90x8 cos(10x9 )
Problem 119
26
Summer Problem Set
119
Rousseau 27
d
8
dx [sin(7x )]
Solution:
56x7 cos(7x8 )
Problem 119
27
Summer Problem Set
120
Rousseau 28
Use the exponential and logarithm rules:
Solution:
ex
121
d
x
dx [15 ]
Solution:
ln(15)15x
122
d
y
dy [7 ]
Solution:
ln(7)7y
123
d
x
dx [29 ]
Solution:
29x ln(29)
124
d
a
da [13 ]
Solution:
13a ln(13)
125
d
p
dp [62 ]
Solution:
62p ln(62)
126
d
x
dx [39 ]
Solution:
39x ln(39)
127
d
y
dy [84 ]
Solution:
84y ln(84)
128
d
y
dy [1 ]
Solution:
0
129
d
x
dx [79 ]
Solution:
79x ln(79)
130
d
x
dx [6e ]
Solution:
6ex
Problem 131
28
d x
dx [e ]
Summer Problem Set
131
Rousseau 29
d
8
dx [6e ]
Solution:
0
132
d
dx [ln(x)]
Solution:
1
x
133
d
a
dx [5 ]
Solution:
0
134
d
3
dx [4 ]
Solution:
4
135
d
2
dx [x ]
Solution:
2x
136
d
dx [log(x)]
Solution:
1
x ln(10)
137
x
d
dx [log( 15 )]
Solution:
x
) :Given
log( 15
log(x) − log(15) :Diff. Log Rule
d
dx [log(x)
− log(15)] :Take
1
x ln(10)
d
dx
−0
1
x ln(10)
138
d
15
dx [log( x )]
Solution:
log( 15
x :Given
log(15) − log(x) :Log Diff. Rule
d
dx [log(15)
− log(x)] :Take
1
x ln(10)
1
− x ln(10)
0−
Problem 139
29
d
dx
Summer Problem Set
139
Rousseau 30
2j
d
dj [log( 5 )]
Solution:
log( 2j
5 ) :Given
log(2j) − log(5) :Log Diff. Rule
log(2) + log(j) − log(5) :Log Sum Rule
d
dx [log(2)
+ log(j) − log(5)] :Take
0+
1
j ln(10)
−0
1
j ln(10)
140
d
dx [log(5x)]
Solution:
x multiplied by n is meaningless
1
x ln(10)
141
d
dx [log(20x)]
Solution:
1
x ln(10)
142
d
32
dx [log(x )]
Solution:
32
x ln(10)
143
d
−12
)]
dx [log(x
Solution:
−12
x ln(10)
Problem 143
30
d
dx
Summer Problem Set
144
Rousseau 31
Use the trig and hyperbolic trig rules:
Solution:
√1
|x| x2 −1
d
−1
dx [15 sec (x)]
145
Solution:
√15
|x| x2 −1
d 2
dy [ 7
146
sec−1 (y)]
Solution:
√2
7|x| x2 −1
d
−1
dx [tan (x)]
147
Solution:
1
1+x2
d
−1
da [15 tan (a)]
148
Solution:
15
1+a2
d 2
dp [ 7
149
tan−1 (p)]
Solution:
2
7(1+x2 )
d
−1
dx [csc (x)]
150
Solution:
−1
√
|x| x2 −1
d
−1
dy [15 csc (y)]
151
Solution:
−15
√
2
|y|
y −1
152
d 2
dy [ 7
csc−1 (y)]
Solution:
−2
√
2
7|y|
153
y −1
d
−1
dx [cot (x)]
Solution:
−1
1+x2
154
d
−1
dx [−15 cot (x)]
Solution:
15
1+x2
Problem 155
31
d
−1
dx [sec (x)]
Summer Problem Set
155
d 2
dx [ 7
Rousseau 32
cot−1 (x)]
Solution:
−2
7(1+x2 )
156
−1
d
dx [sin (x)]
Solution:
√ 1
1−x2
157
d 2
dx [ 7
sin−1 (x)]
Solution:
√2
7 1−x2
158
−1
d
dx [15 sin (x)]
Solution:
√ 15
1−x2
159
d
−1
dx [cos (x)]
Solution:
√ −1
1−x2
160
d 2
dx [ 7
cos−1 (x)]
Solution:
√−2
7 1−x2
161
d
−1
dx [15 cos (x)]
Solution:
√−15
1−x2
162
d
−1
dx [tan (x)]
Solution:
1
1+x2
163
d
−1
dx [15 tan (x)]
Solution:
15
1+x2
164
d 2
dx [ 7
tan−1 (x)]
Solution:
2
7(1+x2 )2
Problem 164
32
Summer Problem Set
Rousseau 33
Chapter 4
Problem 164
33
Summer Problem Set
165
Rousseau 34
4.1, #6: Use the Linear Approximation to estimate
∆f = f (3.02) − f (3):
f (x) = x ln(x)
Solution:
∆f (x) ≈ f ′ (a)∆x
∆f (x) ≈ f ′ (3) × 0.02 :Setup
f ′ (x) = ln(x) + 1 :Differentiate f (x)
ln(3)+1
50
:Expand, Simplest exact value
≈ 0.04197224577 :Approximate
166
#13: Estimate ∆f using the Linear Approximation and use a
calculator to compute both the error and the percentage error:
f (x) = cos(x) a = π4 ∆x = 0.03
Solution:
Approximating ∆f (x):
∆f (x) ≈ −0.03 sin( π4 ) :Setup
√
−3 2
200
:Evaluate
≈ −0.02121320344 :Approximate
Computing ∆f (x):
∆f = f ( π4 + 0.03) − f ( π4 ) :Setup
= cos( π4 + 0.03) − cos( π4 ) :Expand
cos( π4 + 0.03) −
√
2
2
:Evaluate to simplest exact answer
≈ −0.02152819579 :Approximate
Error:
−0.02121320344 + 0.02152819579 :Calculate error
= 0.00041499235 error :Evaluate error
0.00041499235
0.02152819579
× 100 :Calculate % error
≈ 1.39% error :Evaluate and Approximate
Problem 166
34
Summer Problem Set
167
Rousseau 35
#28: A thin silver wire has length L = 18cm when the
temperature is T = 30◦ C. Estimate the the length when
T = 25◦ C if the coefficient of thermal expansion is
◦
k = 1.9 × 10−15 C −1
Solution:
∆L = kL∆T
∆L − 18 × 5 × 1.9 × 10−15 :Use Formula
∆L = 1.71 × 10−13 :Evaluate
Now because silver gets smaller as it cools, the final length at 25◦ C will be smaller than at 30◦ C meaning the
subtraction of ∆L from the initial length will result in the wanted approximation:
18 − 1.71 × 10−13 :Setup
168
#54: approximate Using linerarization and use a calculator to
compute the percent error
1
√
17
Solution:
Define Variables: f (x) =
√1
x
f ′ (x) = − 12 x
−3
2
x = 17 a = 16
Now to Approximate:
f (16) + f ′ (a)(17 − 16) :Set up
16
−1
2
− 12 x
1
4
31
128
Calculator:
−3
2
−
(1) :Expand
1
128
:Simplify
≈ 0.2421875 Linear approximation
1
√ ≈ 0.24253625...
17
Error:
0.2425−0.2421
0.2425
× 100 :Formulate
= 0.16% Error :Simplify
Problem 168
35
Summer Problem Set
169
Rousseau 36
4.2, #1: If f (x) is continuous on (0, 1) then (choose the correct
statement)
Solution:
c f (x) might not have a minimum value on (0, 1)
170
#5: Fermat’s Theorem does not clam that if f ′ (c) = 0, then f (c)
is a local extreme value (this is false). What does Fermat’s
theorem assert
Solution:
Fermat’s theorem states: If f (c) is a local min or max, then c is a critical point on f .
171
#14: Find the critical points of the function:
f (x) = sec−1 (x) − ln(x)
Solution:
Start by taking the derivative
1
1
√
−
|x| x2 − 1 x
Now to find where critical points are when the derivative is equal to zero:
0=
x
√1
− x1
|x| x2 −1
1
√1
x = |x| x2 −1
√
= |x| x2 − 1
x
|x|
=
1=
√
√
:Setup
:add 1x from both sides
:Take 1 over the equation
x2
− 1 :Divide by |x|
x2
n
− 1 : |n|
= 1 provided n > 0
1 = x2 − 1 :Square both sides
2 = x2 :add one to both sides
√
± 2 = x :Take the square root
√
2 = x :Answer must be positive
There is only one critical point of this function, we have the x value, now to find the f (x) value:
√
√
√
f ( 2) = sec−1 ( 2) − ln( 2) :Setup
p
= π4 − ln( (2) :Simplify to exact value
≈ .4388245731... :Approximate for Government work
Putting the values together we get for critical points
√ π
√
( 2, − ln( 2)
4
≈ (1.41, 0.439)
Problem 171
36
Summer Problem Set
172
Rousseau 37
#23: Find the maximum and minimum values of the function
on the given interval
y = 2x2 − 4x + 2, [0, 3]
Solution:
This function is a parabola with a positive a value: it has a minimum at the vertex and a maximum of ∞.
However, given the bound interval, the maximum values would be y of the interval end points. Now to find the
minimum:
x = −b
2a :X value formula
x=
4
2(4)
:Plug in Values
x = 1 :Simplify
y(1) = 2(1)2 − 4(1) + 2 :Plug int to find y value
y(1) = 0 :Simplify
Min = (1, 0) :Put it together
Finding maximums:
y(0) = 2(0)2 − 4(0) + 2 :Plug in endpoint
y(0) = 2 :Simplify
2
]]y(3) = 2(3) − 4(3) + 2 :Plug in other endpoint
y(3) = 8 :Simplify
Maxs = {(0, 2), (3, 8)} :Put it together
173
#42: y =
√
1 + x2 − 2x, [3, 6]
Solution:
This hyperbolic function has no real minimums or maximums. However, given the interval, the minimum and
maximum values are at the interval endpoints:
√
1 + 32 − 2(3) :Plug in value
√
y(3) = −6 + 10 ≈ −2.84 :Simplify and Approximate
y(3) =
√
y(6) = 1 + 62 − 2(6) :Plug in value
√
y(6) = 37 − 12 ≈ −5.92 :Simplify and Approximate
√
Max = (3, −6 + 10) :Max Value
√
Min = (6, 37 − 12) :Min Value
174
#54: y = ex − e2x , [− 21 , 1]
Solution:
This exponential function has a maximum at
−1
2
and the minimum value, given the interval is at 2:
1
1
y(− 21 ) = e− 2 − e2(− 2 ) :Plug in value
y(− 12 ) =
√
e−1
e
≈ 0.2386512185 :Evaluate!
