Carlson, Spring 2003
Name_
MAT 270 Exam 3
Show work on all questions. If you fail to show work, you will not receive credit.
Good Luck!
1. The position of a particle is given by the function s = f (t ) = t 6t + 9t where t is measured in
seconds and s in m
Derivative Practice III
Find the derivative of each of the following functions.
1.
y = x2 2x + 2
()
2. y = arcsin x 2
3. y = 10 5 x
4. y = [arccos( x )]
3
5. y = arctan(e x )
6.
f (x ) =
4 x2 x
3
x
7. g ( x ) = 5 x + 3x 7
8.
f ( x ) = arctan( 5 x )
9. 2 y
MAT 270 - Derivative Practice I Solutions
1. f ( x) = 4 x 3 3x 2 + 2 x
f ' ( x) = 12 x 2 6 x + 2
2.
x2 3
3 x2
f ( x) =
f ' ( x) =
2
6
x+ 3
3
x
(
)
3. f ( x) = 3 2 x 2 5 x + 1
f ' ( x) = 12 x + 15
4.
f ( x) = x
f ' ( x) =
1
2x
+
1
x
1
2x x
x +1
x2
3
f '
MAT 270 - Derivative Practice I
Find the derivative of each of the following functions and simplify.
1.
f ( x) = 4 x 3 3x 2 + 2 x
2.
f ( x) =
3.
f ( x) = 3 2 x 2 5 x + 1
4.
f ( x) = x
5.
f ( x) =
x +1
x2
6.
f ( x) =
x2 2
x2
7.
f ( x) =
x2
x2 2
8.
f ( x)
MAT 270 - Derivative Practice II
Find the derivative of the following functions.
1.
(
f (x ) = 3x 2 4
(
f ( x ) = 5 3 x 2 4
2.
)
5
) (6 x ) = 30 x(3x
4
2
4
)
4
()
f (x ) = 3x 2 2 3 x
()
()
f ( x ) = 6 x 2 3 x + 9 x 2 2 3 x (ln 2)
3.
f ( x ) = e 2 x 1 (3x
Derivative Practice III
Find the derivative of each of the following functions.
1.
y = x2 2x + 2
(
y = 2 x 2 x + x 2 ln 2
)
()
2. y = arcsin x 2
y =
2x
1 x4
3. y = 10 5 x
y =
(
1 (5 x )
10
2
1
2
) (ln 10)( 1)
4. y = [arccos( x )]
3
y =
3(arccos x )
2
1 x
MAT 270 - Derivative Practice II
Find the derivative of the following functions.
(
)
1.
f (x ) = 3x 2 4
2.
f (x ) = 3x 2 2 3 x
3.
f ( x ) = e 2 x 1 (3x + 4)
5
()
3
2
ex
4. g (x ) =
(2 x 1)3
(
)(
5. g ( x ) = e 2 x + x + 3 x 2 2 x + x
(2 3x )
f (x ) =
25
6
Exam #2
1a) The distance s (in feet) of a car moving in a straight line is given by s = 2t t + 3 where t is
measured in seconds. Find the average velocity for the time period from t = 2 to t = 5.
2
(
)(
)
s (5) s (2) 2(5) 2 5 + 3 2(2) 2 2 + 3 48 9 39
=
=
Name_
3/10/2003-Carlson
Derivative Exam
Compute f(x) for each of the following:
1. f(x) = e3x cos (2x)
f ' ( x) = 3e3 x cos(2 x ) 2e3 x sin 2 x
2
- 2
x3
6
f ' ( x) = 20 x 4 + 4
x
2. f(x) = 4x5 -
x3 2 x
3. f(x) =
x2
3
f ' ( x) = 1 + 2
xx
4. f(x) = tan2 (3x
Section 1.2, Select Problems
80 70
10 1
=
=.
9a) Using N in place of x and T in place of y, the slope is T 2 T 1 =
N 2 N1
Then, the linear equation is
Then, T =
1
6
N+
9b) The slope is
1
6
307
6
173 113
60
6
T 80 = ( N 173) .
1
6
.
as shown in part (a). I
Section 2.1, #5, 8b
5a) At t = 2 , y = 40(2) 16(2) 2 = 16. The average velocity between times 2 and 2 + h
40(2 + h) 16(2 + h) 2 16 24h 16h 2
is:
=
= 24 16h, if h 0.
h
h
(i ) h = 0.5,32 ft sec
(ii ) h = 0.1, 25.6 ft sec
(iii ) h = 0.05, 24.8 ft sec
(iv) h
Section 1.3 - 30, 54, 55, 57
30) First note that the domain of f + g is the intersection of the domains of f and g; that is
f + g is only defined where both f and g are defined. Taking the horizontal and vertical
units of length to be the distance between
Section 2.2, Items 2, 10, 40
2) As x approaches 1 from the left, f(x) approaches 3; and as x approaches 1 from the
right, f(x) approaches 7.
NO, the limit does not exist because the left- and right-hand limits are different.
10)
lim f (t ) = 150 mg and li