STUDENTS’ UNDERSTANDING OF THE CONCEPT OF CHAIN RULE IN
FIRST YEAR CALCULUS AND THE RELATION TO THEIR
UNDERSTANDING OF COMPOSITION OF FUNCTIONS
A Thesis
Submitted to the Faculty
of
Purdue University
by
James Franklin Cottrill
In Partial Fulfillment of the
Requirements for the Degree
of
Doctor of Philosophy
May 1999
ii
To Cecelia, who arrived when this work was to begin;
To Stephen, whose arrival delayed its completion;
and
To Julie, my partner in love, life and faith.
iii
ACKNOWLEDGMENTS
I wish to thank my colleagues in RUMEC for their support, encouragement, and
assistance as I learned what it means to conduct research in this field. My special
thanks to Julie Clark, Dave DeVries, George Litman, Karen Thomas, and Draga
Vidaković for their friendship and insights.
I would like to thank Professors Keith Schwingendorf, David Mathews, and Guershon Harel for their willingness to serve on my committee.
I have deep gratitude for Professor Ed Dubinsky and his role in my life. As my
thesis advisor, he helped clarify my thoughts and patiently helped me through the
process. As my colleague, he was a brilliant example of how one can be a great teacher
of mathematics and researcher in education. Also, I thank him for his friendship and
support in my family’s life.
My children’s grandparents have endured years of separation with great patience
and love. I thank them for their support.
Finally, my special thanks to my wife, Julie Cottrill, for her patience and support in a project that involved relocating, transcribing, and months without visible
progress. Thank you.
iv
TABLE OF CONTENTS
Page
LIST OF TABLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
vii
ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
viii
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.1
Purpose . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.2
Historical background . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.2.1
An overview of the development of calculus . . . . . . . . . . .
3
1.2.2
The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.3
The RUMEC study . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.4
An overview of this study . . . . . . . . . . . . . . . . . . . . . . . .
7
2 Theoretical Perspective and Research Paradigm . . . . . . . . . . . . . . .
9
2.1
Theoretical perspective . . . . . . . . . . . . . . . . . . . . . . . . . .
9
2.2
Schema development . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
2.3
Paradigm for research . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
2.4
Initial genetic decomposition . . . . . . . . . . . . . . . . . . . . . . .
11
3 Literature Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
3.1
On functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
3.2
Students’ understanding of function . . . . . . . . . . . . . . . . . . .
13
3.3
Composition of functions . . . . . . . . . . . . . . . . . . . . . . . . .
15
3.4
Difficulties with differentiation . . . . . . . . . . . . . . . . . . . . . .
15
4 Phase 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
4.1
Methods: Subjects . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
4.2
Instrument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
4.3
Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
4.3.1
19
Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v
4.3.2
Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
4.3.3
Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
4.3.4
Chain rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
5 Results: Phase 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
4.4
5.1
Analysis 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
5.2
Analysis 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
5.3
An alternate analysis . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
6 Phase 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
6.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
6.2
Subjects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
6.2.1
Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
6.3
Instrument . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
6.4
Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
6.5
Where the RUMEC study left us . . . . . . . . . . . . . . . . . . . .
34
7 Results: Phase 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
7.1
The triad description revisited . . . . . . . . . . . . . . . . . . . . . .
36
7.1.1
Eli [A125] . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
7.1.2
Tim [A134] . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
7.1.3
Al [A137] . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
7.1.4
Ray [A123] . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
7.1.5
Peg [A132] . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
7.1.6
Jack [A113] . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
Using versus understanding . . . . . . . . . . . . . . . . . . . . . . .
50
7.2.1
Coding the aspects of understanding . . . . . . . . . . . . . .
50
7.2.2
Comparing the codes . . . . . . . . . . . . . . . . . . . . . . .
54
8 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
56
7.2
8.1
Revised genetic decomposition . . . . . . . . . . . . . . . . . . . . . .
56
8.2
Understanding of composition and understanding of the chain rule . .
58
vi
8.3
Comparing instructional methods . . . . . . . . . . . . . . . . . . . .
58
8.3.1
Understanding the chain rule . . . . . . . . . . . . . . . . . .
58
8.3.2
Using the chain rule . . . . . . . . . . . . . . . . . . . . . . .