Max = (− 12 ,
√
e−1
e )
:Max Value
y(1) = e1 − e2 :Plug in value
y(1) ≈ −4.67077427 :Approximate
Min = (1, e1 − e2 ) :Min Value
Problem 175
37
Summer Problem Set
175
Rousseau 38
#64: Verify Rolle’s theorem for the given interval:
1
f (x) = x + x−1 ,
,2
2
Solution:
Verification of three things is necessary for application of Mr. Dinner Rolle’s theorem: continuous on [a, b],
differentiable on (a, b), and f (a) = f (b). NOW!, To Prove or Not to Prove!!!!:
x is continuous on [ 12 , 2] : This function is continuous
1
x−1 is continuous on [ 12 , 2] : This function is continuous given x > 0, which it is on [ 12 , 2]
2
3
x + x−1 is continuous on [ 12 , 2] : f + g is continuous given f and g are continuous, see 1,2
f ′ (x) =
4
5
′
f (x) is continuous on
6
f ( 12 ) =
7
f (2) = 2 +
8
1
2
x2 −1
x2
1
[ 2 , 2]
: Differentiate
: f ′ (x) is discontinuous only when x2 = 0, which will not happen on [ 12 , 2], see 4
+ 2 = 2.5 : Find value of f (a)
1
2
= 2.5 : Find Value of f (b)
f (a) = f (b) : Transitivity see 6,7
9 Rolle’s Theorem can be Applied : All conditions have been met see 3,5,8
176
#65: f (x) = sin(x), [ π4 , 3π
4 ]
Solution:
1
f (x) = sin(x) is continuous : This function is continuous
f ′ (x) = cos(x) : Differentiate
2
3
4
5
6
f ′ (x) is continuous : cos(x) is continuous, see 2
f ( π4 ) = sin( π4 ) =
f ( 3π
4 )
=
sin( 3π
4 )
=
√
2
√2
2
2
: Find f (a)
: Find f (b)
f (a) = f (b) : Transitivity, see 4,5
7 Rolle’s Theorem can be Applied : All Conditions have been met, see 1,3,6
Problem 176
38
Summer Problem Set
177
Rousseau 39
#76: Plot the function using a graphing utility and find its
critical points and extreme values on [5, −5]:
y=
1
1
+
1 + |x − 1| 1 + |x − 4|
Solution:
Begin with a graph:
3
2
1
0
−1
−2
−3
−4
−2
0
2
4
Looking at this graph, there are local maximums at x = {1, 4}, a local min between these points, and finally the
absolute min and maximum at the endpoints of the interval. To find the local minimum, one could use the
derivative, however, looking at the graph, it is exactly between the two local maximums at x = 2.5. Find the
values of all these points:
1
1+|−5−4|
y(−5) = 17
70
End Point = (−5, 17
70 )
1
1
y(1) = 1+|1−1| + 1+|1−4|
y(1) = 54
Local Max = (1, 45 )
1
1
y(2.5) = 1+|2.5−1|
+ 1+|2.5−4|
y(−5) =
1
1+|−5−1|
+
:Plug in −5 to find end point
:Evaluate
:Plug in 1 to find local Max
:Evaluate
:Plug in 2.5 to find local Min
y(2.5) = 0.8 :Evaluate
Local Min = (2.5, 0.8)
1
1+|4−4|
y(4) = 54
Local Max = (4, 45 )
1
1
= 1+|5−1|
+ 1+|5−4|
7
y(5) = 10
7
End Point = (5, 10
)
y(4) =
y(5)
1
1+|4−1|
+
:Plug in 4 to find local Max
:Evaluate
:Plug in 5 to find end point
:Evaluate
:All critical points and extreme values have been found!
I had this function plugged into GeoGebra, so I was easily able to find all critical points by looking at the graph.
Problem 177
39
Summer Problem Set
178
Rousseau 40
4.3, #3: Find a point c satisfying the conclusion of the MVT
for the given interval
y = (x − 1)(x − 3), [1, 3]
Solution:
y = (x − 1)(x − 3) :Given
y = x2 − 4x + 3 :Expand
y ′ = 2x − 4 :Differentiate
y(1) = (1 − 1)(1 − 3) :Setup
y(1) = 0 :Simplify
y(3) = (3 − 1)(3 − 3) :Setup
y(3) = 0 :Simplify
m[1,3] y = 0 :Same y values, has no slope
0 = 2x − 4 :Setup to find f ′ (x)
4 = 2x :Add 4 to both sides
2 = x :Divide by 2
y(2) = (2 − 1)(2 − 3) :Find y(2)
= (1)(−1) :Simplify
= −1
c = (2, −1) :Define point c
Problem 179
40
Summer Problem Set
179
Rousseau 41
#7: y = x ln(x), [1, 2]
Solution:
y(1) = (1) ln(1) :Setup
y(1) = ln(1) :Simplify
y(2) = 2 ln(2) :Setup
= ln(22 ) :Pull in exponent
y(2) = ln(4) :Simplify
m[1,2] (y) =
ln(4)−ln(1)
2−1
ln( 4 )
= 11
:Setup to find slope
:Log difference rule
= ln(4) :Simplify
y = x ln(x) :Given Function
′
y = ln(x) + 1 :Differentiate
ln(4) = ln(x) + 1 :Setup
ln(x) = ln(4) − 1 :Subtract 1 from both sides
eln(x) = eln(4)−1 :Raise e to the equation
x = eln(4)−1 :Simplify
x=
f ( 4e ) =
4
e
4
e
:Exponential quotient rule
ln( 4e ) :Setup
= 4e (ln(4) − ln(e)) :Log Difference rule
= ( 4e )(ln(4) − 1) :Simplify
c = ( 4e , 4e )(ln(4) − 1)) :Put it together
c ≈ (1.4715, 0.5684) :Approximate
Problem 180
41
Summer Problem Set
180
Rousseau 42
#10: y = ex − x, [−1, 1]
Solution:
y(−1) = e−1 + 1 :Setup
=
1
e
+ 1 :Simplify
y(1) = e − 1 :Setup, and simplest answer
m[−1,1] y =
(e−1)−( 1e +1)
1+1
e− 1e −2
=
2
:Find Slope
:Simplify
y = ex − x :Given function
y ′ = ex − 1 :Differentiate
e− 1e −2
2
e−
1
e
= ex − 1 :Thing the do
− 2 = 2ex − 2 :Multiply by 2
e−
1
e
= 2ex :Add 2 to both sides
ln(e − 1e ) = ln(2ex ) :Take the natural log and run
ln(e − 1e ) = ln(ex ) + ln(2) :Log product rule
ln(e − 1e ) = x + ln(2) :Simplify
ln(e − 1e ) − ln(2) = x :Subtract ln(2)
1
y(ln(e − 1e ) − ln(2)) = eln(e− e )−ln(2) − ln(e − 1e ) − ln(2) :Plug in to find y value
= eln(
=
c = (ln(e − 1e ) − ln(2),
e− 1
e
2
)
e− 1e
2
e− 1e
2
e− 1e
2
e− 1
ln( 2 e
− ln(
) :Log difference rule
−
) :Simplify
− ln(
e− 1e
2
)) :Define c
c ≈ (0.1614, 1.0138) :Approximate
181
#12: Determine the intervals on which f ′ (x) is positive and
negative assuming that the figure below is f (x)
y
x
1
2
3
4
5
Solution:
On [0, 1] f ′ (x) Is positive
On [1, 3] f ′ (x) Is negative
On [3, 5] f ′ (x) Is positive
On [1, 5] f ′ (x) Is negative
Problem 182
42
6
Summer Problem Set
182
Rousseau 43
#13: Determine the intervals on which f (x) is increasing or
decreasing assuming that the figure below is f ′ (x)
y
x
1
2
3
4
5
6
Solution:
On [0, 1] f ′ (x) Is negative
On [1, 6] f ′ (x) Is negative
183
#20: Use the First Derivative Test to determine whether the
function attains a local minimum or maximum, (or neither) at
the given critical point:
y = x2 − 27x + 2, c = −3
Solution:
y = x2 − 27x + 2 :Given function
y ′ = 2x − 27 :Differentiate
y ′ (−3) = 2(−3) − 27 :find y ′ (3)
y ′ (3) = −33 :Not a local min or max
Problem 183
43
Summer Problem Set
184
Rousseau 44
#31: Find the critical points and the intervals on which the
function is increasing or decreasing, and apply the First
Derivative Test to each critical point:
1
3
y = x3 + x2 + 2x + 4
3
2
Solution:
Lets begin with a graph:
5
4
3
2
1
0
−1
−4
−3
−2
−1
0
1
2
Now to use the First Derivative rule to find critical points:
y = 13 x3 + 32 x2 + 2x + 4 :Given function
y ′ = x2 + 3x + 2 :Differentiate
0 = x2 + 3x + 2 :Setup to find zeros
0 = (x + 1)(x + 2) Factor quadratic
x = {−1, −2} :Find zeros
y ′ (−1) = (−1)2 + 3(−1) + 2 :Apply First Derivative Test
= 1 − 3 + 2 :Simplify
= 0 :Critical Point at − 1
y ′ (−2) = (−2)2 + 3(−2) + 2 :Apply First Derivative Test
= 4 − 6 + 2 :Simplify
= 0 :Critical point at − 2
y(−1) =
1
3
3 (−1)
+
3
2
2 (−1) + 2(−1)
= − 31 + 32 − 2
+ 4 :Plug in critical point to find y value
+ 4 :Simplify
= 3 16 :y value of critical point
y(−2) = 13 (−2)3 + 32 (−2)2 + 2(−2) + 4 :Find y value of other critical point
= − 83 +
12
2
− 4 + 4 :Simplify
= 8 23 :y value of critical point
Problem 184
44
Summer Problem Set
Rousseau 45
Critical Points = {(−1, 3 16 )(−2, 8 32 )} :Critical Points
x ≤ −2 : y(x) increasing interval
−2 ≥ x ≥ −1 : y(x) decreasing interval
x ≥ −1 : y(x) increasing interval
185
#42: y = cos(θ) + sin(θ), [0, 2π]
Solution:
Graph:
1.5
1
0.5
0
−0.5
−1
−1.5
0
1
2
3
4
5
6
y = sin(θ) + cos(θ) :Given Function
y ′ = cos(θ) − sin(θ) :Differentiate
0 = cos(θ) − sin(θ) :Setup to find zeros
sin(θ) = cos(θ) :Add sin(θ)
θ = { π4 , 5π
4 } :Only two values that satisfy equation
y ′ ( π4 ) = cos( π4 ) − sin( π4 ) :Apply First Derivative test
√
=
2
2
−
√
2
2
:Evaluate
= 0 :Critical point atx =
π
4
5π
5π
y ′ ( 5π
4 ) = cos( 4 ) − sin( 4 ) :Apply First Derivative Test
√
=−
2
2
+
√
2
2
:Evaluate
= 0 :Critical point atx =
5π
4
y( π4 ) = sin( π4 ) + cos( π4 ) :Plug in to find y value
√
=
2
2
+
=
√
2
2
√
:Simplify
2
5π
5π
y( 5π
4 ) = sin( 4 ) + cos( 4 ) :Plug in to find y value
√
=
2
2
+
=
Problem 185
√
2
2
√
:Simplify
2
45
Summer Problem Set
Rousseau 46
√
√
{( π4 , 2), ( 5π
4 , 2)} :Critical Points
π
4
5π
4
186
π
4
5π
4
0≥x≥
: y(x) positive interval
≥x≥
: y(x) negative interval
≥ x ≥ 2π : y(x) positive interval
#57: Sam made two statements that Deborah found dubious
(a) ”Although the average velocity for my trip was 70 mph, st
no point did my speedometer read 70 mph”
(b) ”Although a policeman clocked me at going 70 mph, my
speedometer never read 65 mph”
In each case, which theorem did Deborah apply, to prove
Sam’s statement false: the Intermediate Value Theorem or
Mean Value Theorem. Explain!
