MTH 232 Exam 1 Sample
MTH 232
Exam 1
Prof. Townsend
Sample
Name ______________________________________________________________________
You may use your calculator, your class notes, your book, and your brain as sources of
information. You may not share information or materials with others. If you are using a
smartphone calculator, please resist the temptation to communicate with others. If you are not
using a smartphone calculator, please put your phone out of sight.
1) Show ALL work. Give me something to grade. If you do most of the work in your
head, please tell me what rules you used. You must solve the problems using the rules
given in class, not just using the TI-89/ Nspire. Your solution must clearly show that it
was not done solely on a calculator. Do not erase work you think is incorrect but cross it
off instead. If it was in fact correct, you will earn partial credit.
2) You may use your TI-89/Nspire for algebra and for checking your work. Please write
down any appropriate calculator steps you use so that I could repeat your steps on my
own calculator. If your calculator answer differs from your solution, please tell me.
3) Provide three (3) significant digits for all non-integers.
4) You will receive extra credit for checking your work properly.
5) All six problems bear the same weight (100/6).
Calculator being used: ___________________________________
Note:
1
1
= a −n ,
an
n
a = an ,
m
1
a = a2 ,
n
am = a n ,
Rule #
Rule
(1)
dc
= 0 where c is a constant
dx
d ( cu )
du
=c
dx
dx
n
dx
dx
= nx n −1 with
=1
dx
dx
(3)
(2)
(4)
(5)
(6)
(7)
MTH 232 Exam 1 Sample Exam
a1 = a ,
a0 = 1
d
du dv
(u + v ) = +
dx
dx dx
d
dv
du
(u ⋅ v) = u + v
dx
dx
dx
du
dv
d ⎛ u ⎞ v dx − u dx
⎜ ⎟=
dx ⎝ v ⎠
v2
du n
du
= nu n−1
dx
dx
Page 1
1) Use the table method or the delta (h − > 0) method to find the slope of the line tangent to the
curve y = 5x (1 − 3x ) when x=2.
The table method
Use the following values of x:
x=2.1, 2.01, 2.001, 2.0001.
x = 2, y = −50
x
y
Δx
Δy
m=
Δy
Δx
m → −3(10x −1)
m → −55
2.1
-55.65
0.1
-5.65
2.01
-50.5515
0.01
-0.5515
2.001
-50.055015
0.001
-0.055015
2.0001
-50.00550015
0.0001
-0.00550015
-50.65
-55.15
-55.015
-55.0015
The delta (h − > 0) method
5 ( x + h ) * (1 − 3 * (x + h)) − 5x * (1 − 3x) −30h * x − 5h * ( 3h − 1)
=
=
h
h
−30h * x − 5h 2 + 5h
= −30x − 5h + 5 → −30x + 5
h
Note: for the rest of the problems, use derivative rules in the table below, not the delta
method.
Once you have applied all rules of derivatives to determine
the result further algebraically.
2) Identify the
dy
rule used for each step.
dx
( ) = 5 d ( x ) = 5( 7x ) = 35x
dx
dx
d 5x 7
d
dx
( )
x =
Find
dy
, it is not necessary to manipulate
dx
dy
for
dx
y = 5x 7 − x − 7
7
6
6
d ⎛ 12 ⎞ 1 − 12
1
x ⎟= x =
⎜
dx ⎝ ⎠ 2
2 x
dy
1
= 35x 6 −
dx
2 x
MTH 232 Exam 1 Sample Exam
Rule 3 then rule 2
Rule 2 and algebra rules
Rule 4
Page 2
3) Find the slope of the line tangent to the curve defined by
y = 7x 2 − x − 5 when x=3.
dy
= 14x −1
dx
4) Find
dy
dx
m=
= 14x −1
x=3
(
dy
for
dx
)
y = ( x − 1) x 9 + 1
= 41
x=3
5
It’s a product. Use rule 5.
u = ( x − 1)
(
)
v = x9 + 1
5
(
)
9
4 d x +1
dv
9
= 5 x +1
dx
dx
4
dv
= 5 x 9 + 1 9x 8 = 45x 8 x 9 + 1
dx
du
=1
dx
(
)
(
)
{
(
) } + (x
d(u ⋅ v)
= ( x −1) 45x 8 x 9 +1
dx
(
4
9
)
4
) (1)
+1
5
Tidy (if desired):
4
4
d(u ⋅ v)
= ( x 9 + 1) 45x 8 ( x − 1) + ( x 9 + 1) = ( x 9 + 1) { 45x 9 − 45x 8 + x 9 + 1}
dx
4
d(u ⋅ v)
(TI Answer)
= x 9 +1 46x 9 − 45x 8 +1
dx
{
(
){
}
}
x2 + 1
dy
y= 3
for
x −1
dx
It’s a quotient. Use rule 6.
5) Find
u = x2 + 1
v = x3 − 1
du
dv
= 2x
= 3x 2
dx
dx
u
d( ) ( x 3 − 1) ( 2x ) − ( x 2 + 1) ( 3x 2 )
v =
2
dx
( x 3 − 1)
MTH 232 Exam 1 Sample Exam
Page 3
Tidy (if desired):
u
3
2
2
d( )
4
4
2
4
2
v = x −1 ( 2x ) − x +1 3x = 2x − 2x − 3x − 3x = −2x − x − 3x
2
2
2
dx
x 3 −1
x 3 −1
x 3 −1
(
)
(
(
)( )
)
u
d( ) −x x 3 + 3x + 2
v =
2
dx
x3 − 1
(
(
6) Find
d2y
for
dx 2
)
(
)
)
(
)
(TI Answer)
y = 2x 8 − 3x 2 + 9
dy
= 16x 7 − 6x
dx
d2y
= 112x 6 − 6
2
dx
dy
in terms of x and y using the method of implicit derivatives of
dx
x 3 y − 2x 3 y 4 = 5
Extra Credit) Find
Analytical (10 points) , TI-89/Nspire only (3 points)
dy
dy
− 6x 2 y 4 − 8x 3 y 3
=0
dx
dx
dy
x 3 − 8x 3 y 3
= 6x 2 y 4 − 3x 2 y
dx
2
3
3y 2y 3 − 1
dy 6x 2 y 4 − 3x 2 y 3x y 2y − 1
= 3
=
=
dx
x − 8x 3 y 3
−x 3 8y 3 − 1
−x 8y 3 − 1
3x 2 y + x 3
(
)
(
(
)
)
(
(
)
)
TI Answer:
impDif(x^3*y-2*x^3*y^4=5,x,y)
(
)
3
dy −3y 2y − 1
=
dx
x 8y 3 − 1
(
)
MTH 232 Exam 1 Sample Exam
Page 4
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