1. y  x
7
dy
d 7

x
dx dx
 7x 7 1
Theorem 1
 7x 6
2.
y  x8
dy d 8

x  8 x 8 1 8 x
dx dx
7
3. y  3 x
dy
d
 (  3x )
dx dx
dy
 3 x
dx
 3(1x11 )
Theorem 3
Theorem 1
 3( x 0 )
 3
[ a 1]
0
4. y  0.5 x
dy
 0.5
dx
5. y  12
Constant function
dy d
 12
dx dx
=0
9. y  x
Theorem 2
6
dy
d 6

x
dx dx
 6x 61
Theorem 1
 6x 7
10. y  x
8
dy
 8 x 81  8 x 9
dx
11. y  4 x 2
dy
d
 ( 4x 2 )
dx dx
dy
 4 ( x 2 )
dx
 4(2 x 21 )
Theorem 3
Theorem 1
 4(2 x 21 )
 8x 3
7
 7 x 3
x3
dy
d
 ( 7x 3 )
dx dx
dy
 7 ( x 3 )
dx
21 y 
 7(3x 31 )
Theorem 3
Theorem 1
 21x 4

22. y 
21
x4
6
 6 x 4
x4
dy
24
 6(4 x 41 )  24 x 5   5
dx
x
d 4
3
( x )
dx
x
d 4
d 3

x
dx
dx x
25.
Theorem 4
d 14 d 1
x  3x
dx
dx
d 14
d

x  3 x 1
dx
dx
1 1 1
 x 4  3(1x 11 )
4
1 3
 x 4  3 x 2
4


1
4 4 x3
27.


Properties of exponents
Theorem 3
Theorem 1
3
x2
d
2
( x
)
dx
x
d
d 2
x
dx
dx x
Theorem 4
d 12 d
1
x  2x 2
dx
dx
d 12
d 1

x 2 x 2
dx
dx
1 12 1
1  1 1
 x
 2( x 2 )
2
2
1 1
3
 x 2x 2
2


1
2 x

Properties of exponents
Theorem 3
Theorem 1
1
x3
33. f ( x)  0.6 x
1.5
d
( 0 . 1x6. 5 )
dx
dy
 0.6 ( x1.5 )
dx
f ( x)
 0.6(1.5 x1.51 )
 0.9x 0.5
Theorem 3
Theorem 1
37. f ( x) 
4
4 x 3

7 x3
7
d 4 3
( x )
dx 7
4 dy 3

(x )
7 dx
4
 ( 3 x 31 )
7
12
12
  x 4   4
7
7x
f ( x)
45. f ( x)  0.01x  0.5 x  70
2
d
( 0 . 0x2 1
0
x. 5
70)
dx
d
d
d
 (0.01x 2 )  (0.5 x)  (70)
dx
dx
dx
f ( x)
 0.01(2 x 21 )  0.5(1x11 )  0
 0.02x  0.5
51. f ( x)  x  4 x  5
2
First, we find f ( x )
d 2
( x  4 x  5)
dx
d
d
d
 ( x 2 )  4 ( x)  5
dx
dx
dx
 2x  4
f ( x) 
Therefore, f (10)  2(10)  4  24
52. f ( x) 
f ( x) 
xx
1
2
1 12 1 1  12
1
1
x
 x  1 
2
2
2x 2 2 x
Therefore,
f (4) 
1
2 4

1
1

2 2 4
53. y 
4
 4 x 2
2
x
dy
first
dx
dy d
 (4 x 2 )
dx dx
Find
 4(2 x 21 )  8 x 3  
Therefore,
dy
dx

x 2
8
x3
8
1
(2)3
55. We will need the derivative to find the slope of the tangent line at each of the indicated points.
We find the derivative first.
f ( x)  x 3  2 x  1
f ( x) 
d 3
( x  2 x  1)  3 x 2  2
dx
a) using the derivative, we find the slope of the line tangent to the curve at point (2, 5) by
evaluating the derivative at x=2. f (2)  3(2)  2  10 . Therefore the slope of the tangent
2
line is 10. We use the point-slope equation of find the equation of the tangent line on the next
page.
y  y1  m( x  x1 )
y  5  10( x  2)
y  5  10 x  20
y  10 x  15
b) Using the derivative, we find the slope of the line tangent to the curve at point (-1, 2) by
evaluating the derivative at x=-1. f (1)  3(1)  2  1 . We use the point-slope equation
2
to find the equation of the tangent line.
y  y1  m( x  x1 )
y  2  1( x  (1))
y  2  x 1
y  x3
c) Using the derivative, we find the slope of the line tangent to the curve at point (0, 1) by
evaluating the derivative at x=0. f (0)  3(0)  2  2 . We use the point-slope equation
2
to find the equation of the tangent line.
y  y1  m( x  x1 )
y  1  2( x  0)
y  1  2 x
y  2 x  1
76. y  6 x  x
2
dy
 6  2x
dx
dy
1
solve:
dx
6  2x  1  x 
For x 
5
2
5
,
2
5
5
25 35
y  6( )  ( ) 2  15 