Solution:
For the first statement, Deborah used the mean value theorem: if the average from the start to the end was
70mph, then sometime between then the instantaneous rate of change must have been 70mph, meaning that at
some point, the speedometer must have read 70 mph. Also given that a car cannot change velocity instantly, it
provides a continuous and differentiable function.
For the second statement, Deborah also used the intermediate value theorem, having been clocked at 70, and
having to have start at zero, Deborah must have been going 65 at some point between that time, because a car
cannot instantly change velocity.
Problem 186
46
Summer Problem Set
187
Rousseau 47
4.4, #2: Match each statement with a graph in the figure that
represents company profits as a function of time:
(a) The outlook is great: The growth rate keep increasing
(b) We’re loosing money, but not a quickly as before
(c) We’re loosing money, and its getting worse as time goes on
(d) We’re doing well, but our growth rate is leveling off
(e) Business had been cooling off, but now its picking up
(f) Business had been picking up, but now its cooling off
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Solution:
Parings are: (a, ii) (b, iv) (c, i) (d, iii) (e, vi) (f, v).
188
#7: Determine what intervals on which the function is concave
up or down and find the points of inflection
y = x2 + 7x + 10
Solution:
This function is a parabola, with a positive a value, meaning that it opens upward, and never downward, in
the real plane. In other words, this function is concave up, convex down, for all real values of x. This also
means that there are no points of inflection, only a local minimum.
189
#25: Find the critical points of f (x) and use the second
derivative test (if possible) to determine weather wether each
corresponds to a local minimum or maximum
f (x) = 3x4 − 8x3 + 6x2
Solution:
To find critical points, use the zeros of the derivative:
f (x) = 3x4 − 8x3 + 6x2 :Given function
f ′ (x) = 12x3 − 24x2 + 12x :Differentiate
0 = 12x3 − 24x2 + 12x :Setup to find zeros
0 = 12x(x2 − 2x + 1) :Factor out 12x
0 = 12x(x − 1)2 :Factor perfect square trinomial
x = (0, 1) :These are the zeros
f ′′ (x) = 36x2 − 48x + 12 :Differentiate the Derivative
f ′′ (0) = 36(0)2 − 48(0) + 12 :Apply Second Derivative test atx = 0
f ′′ (0) = 12 = + :Local Minimum at x = 0
f ′′ (1) = 36(1)2 − 48(1) + 12 :Apply Second Derivative test atx = 1
= 36 − 48 + 12 :Simplify
f ′′ (1) = 0 :Neither local min or max at x = 1
Problem 190
47
Summer Problem Set
190
Rousseau 48
#42 Find the intervals on which f is concave up or down, the
points of inflection, and the critical points, and determine
wether each critical point corresponds to a local minimum or
maximum (or neither)
x
f (x) = 2
x +2
Solution:
Use the zeros of the derivative to find critical points.
x
x2 +2
2
−x +2
x4 +4x2
−x2 +2
x4 +4x2
2
f (x) =
′
f (x) =
0=
:Given function
:Differentiate
:Setup to find zeros
0 = −x + 2 :Equation will be zero when numerator is zero
0 = x2 − 2 :Multiply by − 1
√
0 = (x + 2)(x − 2) :Factor, Diff. of 2 squares
√ √
x = {− 2, 2} :Zeros
√
I could use the second derive test to see weather they are a local min or√max, however, this function is + √
for all
+x, and − for all −x, meaning that there is a local minimum at x = − 2, and a local maximum at x = 2.
191
#57 Sketch a graph of a function f (x) satisfying all the given
conditions:
i f ′ (x) > 0 for all x
ii f ′′ (x) < 0 for all x < 0
iii f ′′ (x) > 0 for all x > 0
Solution:
To begin, I thought about function that are + for all +x and − for all −x, I came up with the most basic of
nx, meaning f ′′ (x) = nx, the n here must be > 0, but beyond that it is arbitrary. Now, f ′ (x) need be greater
than zero, a simple function, that has one more x value than the previous is nx2 , n with the same parameters of
f ′′ (x). Continuing the pattern, this means that f (x) will be a cubic function, f (x) = nx3 , lets see how it works
with f (x) = x3 :
f ′ (x) = 3x3 :Differentiate
f ′′ (x) = 6x :Differentiate again
f ′ (x) and f ′′ (x) both fit within the given constraints. Here is the graph:
Problem 191
48
Summer Problem Set
Rousseau 49
2
1
0
−1
−2
−3
−2
−1
0
1
2
3
Another reason that this function works, is that even though it has the potential to have local min’s and max’s,
this function is never decreasing.
Problem 191
49
Summer Problem Set
192
Rousseau 50
4.5, #2: State the change at each transition point A-G in
figure 22. Example: f ′ (x) goes from + to − at A
6
5
4
3
2
1
0
A
0
B
2
1
C
3
D
E
4
F
5
G
6
7
Solution:
A +−
B −−
C −+
D ++
E +−
F −−
G −+
193
#5: Draw the graph of a function for which f ′ and f ′′ take on
the given sign combinations:
−+, − −, − +
Solution:
4
2
0
−2
−4
−3
Problem 194
−2
−1
0
50
1
2
3
Summer Problem Set
194
Rousseau 51
#14: Sketch a graph of the function. Indicate the transition
points, (local extrema and points of inflection)
y = x3 − 3x + 5
Solution:
8
L Max
6
Inflection
4
L Min
2
0
−4
195
−3
−2
−1
0
1
2
3
4
1
#30: y = (x3 − 4x) 3
Solution:
In the graph above, I had to use GeoGebra, due to some funky error in PgfPlots. Point B corresponds to a
local maximum and point A corresponds to a local minimum. Points D through F are the points of inflection.
Problem 195
51
Summer Problem Set
196
Rousseau 52
2
#35: y = (2x2 − 1)e−x
Solution:
2
1
L Max
L Max
Inflection
Inflection
0
Inflection
−1
L Min
−2
−3
197
Inflection
−2
−1
0
1
2
3
#42: Sketch the graph over the given interval. Indicate the
transition points:
x + sin(x), [0, 2π]
Solution:
8
6
4
2
0
Problem 197
0
1
2
3
52
4
5
6
Summer Problem Set
198
Rousseau 53
#52: Calculate the following limits (divide the numerator and
denominator by the highest power of x appearing in the
denominator)
2
3x + 20x
lim
x→∞
4x2 + 9
lim f (x) =
x→±∞
lim
x→∞
199
3x2 +20x
4x2 +9
=
an
× lim xn−m
bm x→±∞
lim x2−2 :Apply Rule
3
4 x→∞
= 34 lim (1)
x→∞
= 34
:Simplify
2
+20x
#53: lim ( 2x3x
4 +3x3 −29 )
x→∞
Solution:
2
+20x
lim ( 3x
)
4
3
x→∞ 2x +3x −29
=
3
2
× lim (x2−4 ) :Apply Rule
x→∞
=
3
2
× 0 :Simplify
=0
200
2
−9
#56: lim ( 7x
4x+3 )
x→∞
Solution:
2
−9
lim ( 7x
4x+3 ) =
x→∞
7
4
× lim (x) :Apply Rule
x→∞
7
4 ×
=
∞ :Evaluate
=∞
Problem 200
53
Summer Problem Set
201
Rousseau 54
#73: Match the functions to their graphs in the figure below:
(a) y =
(c) y =
1
x2 −1
1
x2 +1
3
3
2
2
1
1
0
0
−1
−1
−2
−2
−3
−4
−3
−2
−1
0
1
2
3
4
−3
−4
(b) y =
(d) y =
−3
−2
−1
(A)
3
2
2
1
1
0
0
−1
−1
−2
−2
−3
−2
−1
0
1
2
3
4
−3
−4
(C)
−3
−2
−1
0
(D)
Solution:
Parings are:
(A, b)
(B, c)
(C, a)
(D, d)
Problem 201
1
2
3
4
1
2
3
4
(B)
3
−3
−4
0
x2
x2 +1
x
x2 −1
54
Summer Problem Set
202
Rousseau 55
4.6, #21: Kepler’s Wine Barrel Problem: The following
problem was sated and solved in the work Nova stereometeria
doliorium vinariorum (New Solid Geometry of a Wine
Barrel), published in 1615 by the astronomer Johannes Kepler
(1571-1630). What are the the dimensions of the cylinder of
largest volume that can be inscribed in the sphere of radius R.
Hint: show that the volume of an inscribed cylinder is
2πx(R2 − x2 ) where x is one half the height of the cylinder
Solution:
The volume of a cylinder, or any extruded shape, is the area of the base by the height. In this case, it would
be the area of the circle by the height:
πr2 × h
Now, r and h are related by the equation of a circle, with the same radius as the sphere:
x2 + y 2 = R 2
y is directly related to r and x is half the height. To make this into a function of one variable, y needs to be in
terms of x:
x2 + y 2 = R2 :Given function
y 2 = R2 − x2 :Subtract x2 from both sides
Looking at this function, all that is actually needed is y 2 , time to assemble the volume function:
V (x) = 2x × π(R2 − x2 )
Graph time:
4
2
0
−2
−4
−4
−3
−2
−1
0
1
2
3
4
The point indicated is the one needed, I will use the zeros of the derivative to find this critical point:
V (x) = 2x × π(R2 − x2 ) :Given Function
= −2πx3 + 2πR2 x :Distribute
V ′ (x) = −6πx2 + 2πR2 :Differentiate
0 = −6πx2 + 2πR2 :Setup to find zeros
6πx2 = 2πR2 :Add 6πx2 to both sides
2πR2
6π
2
x2 = R3
q
2
x = R3
x2 =
Problem 202
:Divide by 6π
:Simplify
:Take the square root, and leave no evidence
55
Summer Problem Set
Rousseau 56
knowing what one half the height is, is not enough to solve this problem, next is to find the radius:
y 2 = R2 − x2 : y in terms of x
R2
3
2
y 2 = 2R
3
q
2
y = 2R
3
y 2 = R2 −
:Substitute x2
:Simplify
:Take the square root, with extreme prejudice
Now to make these into dimensions of a cylinder, y is the radius of the base, x however is only one half the
height, making the dimensions:
p
√
R (6)
R2 3
, Height =
Radius =
3
3
Problem 202
56
Summer Problem Set
203
Rousseau 57
#42: The monthly output P of a light bulb factory is given by
the formula P = 350LK where L is the amount invested in
labour and K the amount invested in equipment (in thousands
of dollars). If the company needs to produce 10,000 units per
month, how should the investment be divided among labour
and equipment to minimise cost of production? Total cost of
production being L + K
Solution:
10, 000 = 350LK :Defined function
200
7 = LK
200
7K = L
:Divide by 350
:Divide by K to get L in terms of K
Cost = L + K :Cost Function
Cost(K) = K +
200
7K
:Substitute L
Lets look at a graph:
30
20
10
0
−10
−20
−30
−40
−30
−20
−10
0
10
20
30
40
Given that one cannot invest negative money, the solution must be positive, at the point indicated. To find this
critical point, look at the zeros of the derivative:
C(K) = K +
′
C (K) = 1 −
0=1−
1=
200
7K
200
7K 2
200
7K 2
200
7K 2
:Cost Function
:Differentiate
:Find zeros
:Add
200
7K 2
to both sides
2
7K = 200 :Multiply both sides by 7K 2
K 2 = 200
:Divide by 7
q 7
K = 200
:Take the positive square root
7
K=
√
10 14
7
:Simplify
K ≈ 5.345224838 :Approximate
Problem 203
57
Summer Problem Set
Rousseau 58
Now that the amount to invest in equipment is known, now to find the amount to invest in labor:
200
7K
200
√
L=
L=
L
7 10 7 14
√
= 10200
14
= √2014
√
= 201414
√
= 10 7 14
:Setup
:Plug in K value
:Simplify
:Simplest Exact Answer
L ≈ 5.345224838 :Approximate
So the best way to divide the investment is equally between labour and equipment with the investment of:
$5345.22
Problem 203
58
Summer Problem Set
204
Rousseau 59
4.7, #2: Show that L’Hôptital’s rule is applicable and use it to
evaluate the limit:
2x2 + x − 10
lim
x→2 2x5 − 40x + 16
Solution:
These functions are continuous and differentiable at x = 2
2x2 − x + 10 :Given
4x − 1 :Differentiate
5
2x − 40x + 16 :Given
10x4 − 40 :Differentiate
)
lim ( 4x−1
4
x→2 10x −40
4(2)−1
10(2)4 −40
7
= 120
205
:Apply L’Hôptital’s rule
:Use Substitutions Method
:Evaluate
#17: Apply L’Hôptital’s rule to evaluate the limit, it may need
to be applied more than once:
sin(4x)
lim
x→0 sin(3x)
Solution:
sin(4x) :Given
4 cos(4x) :Differentiate
sin(3x) :Given
3 cos(3x) :Differentiate
lim ( 4 cos(4x) )
x→0 3 cos(3x)
4 cos(4(0))
3 cos(3(0))
4
3
206
x2
:Apply L’Hôptital’s rule
:Use Substitutional Method
:Evaluate
4
−e
#22: lim ( e x−2
)
x→2
Solution:
2
ex − e4 :Given
2xex
2
:Derive
x − 2 :Given
1 :Derive
x2
lim (2xe ) :Apply L’Hôptital’s rule
x→2
2
2(2)e2
:Use Substitutional Method
4
4e ≈ 218.3926001 :Evaluate and Approximate
Problem 207
59
Summer Problem Set
207
Rousseau 60
#42: limπ (sec(x) − tan(x))
x→ 2
Solution:
sec(x) − tan(x) :Given
sin(x)
1
cos(x) − cos(x)
− sin(x)+1
cos(x)
:Put in terms of sin and cos
:Subtract Fractions
− sin(x) + 1 :See Above
− cos(x) :Differentiate
cos(x) :Denominator
− sin(x) :Differentiate
cos(x)
lim ( −
− sin(x) )
x→ π
2
:Apply L’Hôptital’s rule
lim (cot(x)) :Simplify
x→ π
2
cot( π2 ) :Use Substitutional Method
0 :Evaluate
208
#52: Evaluate the limit using L’Hôptital’s rule is necessary
lim ln(x) tan−1 (x)
x→0
Solution:
There is no need for L’Hôptital’s rule, this can be evaluated using the substiutional method:
ln(0) tan−1 (0) :Substitute
ln(0)(0) :Ignore the errors and evaluate where possible
= 0 :ANYTHING by 0 is 0
Problem 208
60
Summer Problem Set
209
Rousseau 61
4.8, #7: Use Newton’s Method to find the two solutions of
ex = 5x
25
y = ex
20
15
y = 5x
10
5
0
0
0.5
1
1.5
2
2.5
3
Solution:
xn+1 = xn −
f (xn )
f ′ (xn )
f (x) = ex − 5x :Our current Function
f ′ (x) = ex − 5 :Take Derivative
x0 = 3 :Define x0
x1 = 3 −
e3 −15
e3 −5
:Function Setup
x1 ≈ 2.662886581 ≈ 2.66 :Approximate
x2 = 2.66 −
e2.66 −13.5
e2.66 −5
:Setup
x2 ≈ 2.574343337 ≈ 2.57 :Approximate
x3 = 2.57 −
e2.57 −5(2.57)
e2.57 −5
Setup
x3 ≈ 2.54324211 ≈ 2.54 :Approximate
Now to test the solution:
f (2.54) = e2.54 − 5(2.54) :Plug in to test Value
≈ −0.020... :Close Enough
Finding Another Solution:
x1 = 0 −
x0 = 0
e0 −5(0)
e0 −5
:Define x0
:Formulate
x1 = 0.25 :Simplify
x2 = 0.25 −
e0.25 −1.25
e0.25 −5
:Formulate
x2 ≈ 0.2591565257 ≈ 0.26 :Approximate
Testing:
e0.26 − 5(0.26) :Function
≈ −0.003069913334 ≈ 0.00 :Close Enough
The two solutions have bee found.
Problem 210
61
Summer Problem Set
210
Rousseau 62
#12: Use newton’s method to approximate the root to three
decimal places, compare with value obtained from a calculator
1
2− 2
Solution:
1
f (0) = 2− 2 :Starting to Define function
1
x = 2− 2 :Define x equation
1
x−1 = 2 2 :Take 1 over the equation
x−2 = 2 :Square the equation
x
−2
− 2 = 0 :Subtract 2
f (x) = x−2 − 2 :Now we have our function
f ′ (x) = −2(x)−3 :Differentiate
Graph:
2
1
0
−1
−2
−1
0
1
2
3
4
Now to Approximate:
x1 = 1 −
x0 = 1
(1)−2 −2
−2(1)−3
:Define x0
:Setup
x1 = 0.5 :Evaluate
x2 = 0.5 −
(0.5)−2 −2
−2(0.5)−3
:Setup
x2 = 0.625 :Evaluate
x3 = 0.625 −
(0.625)−2 −2
−2(0.625)−3
:Setup
x3 ≈ 0.693359375 ≈ 0.693 :Approximate
x4 = 0.693 −
(0.693)−2 −2
−2(0.693)−3
:Setup
x4 ≈ 0.706687443 ≈ 0.707 :Approximate
Comparing with value from a calculator
0.7071067812...
My value is close enough for three decimal places.
Problem 211
62
5
Summer Problem Set
211
Rousseau 63
#17: Estimate the smallest positive solution of sin(θ)
= 0.9 to
θ
three decimal places, use graphing calculator to choose initial
guess.