2
2
4
4
5 35
The tangent line has slope 1 at ( , )
2 4
77. y  0.025x  4 x
2
To find the tangent line that has slope equal to 1, we need to find the values of x that make
First, we find the derivative.
dy d
 (0.025 x 2  4 x)  0.025(2 x)  4  0.05 x  4
dx dx
dy
1
solve:
dx
0.05x  4  1  0.05x  3  x  60
So the tangent will occur when x=60. Next we find the point on the graph.
For x  60 ,
y  0.025(60)2  4(60)  90  240  150
The tangent line has slope 1 at (60,150)
dy
 1.
dx
79. y 
1 3
x  2x2  2x
3
To find the tangent line that has slope equal to 1, we need to find the values of x that make
dy
 1.
dx
First, we find the derivative.
dy d 1 3
 ( x  2 x 2  2 x)  x 2  4 x  2
dx dx 3
dy
1
solve:
dx
x2  4x  2  1
x2  4 x  1  0
This is a quadratic equation, not readily factorable, so we use the quadratic formula where a=1,
b=4, and, c=1.
x
b  b2  4ac
2a
(4)  (4)2  4(1)(1) 4  12
x

 2  3
2(1)
2
There are two tangent lines that have slope equal to 1. The first one occurs at x  2  3 and
the second one occurs at x  2  3 . We use the original equation to find the point on the
graph.
For x  2  3 ,
1
y  (2  3)3  2(2  3) 2  2(2  3)
3
1
 (26  15 3)  2(7  4 3)  4  2 3
3
26
4
   5 3  14  8 3  4  2 3   3
3
3
for x  2  3
1
y  (2  3)3  2(2  3) 2  2(2  3)
3
1
 (26  15 3)  2(7  4 3)  4  2 3
3
26
4
   5 3  14  8 3  4  2 3   3
3
3
The tangent line has slope 1 at the points ( 2  3,
4
4
 3) and (2  3,  3)
3
3
81. a) In order to find the rate of change of the area with respect to the radius, we must find the
derivative of the function with respect to r.
A(r ) 
d
(3.14r 2 )  3.14(2r )  6.28r
dx
b) Answers will vary. A(r )  6.28r means that the area of the wound with a radius r cm will
2
increase at a rate of 6.28r cm for each centimeter increase in the radius.
82. a) C (r )  6.28r
C (r )  6.28
b) Answers will vary. C (r )  6.28 means that the circumference of the wound with a radius r
cm will increase at a rate of 6.28 cm for each centimeter increase in the radius.
84. T (t )  0.1t  1.2t  98.6
2
a) T (t )  0.2t  1.2
b) Evaluate T when t=1.5
T (1.5)  0.1(1.5)2  1.2(1.5)  98.6  100.175
The temperature of the ill person after 1.5 days is 100.175 degrees Fahrenheit.
d) Evaluate T (t ) when t=1.5
T (1.5)  0.2(1.5)  1.2  0.9
The ill person’s temperature is increasing 0.9 degrees Fahrenheit per day after 1.5 days.
87. a) Using the power rule, we find the growth rate
dP
.
dt
dP d
 (100000  2000t 2 )  0  2000(2t )  4000t
dt dt
b) Evaluate the function P when t=10
P(10)  100000  2000(10)2  300000
The population of the city will be 300000 people after 10 years.
c) Evaluate the derivative P(t ) when t=10.
dP
 P(10)  4000(10)  40000
dt t 10
The population’s growth rate after 10 years is 40000 people per year.
e) Answer will vary.
P(10)  40000 means that after 10 years, the cities population is growing at a rate of 40000
people per year.
94. y  x 
4
4 2
x 4
3
dy
8
 4 x3  x
dx
3
dy
0
Solve
dx
8
4 x3  x  0
3
4x  0
or
x  0 or x  
x2 
2
0
3
2
3
For x  0
4
y  (0) 4  (0) 2  4  4
3
2
3
For x 
y(
2 4 4 2 2
40
)  ( ) 4  
3
3 3
9
2
3
For x  
y  (
2 4 4
2
40
)  ( ) 2  4  
3
3
3
9
There are three points on the graph for which the tangent line is horizontal
2 40
2 40
,  ) , and , ( ,  )
3
9
3 9
(0, 4) , (
95. y  2 x  x  2
6
4
A horizontal tangent line has slope equal to 0, so we need to find the values of x that make
dy
0
dx
First, we find the derivative
dy d
 (2 x 6  x 4  2)
dx dx
 12 x5  4 x3
Next we set the derivative equal to zero and solve for x
Solve
dy
0
dx
12 x 5  4 x 3  0
4 x3  0
or
x  0 or x  
x2 
1
0
3
1
3
For x  0
y  2(0)6  (0)4  2  2
1
3
For x 
y  2(
1 6
1
1
1
55
)  ( )4  2  2( )   2  
3
3
27 9
27
For x  
y  2(
1
3
1 6
1
55
)  ( ) 4  2  
3
3
27
There are three points on the graph for which the tangent line is horizontal
(0, 2) , (
1 55
1 55
,  ) , and , ( ,  )
3 27
3 27