Solution:
Define our function:
f (x) =
sin(θ)
− 0.9
θ
Graph:
2
1
0
−1
−2
−1
0
Using the graph: x0 = 1. And f ′ (x) =
1
2
x cos(x)−sin(x)
.
x2
x−
x1 = 1 −
3
4
After some simplify, x −
5
f (x)
f ′ (x)
is:
x sin(x) − 9x2
x cos(x) − sin(x)
sin(1)−9
cos(1)−sin(1)
:Setup
x1 ≈ 0.8056603515 ≈ 0.806 :Approximate
x2 = 0.806 −
(0.805) sin(0.806)−0.9(0.806)2
(0.806) cos(0.806)−sin(0.806)
:Setup
≈ 0.7868846085 ≈ 0.787 :Approximate
X3 = 0.787 −
(0.787) sin(0.787)−0.9(0.787)2
(0.787) cos(0.787)−sin(0.787)
:Setup
x3 ≈ 0.7866831278 :Approximate
≈ 0.787 :Solution to three decimal places
Problem 211
63
Summer Problem Set
212
Rousseau 64
4.9, #9: Match the functions in 9 − 12 with their
antiderivatives A − D
A F (x) = cos(1 − x)
B F (x) = − cos(x)
C F (x) = − 12 cos(x2 )
D F (x) = sin(x) − x cos(x)
f (x) = sin(x)
Solution:
B F (x) = − cos(x)
213
#10: f (x) = x sin(x2 )
Solution:
C F (x) = −1 21 cos(x2 )
214
#11: f (x) = sin(1 − x)
Solution:
A F (x) = cos(1 − x)
215
#12: f (x) = x sin(x)
Solution:
D F (x) = sin(x) − x cos(x)
216
#14: Evaluate the Indefinite integral:
Z
(9 − 5x)dx
Solution:
5 2
2 x + 9x + C
217
#16:
R
8s−4 ds
R
(5x3 − x−2 − x 5 )dx
Solution:
− 83 s−3 + C
218
#18:
3
Solution:
8
5 4
−1
− 58 x 5 + C
4x + x
219
#20:
R
√1 dx
x
Solution:
√
2 x+C
Problem 220
64
Summer Problem Set
220
#22:
R
Rousseau 65
(x3 + 4x−2 )dx
Solution:
1 4
−1
+C
4 x − 4x
221
#24:
R √
( x)dx
Solution:
2 32
3x
222
#26:
R
(4t − 9)−3 dt
Solution:
We need to transform this equation:
1
(4t − 9)3
Now let u = 4t − 9, to integrate by substitution.
du
=4
dx
dx =
Substitution:
Z
−
1
du
4
1 1
du
u3 4
1
1
×
2u2
4
−1
8u2
Replacing u:
−8−1 (4t − 9)−1 + C
223
#28:
R
3
3 dx
x2
Solution:
3
3
x2
3x
R
3x
−3
2
:Given
−3
2
:Re-arrange
R
dx :Take
1
= −6x− 2 + C
224
#30:
R
(x + x−1 )(3x2 − 5x)dx
Solution:
(x + x−1 )(3x2 − 5x) :Given
3x3 − 5x2 + 3x − 5 :Foil it out
R
(3x − 5x2 + 3x − 5)dx :Take the integral
3 4
4x
Problem 225
3
− 53 x3 + 32 x2 − 5x + C :Simplify
65
Summer Problem Set
225
#32:
R
Rousseau 66
x2 −2x−3
dx
x4
x2 −2x−3
x4
x2
2x
3
4 −
4 −
4
R 1 x 2 x 3 x
( x2 − x3 − x4 )dx
− x1 + x12 + x14 + C
226
#34:
R
:Given
:Separate fraction
:Simplify and take
R
sin(9x)dx
Solution:
let u = 9x
du
dx
dx =
R
1
9
227
#36:
R
= 9 :Find dx in terms of u
1
9 du
:Divide by 9 and multiply by dx
1
9 sin(u)du
1
9 cos(u)du
:Substitute
cos(9x) + C :Un-substitute
(4θ + cos(8θ)dθ
Solution:
The integral will need to be spit up to best evaluate
cos(8θ) :Given
letu = 8θ :Define u
du
dθ
dθ =
R
= 8 Derive u
1
8 du
:Divide by 8 and multiply by dθ
cos(u) 18 du :Substitute
− 18 sin(u)
− 18 sin(8θ) :Un-substitute and be done
Now to evaluate the other part of the integral:
R
(4θ)dθ :Given
2θ2
Now to add them together:
2θ2 −
228
#38:
R
1
sin(8θ) + C
8
18 sin(3z + 8)dz
Solution:
18 sin(3z + 8) :Given
u = 3z + 18 :Define u
du
dz
dz =
R
= 3 :Differentiate u
1
3 du
:Divide by 3, multiply by du
18 sin(u) 13 du :Substitute
−6 cos(u) + C :Simplify
−6 cos(3z + 8) + C :Un-substitute
Problem 229
66
Summer Problem Set
229
#40:
R
Rousseau 67
( x8 + 3ex )dx
Solution:
First we need to transform this function
8
x
+ 3ex :Given
8 + x3ex :Multiply by x
Now to integrate by parts:
Z
Z
udv = uv −
vdu
x3ex :Start here
u = x, dv = 3ex :Define u and dv
du = 1, v = 3ex :Find du and v
x3ex −
R
3ex (1) Plug it in
x3ex − 3ex :Simplify
Now to integrate the other part
R
8dx :The other part of f (x)
8x
Putting them together:
8x + x3ex − 3ex
230
#42:
R
(2x + e14−2x )dx
Solution:
e14−2x :Start Here
u = 14 − 2x :Defineu
du
dx
= −2 :Derive u
dx = − 12 du :Divide by -2, multiply by dx
R
R
2xdx + (− 21 eu )du :Plug it all in
2 − 21 eu
2 − 12 e14−2x + C :Un-substitute
231
#52: Solve the differential equation with initial condition:
dy
= 8x3 + 3x2 − 3, y(1) = 1
dx
Solution:
Start by integration:
R
dy
( dx
= 8x3 + 3x2 − 3)dx :Start Here
y(x) = 2x4 + x3 − 3x + C :Integrate dx and add constant
y(1) = 2(1)4 + (1)3 − 3(1) + C = 1 :Plug in the given information to find the constant
1 = 2 + 1 − 3 + C :Simplify
1=C
4
3
y(x) = 2x + x − 3x + 1 :Use the known information to find function
Now to check answer:
d
4
dx 2x
dy
dx =
Problem 232
+ x3 − 3x + 1 :Check by deriving
8x3 + 3x2 − 3 :And we’re back
67
Summer Problem Set
232
Rousseau 68
#65: First find f ′ then find f
f ′′ (x) = x3 − 2x + 1, f ′ (1) = 0, f (4) = 4
Solution:
Start by integration:
R
f ′ (x) = (x3 − 2x + 1)dx :Set it up
f ′ (x) = 14 x4 − x2 + x + C :Simplify
f ′ (1) = 41 (1)4 − (1)2 + (1) + C = 0 :Plug in given information
0=
1
4
− 1 + 1 + C :Simplify
− 41 = C :Solve for C
f ′ (x) = 41 x4 − x2 + x −
1
4
Plug into f ′ (x)
Now to integrate again:
1 4
1
2
4x − x + x − 4
1 3
1 2
1
1 5
= 20 x − 3 x + 2 x − 4 x + C
1
1
1
1
5
3
2
20 (4) − 3 (4) + 2 (4) − 4 (4) + C = 4
4 = 51.2 − 21 13 + 8 − 1 + C
−32 13
15 = C
1 3
1 2
1
13
1 5
20 x − 3 x + 2 x − 4 x − 32 15
f (x) =
f (4) =
Problem 232
R
68
:Set it up
:Simplify
:Plug in known information
:Simplify4 = 36 13
15 + C
:Solve for C
The Final Function
Summer Problem Set
Rousseau 69
Chapter 5
Problem 232
69
Summer Problem Set
233
Rousseau 70
5.1, #13: Calculate the approximation for the given function
and interval:
R8 , f (x) = 7 − x, [3, 5]
Solution:
∆x =
∆x
8
P
1
4
j=1
:Calculate∆x
:Simplify
(7 − (3 + 41 j)) :Set Up
1
4
1
4
5−3
8
= 41
8
P
P
(4 − 4j ) :Simplify
j=1
× 4((4 − 41 ) + (4 − 84 )) :Expand
15
4
+2
= 5 34
234
#15: M4 , f (x) = x2 , [0, 1]
Solution:
∆x =
1
4
8
P
=
1
4
:Calculate ∆x
((0 + (j − 12 ) × 14 )2 ) :It’s all a set up!
j=1
8
P
1
4
235
0−1
4
j=1
2
j
( 16
−
1
64 )
:Simplify
1
4
101
8
101
32
:Sires is neither arithmetic nor geometric
×
#19: L5 , f (x) = x−1 , [1, 2]
Solution:
∆x =
1
5
4
P
=
1
5
:Find the value of ∆x
((1 + 4j )−1 ) :Plug in values
j=0
1
5
Problem 235
1−2
5
4
P
( 1+1 j )
4
j=0
1
743
×
5
210
743
= 1050
:Sum is neither Arithmetic or geometric
70
Summer Problem Set
236
Rousseau 71
5.2, #13: Evaluate the integrals shown in the figure below, the
two parts are semicircles:
y
x
2
236.1
a.
R2
0
4
6
f (x)dx
Solution:
Acircle = πr2 :Area of a circle formula
1
2 Acircle |r
= 1 :Area of a Semi Circle is
1
2
Acircle . From the picture r = 1
1
2
2 π(1)
= − π2
:Plug in values
:Area is below the x axis making it negative
≈ −1.570796327 :Approximation
236.2
b.
R6
0
f (x)dx
Solution:
R2
0
R6
R
f (x) dx :Re-write
R6
− π2 + 2 f (x) :Plug in answer from last part
f (x) dx +
R6
2
2
f (x) = 12 π(2)2 :Use the area formula
= 2π :Simplify
− π2 + 2π :Add it together
= 1.5π :Simplify
≈ 4.71238898 :Approximation
236.3
c.
R4
1
f (x)dx
Solution:
R
R
1 2
1 6
2 0 +2 2
1
π
1
2 (− 2 ) + 2 (2π)
π − π4
3
4π
:re-write
R
:Plug in parts from other questions
:Simplify
≈ 2.35619449 :Approximation
Problem 236
71
Summer Problem Set
236.4
d.
R6
1
Rousseau 72
f (x)dx
Solution:
1
2
R2
0
f (x)dx +
π
1
2 (− 2 )
R6
2
R
:Re-write
+ 2π :Plug in values from previous parts
2π −
π
4
:Simplify
1.75π
≈ 5.497787144 :Approximation
237
#27: Calculate the Riemann sum R(f, P, C) for the given
function partition and choice of intermediate points. Also,
sketch the graph of f and the rectangles corresponding to
R(f, P, C)
f (x) = x, P = {1, 1.2, 1.5, 2}, C = {1.1, 1.4, 1.9}
Solution:
4
P
f (ci )∆xi :Setup
P
i=1
f (1.1)(1.2 − 1) + f (1.4)(1.5 − 1.2) + f (1.9)(2 − 1.5) :Expand with the given sets
(1.1)(.2) + (1.4)(.3) + (1.9)(.5) :Simplify
0.22 + 0.42 + 0.95
1.59
Graph:
2
1.5
1
0.5
0
Problem 237
0
0.5
1
1.5
2
72
2.5
3
3.5
4
Summer Problem Set
238
Rousseau 73
#31: Use the basic properties of the integral and the formulas
in the summary to calculate the integrals:
Z 4
x2 dx
0
Solution:
Rb
R 04
0
x2 dx = 31 b3 :General Formula
x2 dx = 13 43 :Plug in Values
21 13 :Simplify
239
#37:
R1
−a (x
2
+ x)dx
Solution:
R0
R1
(x2 + x)dx + 0 (x2 + x)dx
−a
R −a
R1
− 0 (x2 + x)dx + 0 (x2 + x)dx
R −a
R −a
R1
R1
− 0 (x2 )dx − 0 (x)dx + 0 (x2 )dx + 0 (x)dx
− 31 (−a)3 − 12 (−a)2 + 31 (1)3 + 21 (1)2
5
1
1
3
2
6 − 3 (−a) − 2 (−a)
240
:Split
R
R
:Negate
R
:Split s again
:Integrate
:Simplify
#49: Use the formulas in the summary and Eq. (9) to evaluate
the integral:
Z 2
(x − x3 )dx
1
Solution:
R2
R1
(x − x3 )dx − 0 (x − x3 )dx
0
R2
R2
R1
R1
( 0 (x)dx − 0 (x3 )dx) − ( 0 (x) − 0 (x3 ))
( 21 (2)2 − 14 (2)4 ) − ( 21 (1)2 − 14 (1)4 )
(2 − 4) − ( 12 − 14 )
−2 − 41
:Split Integral
:Split Again
:Integrate
:Simplify
−2.25
241
#59: Express the integrals as a single integral:
Z 3
Z 7
f (x)dx +
f (x)dx
0
3
Solution:
Z
7
f (x)dx
0
Integrals are a summation, the 3 gets included when adding both together.
Problem 241
73
Summer Problem Set
242
Rousseau 74
5.3, #7: Evaluate using FTC 1
Z 2
u2 du
−3
Solution:
R
u2 ud :Find Indefinite integral
1 3
3u
1
3
3 (2)
+ C :Integrate
− 13 (−3)3 :Use FTC 1
8
3
+ 9 :Simplify
= 11 32
243
#16:
R4
0
(3x5 + x2 − 2x)dx
Solution:
R
(3x5 + x2 − 2x)dx :Take Indefinite
3 4
4x
+ 13 x3 − x2 + C :Integrate
( 43 (4)4 + 13 (4)3 − (4)2 ) − ( 34 (0)4 + 31 (0)3 − (0)2 ) :Use FTC 1
193 13 :Simplify
244
#21:
R4
1
1 t2 dt
Solution:
t−2 dt :Indefinite
R
R
− 1t + C :Integrate
− 14 +
=
245
#33:
R π2
− π2
1
1
3
4
:Apply FTC 1
:Simplify
cos(x) dx
Solution:
R
cos(x) dx :Indefinite
R
sin(x) + C :Integrate
sin( π2 ) − sin(− π2 ) :Apply FTC 1
1 + 1 :Simplify
=2
Problem 246
74
R
Summer Problem Set
246
#37:
Rπ
4
0
Rousseau 75
sec(t)2 dt
Solution:
R
sec(x)2 dx :Indefinite
R
tan(x) + C :Integrate
tan( π4 ) − tan(0) Apply FTC 1
√
2
2
247
:Simplify
#51: Evaluate in terms of constants:
Z 5a
dx
x
a
Solution:
R
1
x
dx :Indefinite
R
ln(x) :Integrate
ln(5a) − ln(a) :Apply Constants
ln( 5a
a ) :Log Difference Rule
ln(5) :Simplify
≈ 1.60943912 :Approximate
Problem 247
75
Summer Problem Set
248
Rousseau 76
5.4, #11: Find formula’s for the Function represented by the
integral:
Z 5
et dt
x
Solution:
Z
b
f (x) = F (b) − F (a)
a
R
et dt :Start by calculating indefinite integral to find F (x)
et + C :Integrate
e5 − ex :Calculate Definite integral
249
#29: Calculate the Derivative:
Z x2
d
sin(t)2 dt
dx 0
Solution:
R
sin(t)2 dt :Indefinite Integral
− sin(2t)−2t
+ C :Integrate
4
2
2
sin(2(0))−2(0)
4
2
2
− sin(2x4)−2x
sin(2x2 )−2x2
d
)
dx (−
4
1 d
2
2
− 4 dx sin(2x ) − 2x
− 14 (4x cos(2x2 ) − 4x)
2
− sin(2x4)−2x +
:Find Definite integral
:Simplify
:Setup
:Algebrate
:Differentiate
x − x cos(2x ) :Simplify
250
#31:
d
ds
R cos(s)
−6
(u4 − 3u)du
Solution:
R
(u4 − 3u)du :Indefinite Integral
1 5
5u
( 15 cos(s)5 −
3
2
− 32 u2 + C :Integrate
cos(s)2 ) − ( 15 (−6)5 − 32 (−6)2 ) :Definite Integral
1
5
d 1
ds 5
cos(s)5 −
cos(s)5 −
3
2
3
2
cos(s)2 + 1609.2 :Simplify
cos(s)2 + 1609.2 :Differentiate
4
− cos(s) sin(s) + 3 sin(s) cos(s) :Simplify
251
251.1
#41: Area Functions and Concavity Explain why the
following statements are true. Assume f (x) is Differentiable
a. If c is an inflection point of A(x) then, f (c) = 0
Solution:
A point of inflection is where the gradient, rate of change, of A(x) flips from positive to negative, or negative
Problem 251
76
Summer Problem Set
Rousseau 77
to positive. meaning that the derived function of A(x), that being f (x), will have to cross the x-axis at that x
value, meaning that at that value it must be zero.
251.2
b. A(x) is concave up then f (x) is increasing
Solution:
When A(x) is concave up, the values of A(x) is increasing drastically more than the value of x meaning that
will be increasing, and so will the derivative f (x).
the gradient, A(x)
x
251.3
c. A(x) is concave down then f (x) is decreasing
Solution:
When A(x) is concave down, then A(x) is not increasing as much as x is. This leads to decreasing gradient
values, that being decreasing derivative values in f (x).
Problem 251
77
Summer Problem Set
Rousseau 78
252
5.5, #23: The heat capacity C(T ) of a substance is the amount
of energy, (in joules) required to raise the temperature of 1g by
1◦ C at temperature T
252.1
a. Explain why the energy required to raise the temperature from T1 to T2
is the area under the graph of C(T ) over [T1 , T2 ]
Solution:
The area under the curve represents the total change of energy for that temperature. This expression of total
change manifests as a multiplication, which is then viewed on the graph as being an area. C(T ) is a graph of
how energy changes over a given temperature, is one is given the ratio of energy to temperature, and need find
the energy, the operation that should be performed is multiplication, by looking at the graph of C(T ) one can
see the mapped relation of temperature over energy for temperature, these values on a graph represent
dimensions of an area, the area under the function C(T ).
252.2
b. How much energy
is required to raise the temperature from 50 to 100◦ C
√
if C(T ) = 6 + 0.2 T
Solution:
√
(6 + 0.2 T )dT :Take Integral to find energy
√
R
(6 + 0.2 T )dT :Take Indefinite Integral to find definite integral
R 100
50
3
6x + 0.3T 2 :Integrate
R
R
3
3
(6(100) + 0.3(100) 2 − (6(50) + 0.3(50) 2 ) Use indefinite to find Def.
√
900 − 600 + 75 2 :Simplify
√
300 + 75 2 Joules :Simplest Exact Answer
≈ 406.0660172 Joules :Approximation for Government Work
Problem 252
78
Summer Problem Set
253
Rousseau 79
5.6, #13: Write the integral in terms u and du. Then Evaluate.
Z
x+1
dx, u = x2 + 2x
2
3
(x + 2x)
Solution:
R
du
dx
1
2x+2 du
x+1
u3 (2x+1) du
R 1 −3
du
2u
− 14 u−2 + C
−2
− 14 (x2 + 2x)
254
#24:
R
:Substitute
= 2x + 2 :Find dx in terms of du
dx =
R
x+1
u3 dx
:Divide by 2x + 2 and multiply by du
:Substitute in terms of du
:Factor out x + 1
:Simplify
+ C :Un-substitute
(sec(θ)2 )etan(θ) dθ, u = tan(θ)
Solution:
R
(sec(θ)2 )eu dθ :Substitute integral
du
dx
= sec(θ)2 :Find the value of du
du =
R
1
sec(θ)2 dx
:THE ALGEBRA!!!
1
sec(θ)2 eu sec(θ)
2 du :DO THE DO!!!
R u
e du :BURR
eu :cAlC
e
tan(θ)
+ C :The u
I’m sorry, it is too late at night, my answers should be right, but yeah.
Problem 254
79
Summer Problem Set
255
Rousseau 80
#37: Evaluate the Indefinite integral:
Z p
x x2 − 4 dx
Solution:
u = x2 − 4 :Define u
R √
x u dx :It is ...
du
dx
= 2x :Derivitive!
dx =
1
2x du
:Algebra!
R √ 1
x u du :Substitution!
R 1 √2x
2 u du :Simplify!
3
1
2
3 (u)
1
2
3 (x
256
#57:
R
+C
3
2
− 4) + C :Un-Substitute!
cos(2x)
1+sin(2x) dx
Solution:
u = sin(2x) :Define u
R cos(2x)
1+u dx :What is is ...
du
dx
= 2 cos(2x) :Derivitive
dx =
R
1
2
#61:
R
:Do Algebra
1
( 2 cos(2x)
)( cos(2x)
)du :Put in du
R 1+u
1
( 2+2u )du :Simplify
R 1
( 2 (u + 1)−1 )du
1
2
257
1
2 cos(2x) du
ln(u + 1) + C :Integrate
ln(sin(2x) + 1) + C :Un-substitute
sec(x)2 (4 tan(x)3 − 3 tan(x)2 ) dx
Solution:
u = tan(x) :Define u
du
dx
= sec(x)2 :Derive
dx =
R
1
sec(x)2
:Algebrate
1
sec(x)2 (4u3 − 3u2 ) sec(x)
:Setup
2
R
3
2
(4u − 3u )dx :Simplify
u4 − u3 + C
tan(x)4 − tan(x)3 + C :Un-substitute
Problem 258
80
Summer Problem Set
258
Rousseau 81
#87: Use the Change of Variables Formula to evaluate the
definite integral:
Z 1
θ tan(θ2 ) dθ
0
Solution:
Z
b
f (u(x))u′ (x) dx =
a
Z
u(b)
f (u)du
u(b)
u = θ2 :Define u
u′ = 2θ :Find u′
1 1
′
2 0 tan(u)u
2
R
1
1
2 02 tan(u)
R
1 1
2 0 tan(u)
R
1
2 ln(| sec(u)|)
1
2
2 ln(| sec(θ )|)
Problem 258
dθ :Setup
du :Apply Formula
du :Simplify
+ C :Integrate
+ C :Un-substitute
81
Summer Problem Set
259
Rousseau 82
5.7, #21: Calculate the integral
Z
(x + 1)dx
√
1 − x2
Solution:
Z
Z
udv = uv −
vdu
u = x + 1, dv =
√ 1
1−x2
−1
du = 1, v = sin
:Define u and dv
(x) :Find v and du
R
(x + 1)(sin−1 (x)) − sin−1 (x)(1) :Integrate by parts
√
(x + 1)(sin−1 (x)) − ((x) sin−1 (x) + −x2 + 1) + C Simplify
260
#25:
R
tan−1 (x)dx
1+x2
Solution:
u = tan−1 (x) :Define u
du
dx
1
1+x2
2
=
:Find dx int terms of du
du = (1 + x )dx :Multiply both sides by dx and (1 + x2 )
R
udu :Substitute
−1
tan
u2
2
(x)
:Integrate
:Un-substitute
2
261
#47:
R
√ dx
x 25x2 −1
Solution:
u=
du
dx
dx =
√
25x2 − 1 Define u
√ 25
25x2 −1
25
( √25x
)du
2 −1
=
R
tan
Problem 261
−1
1
u2 +1 du
−1
:Find du in terms of dx
:Do algebra to it
:Set it up
tan (u) :Simplify
√
( 25x2 − 1) :Un-substitute
82
Summer Problem Set
262
262.1
Rousseau 83
5.8, #33: A bank pays interest at a rate of 5%. What is the
yearly multiplier of interest compounded (A-C)?
A: Yearly
Solution:
A = P (1 +
0.05 (1)(t)
1 )
t
:Compounded Interest Formula
A = P (1.05) :Simplify
Yearly Multiplier = 1.05 :Locate Yearly Multiplier
262.2
B: Three times a Year
Solution:
0.05 3(t)
3 )
61 3t
A = P ( 60
)
226981 t
= P ( 216000 )
A = P (1 +
A
Yearly Multiplier =
262.3
226981
216000
:Plug into Formula
:Simplify
≈ 1.050837963 :Find Wanted Answer
C: Continuously
Solution:
A = P e(0.05)t :Continually Compounded Interest Formula
Yearly Multiplier = e0.05 ≈ 1.05127196 :The Answer
263
#34: How long will it take for $4, 000 to double in value if it is
deposited in an account bearing %7 interest, continuously
compounded.
Solution:
8, 000 = 4, 000e0.07×t :Setup Equation
2 = e0.07×t :Divide both sides by 4, 000
ln(2) = ln(e0.07×t ) :Take the ln of both sides
ln(2) = 0.07 × t :Simplify
100 ln(2)
7
= t :Divide by 0.07
≈ 9.902102579 Years :Approximate to find how long
Problem 263
83
Summer Problem Set
Rousseau 84
Further Practice
Problem 263
84
Summer Problem Set
264
Rousseau 85
Evaluate the definite integral:
Z 3
x2 dx
0
Solution:
R
x2 dx = 31 x3 :Take Indefinite
R
− 13 (0)3 :Setup Definite
R
1
3
3 (3)
=9
265
Rπ
0
cos(x)dx
Solution:
R
cos(x)dx = sin(x) :Indefinite
R
sin(π) − sin(0) Definite
R
=0
266
Rπ
2
0
cos(x)dx
Solution:
R
cos(x)dx = sin(x) :Indefinite
R
sin( π2 ) − sin(0) Definite
R
=1
267
Rπ
2
0
sin(x)dx
Solution:
R
sin(x)dx = − cos(x) :Indefinite
R
cos(0) − cos( π2 ) :Definite
R
=1
268
Rπ
0
sin(x)dx
Solution:
R
sin(x)dx = − cos(x) :Indefinite
R
cos(0) − cos(π) :Definite
=2
269
R3
−3 x
3
dx
Solution:
x3 dx = 41 x4 :Indefinite
R
1
1
4
4
4 (3) − 4 (−3) :Definite
R
=0
Problem 270
85
R
R
Summer Problem Set
270
R1
−1 x
6
Rousseau 86
dx
Solution:
x6 dx = 17 x7 :Indefinite
R
1
1
7
7
7 (1) − 7 (−1) :Definite
R
=
271
R1
−1 x
7
R
2
7
dx
Solution:
x7 dx = 81 x8 :Indefinite
R
1
1
8
8
8 (1) − 8 (−1) :Definite
R
R
=0
272
R2
0
x4 dx
Solution:
x4 dx = 15 x5 :Indefinite
R
1
1
5
5
5 (2) − 5 (0) :Definite
R
=
273
R4
0
R
32
5
x4 dx
Solution:
x4 dx = 15 x5 :Indefinite
R
1
1
5
5
5 (4) − 5 (0) :Definite
R
R
= 204 45
274
Rπ
−π
cos(x)dx
Solution:
R
cos(x)dx = sin(x) :Indefinite
R
sin(π) − sin(−π) :Definite
R
=0
275
R 3π
2
0
sin(x)dx
Solution:
R
sin(x)dx = − cos(x) :Indefinite
R
cos(0) − cos( 3π
2 ) :Definite
=1
Problem 276
86
R
Summer Problem Set
276
R 3π
2
0
Rousseau 87
cos(x)dx
Solution:
R
R
cos(x)dx = sin(x) :Indefinite
R
sin( 3π
2 ) − sin(0) Definite
= −1
277
Rπ
−π
sin(x)dx
Solution:
R
sin(x)dx = − cos(x) :Indefinite
R
cos(−π) − cos(π) :Definite
=0
278
R3
0
5x4 dx
Solution:
R
5x4 dx = x5 :Indefinite
R
(3)5 − (0)5 :Definite
R
= 243
279
R1
0
82x81 dx
Solution:
R
82x81 dx = x8 2 :Indefinite
R
(1)82 − (0)82 :Definite
R
=1
280
R1
81
−1 82x dx
Solution:
R
82x81 dx = x8 2 :Indefinite
R
(1)82 − (−1)82 :Definite
R
=0
281
R1
0
8x15 dx
Solution:
R
8
16
16 (1)
Problem 282
8 16
16 x
8
16
16 (0)
8
16
8x15 dx =
−
87
:Indefinite
R
:Definite
R
R
Summer Problem Set
282
R2
0
Rousseau 88
x15 dx
Solution:
R
1 16
16 x
1
16
16 (0)
x15 dx =
1
16
16 (2)
−
:Indefinite
R
:Definite
= 4096
283
R3
0
15x4 dx
15x4 = 3x5 :Indefinite
R
3(3)5 − 3(0)5 :Definite
R
= 279
Problem 283
88
R
R
Summer Problem Set
284
Rousseau 89
Find the following derivatives, use rules as necessary
d ex − e−x
dx
2
Solution:
x
−x
d e −e ]
]
dx [
2
d ex
e−1
dx [ 2 − 2
d ex
d e−x
dx [ 2 ] − dx [ 2 ]
x
e
e−x
2 + 2
:Given
:Split fraction
:Split
d
dx
:Differentiate
d ex +e−x
dx [
2
285
Solution:
ex −e−x
2
286
d ex −e−x
dx [ ex +e( −x) ]
Solution:
x
−x 2
(e +e
287
) −(ex −e−x )2
(ex +e−x )2
d
2
dx [ ex −e−x ]
Solution:
x
2
ex −e−x
−x −1
x
−x −2
x
2
ex +e−x
−x −1
x
−x −2
2(e − e
x
−2(e + e
288
−x
)(e − e
)
)
:Given Function
:Algebrate
:Differentiate
d
2
dx [ ex +e−x
Solution:
2(e + e
x
−2(e − e
289
−x
)(e + e
)
)
d ex +e−x
dx [ ex −e( −x) ]
Solution:
x
−x 2
(e −e
) −(ex +e−x )2
(ex −e−x )2
Problem 289
89
:Given Function
:Algebrate
:Differentiate
Summer Problem Set
290
Rousseau 90
Chain Rule derivatives with exponential functions:
d x2
[e ]
dx
Solution:
d
f (g(x)) = g ′ (x)f ′ (g(x))
dx
2xex
291
2
d x3
dx [e ]
Solution:
3
3xex
292
d sin(x)
]
dx [e
Solution:
cos(x)esin(x)
293
d
dx [ln(sin(x))]
Solution:
cos(x)
sin(x)
cot(x)
294
d
dx [ln(cos(x))]
Solution:
− sin(x)
cos(x)
− tan(x)
295
d
dx [sin(ln(x))]
Solution:
cos(ln(x))
x
296
d
dx [cos(ln(x))]
Solution:
− sin(ln(x)
x
297
d
sin(x)
]
dx [15
Solution:
cos(x) ln(15)15sin(x)
298
d
cos(x)
]
dx [15
Solution:
− sin(x) ln(15)15cos(x)
Problem 299
90
Summer Problem Set
299
Rousseau 91
d
ex
dx [10 ]
Solution:
x
ln(10)ex 10e
300
d
ln(x)
]
dx [10
Solution:ln(x)
ln(10)10
x
301
d
x
dx [log10 (e )]
Solution:
ex
ln(10)x
302
d
10
dx [log10 (x )]
Solution:
1
303
d
dx [log10 (sin(x))]
Solution:
cos(x)
sin(x) ln(10)
304
d
x
dx [log10 (12 )]
Solution:
ln(12)12x
ln(10)12x
ln(12)
ln(10)
Problem 304
91
Summer Problem Set
305
Rousseau 92
Antiderivatives –exponential and log rules
Z
ex dx
Solution:
ex + C
306
1
x dx
R
Solution:
ln(x) + C
307
8
x dx
R
Solution:
8 ln(x) + C
308
1
8x dx
R
Solution:
1
8 ln(x) + C
309
R
5x dx
Solution:
5x
ln(5) + C
310
R
6x dx
Solution:
6x
ln(6) + C
311
R
10x dx
Solution:
10x
ln(10) + C
312
R
ln(15) · 15x
Solution:
ln(15) ×
15x
ln(x)
x
+ C :Apply Exponent Rule
15 + C :Simplify
313
ln(x)dx
Solution:
x ln(x) − x + C
314
R
30ex dx
Solution:
30ex + C
Problem 315
92
Summer Problem Set
315
R
Rousseau 93
15 ln(x)dx
Solution:
15(x ln(x) − x) + C
316
24x
Solution:
24x
ln(24) + C
317
R
55x dx
Solution:
55x
ln(55) + C
318
R
7x dx
Solution:
7x
ln(7) + C
319
R
16x dx
Solution:
16x
ln(16) + C
320
R
ln(24) · 24x
Solution:
24x + C
Problem 320
93
Summer Problem Set
321
Rousseau 94
Chain Rule with Trig and Hyperbolic Trig
d
[tan(csc(x))]
dx
Solution:
(− cos(x)over sin(x)2 ) sec(csc(x))2
322
d
dx [csc(tan(x))]
Solution:
cos(tan(x))
sec(x)2 ( −sin(tan(x))
2 )
323
d
dx [sec(cot(x))]
Solution:
sec(x)2
sin(cot(x))
( cos(cot(x))
2 )(− tan(x)2 )
324
d
dx [cot(sec(x))]
Solution:
sin(x)
sec(sec(x))2
( cos(x)
2 )(− tan(sec(x))2 )
325
d
dx [tan(sinh(x))]
Solution:
(cosh(x))(sec(sinh(x)2 )
326
d
dx [sec(tanh(x))]
Solution:
sin(tanh(x))
(sech(x)2 )( cos(tanh(x))
2)
327
d
dx [tanh(csch(x))]
Solution:
(sech(csch(x))2 )( −cosh(x)
sinh(x)2 )
328
d
dx [csch(tanh(x))]
Solution:
(sech(x)2 )( −cosh(tanh(x))
sinh(tanh(x))2 )
329
d
dx [sech(coth(x))]
Solution:
sinh(coth(x))
sech(x)2
( cosh(coth(x))
2 )(− tanh(x)2 )
330
d
dx [coth(sech(x))]
Solution:
sinh(x)
sech(sech(x))2
( cosh(x)
2 )(− tanh(sech(x))2 )
Problem 331
94
Summer Problem Set
331
Rousseau 95
d
dx [csch(cosh(x))]
Solution:
cosh(cosh(x))
sinh(x)2 ( −sinh(cosh(x))
2 )
332
d
dx [cot(csch(x))]
Solution:
2
cosh(x)
( −sec(csch(x))
tan(csch(x))2 )(− sinh(x)2 )
333
d
dx [cot(csch(x))]
Solution:
2
csc (csch(x)) coth(x)
sinh(x)
334
d
dx [tan(tan(x))]
Solution:
(tan(tan(x))2 + 1)(tan(x)2 + 1)
335
d
dx [sec(sec(x))]
Solution:
tan(x)
( tan(sec(x))
cos(sec(x)) )( cos(x) )
336
d
x
dx [sinh(e )]
Solution:
ex cosh(ex )
337
d
tanh(x)
]
dx [20
Solution:
ln(20)sech(x)2 20tanh(x)
338
d
dx [csc(log20 (x))]
Solution:
− cot(log20 (x))
x ln(20) sin(log20 (x)
339
d
16
dx [coth(x )]
Solution:
−16x15 csch(x16 )2
340
d
dx [csc(csc(x))]
Solution:
cot(x) cot(csc(x))
sin(x) sin(csc(x))
341
d
dx [cot(cot(x))]
Solution:
csc(cot(x)) csc(x)
Problem 342
95
Summer Problem Set
342
Rousseau 96
d cosh(x)
]
dx [e
Solution:
ecosh(x) sinh(x)
343
d
x
dx [tanh(20 )]
Solution:
ln(20)20x sec(20x )2
344
d
dx [log20 (csc(x))]
Solution:
− cot(x)
ln(20) sin(x) csc(x)
345
d
16
dx [coth(x) ]
Solution:
−16csch(x)2 coth(x)15
Problem 345
96
Summer Problem Set
346
Rousseau 97
Product Rule with Trig and Hyperbolic Trig:
d
[tan(x) sec(x)]
dx
Solution:
(f (x)g(x))′ = f ′ (x)g(x) + g ′ (x)f (x)
sec(x)3 +
347
tan(x)2
cos(x)
d
dx [tan(x) tanh(x)]
Solution:
tanh(x) sec(x)2 + tan(x)]sech(x)2
348
d
dx [− csc(x) sec(x)]
Solution:
1
sin(x)2 −
349
1
cos(x)2
d
dx [csc(x) sec(x)]
Solution:
1
− sin(x)
2 +
350
1
cos(x)2
By the way, this is a repeat question.
d
dx [sin(x) sinh(x)]
Solution:
sinh(x) cos(x) + sin(x) cosh(x)
351
d
dx [sec(x)sech(x)]
Solution:
sinh(x)
sin(x)
− sec(x)( cosh(x)
2 ) + sech(x)( cos(x)2 )
352
d
dx [tan(x) cot(x)]
Solution:
0
353
d
dx [tan(x) sin(x)]
Solution:
sec(x)2 sin(x) + tan(x) sin(x)
354
d
dx [csc(x) cot(x)]
Solution:2
− cos(x)
sin(x)3
355
+
sin(x)
cos(x)2
d
dx [cot(x) coth(x)]
Solution:
− cot(x)csch2 (x) − coth(x) csc2 (x)
Problem 356
97
Summer Problem Set
356
Rousseau 98
d
dx [cos(x) cosh(x)]
Solution:
sinh(x) cos(x) − cosh(x) sin(x)
357
d
dx [csc(x)csch(x)]
Solution:
cosh(x)
cos(x)
csch(x)( sin(x)
2 ) − csc(x)( sinh(x)2 )
358
d
dx [csch(x)sech(x)]
Solution:
1
− cosh(x)
2 −
359
1
sinh(x)2
d
dx [sinh(x) cosh(x)]
Solution:
sinh(x)2 + cosh(x)2
360
d
dx [cot(x) sec(x)]
Solution:
− cos(x)
sin(x)2
361
d
dx [ln(x) sinh(x)]
Solution:
sinh(x)
+ cosh(x) ln(x)
x
362
d
dx [log5 (x) tanh(x)]
Solution:
tanh(x)
2
x ln(5) + sech(x) log5 (x)
363
d sin(x)
csch(x)]
dx [e
Solution:
cos(x)csch(x)esin(x) −
364
d
dx [csc(x) sec(x)]
Solution:
1
cos(x)2 −
365
cosh(x)esin(x)
sinh(x)2
1
sin(x)2
By the way, this is a repeat question.
d
dx [sin(x) cos(x)]
Solution:
cos(x)2 − sin(x)2 ’
366
d
dx [coth(x)sech(x)]
Solution:
−csch(x)2 sech(x) − sech(x)
Problem 367
98
Summer Problem Set
367
Rousseau 99
d x
dx [e cosh(x)]
Solution:
ex cosh(x) + ex sinh(x)
368
d
x
dx [5 sech(x)]
Solution:
sinh(x)
ln(5)sech(x)5x − 5x cosh(x)
2
369
d
dx [ln(cos(x)) coth(x)
Solution:
− tan(x) coth(x) − csch(x)2 ln(cos(x))
Problem 369
99
Summer Problem Set
370
Rousseau 100
Antiderivatives –Trig and Hyperbolic Trig Functions:
Z
sinh(x)dx
Solution:
cosh(x) + C
371
R
sec(x)2 dx
Solution:
tan(x) + C
372
R
csc(x)2 dx
Solution:
− cot(x) + C
373
R
tan(x)dx
Solution:
ln | sec(x)| + C
374
R
cosh(x)dx
Solution:
sinh(x) + C
375
R
sech(x)2 dx
Solution:
tanh(x)
376
R
csch(x)2 dx
Solution:
− coth(x) + C
377
R
tanh(x)dx
Solution:
ln | cosh(x)| + C
378
R
6 sec(x) tan(x)dx
Solution:
Z
1 sin(x)
dx
cos(x) cos(x)
Z
sin(x)
6
dx
cos(x)2
6
6 sec(x) + C
Problem 379
100
Summer Problem Set
379
R
Rousseau 101
cot(x)dx
Solution:
ln | sin(x)| + C
380
R
sec(x)dx
Solution:
ln | tan(x) + sec(x)|
381
R
sin(x)2 dx
Solution:
Z
Z
n−1
sin(x) =
n
Z
sinn−2 (x)dx −
cos(x) sin(x)n−1
+C
n
2−1
2
Z
sin2−2 (x)dx −
cos(x) sin(x)2−1
+C
2
n
sin(x)2 =
1
cos(x) sin(x)
x−
+C
2
2
382
R
11sech(x) tanh(x)dx
Solution:
−11sech(x) + C
383
R
coth(x)dx
Solution:
ln | sinh(x)| + C
384
R
sech(x)dx
Solution:
tan−1 (sinh(x))
385
R
cot(x)2 dx
Solution:
Z
csc(x)2 − 1dx
− cot(x) − x + C
Problem 385
101