59
8.3.3
Finding a balance . . . . . . . . . . . . . . . . . . . . . . . . .
60
8.4
Quantitative versus qualitative data . . . . . . . . . . . . . . . . . . .
60
8.5
Limitations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61
8.6
Instructional strategies . . . . . . . . . . . . . . . . . . . . . . . . . .
61
LIST OF REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
63
APPENDICES
A Questionnaire from Phase 1 . . . . . . . . . . . . . . . . . . . . . . . . . .
69
B Phase 1 Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
80
C Interview Guide for Phase 2 . . . . . . . . . . . . . . . . . . . . . . . . . .
84
C.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84
C.2 Review of function and composition . . . . . . . . . . . . . . . . . . .
84
C.3 Classifying the 10 derivative problems . . . . . . . . . . . . . . . . . .
85
C.4 Establishing the notion of chain rule . . . . . . . . . . . . . . . . . .
85
C.5 Extending the notion . . . . . . . . . . . . . . . . . . . . . . . . . . .
86
VITA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88
vii
LIST OF TABLES
Table
Page
5.1
Descriptive statistics for variables . . . . . . . . . . . . . . . . . . . .
28
5.2
Correlation statistics for variables . . . . . . . . . . . . . . . . . . . .
28
5.3
Regression model with PGPA . . . . . . . . . . . . . . . . . . . . . .
29
5.4
Regression model without PGPA . . . . . . . . . . . . . . . . . . . .
30
5.5
Comparison of courses . . . . . . . . . . . . . . . . . . . . . . . . . .
30
5.6
Comparison of Traditional and any C4 L . . . . . . . . . . . . . . . . .
31
5.7
Comparison of first quarter course . . . . . . . . . . . . . . . . . . . .
31
6.1
Matrix of interviews . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
7.1
Aspects of understanding . . . . . . . . . . . . . . . . . . . . . . . . .
55
APPENDICES
Table
B.1 Distribution of codes by item . . . . . . . . . . . . . . . . . . . . . .
81
B.2 Student codes in tuples . . . . . . . . . . . . . . . . . . . . . . . . . .
82
B.3 Combined code data . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
viii
ABSTRACT
Cottrill, James Franklin, Ph.D., Purdue University, May, 1999. Students’ Understanding of the Concept of Chain Rule in First Year Calculus and the Relation to
Their Understanding of Composition of Functions. Major Professor: Ed Dubinsky.
The present study is a follow-up to a study conducted by this author with seven
others. That study (Clark et al., 1997) proposed the use of Piaget and Garcia’s
triad mechanism to describe the development of a schema, specifically with respect
to students’ understanding of the chain rule. This study was designed to collect
data to test this theory. Both quantitative and qualitative methods were employed.
Data were collected via a questionnaire on a group of calculus students’ (n = 34)
knowledge of and skill with function, composition of functions, differentiation and
chain rule. The data were analyzed to investigate how their performance on the
composition of functions items related to that of the chain rule. No strong correlation
is reported. Some of the subjects had experienced a reform calculus course which used
computer experiences and cooperative learning. A comparison was made between the
two groups, although no overall differences are reported. Follow-up interviews based
on the questionnaire responses were conducted with six subjects. These results are
positive in the sense that the triad description fits with the data. We present detailed
descriptions of the Intra-, Inter-, and Trans- levels of the development of the chain
rule schema. Some evidence is presented to support the notion that understanding
of composition of functions is key to understanding the chain rule. The type of
instruction was a factor in how a student performed on these tasks. The differences
between the types seem to be related to the difference between using the chain rule
and understanding the chain rule, and an explanation of these differences is offered.
The need for collecting differing types of data is established. Two students scored
high quantitatively but did not perform well in the interview, while another student
failed the quantitative work but demonstrated a high level in the interview. Finally,
the use of writing as a pedagogical tool is recommended based on the results.
Chapter 1: Introduction
This study looks at a particular concept in calculus, the chain rule, in the first
year calculus course. The chain rule is the underlying concept in many applications
of calculus: implicit differentiation, solving related rate problems, and solving differential equations. The rule states that if g is differentiable at c and f is differentiable
at g(c), then the composite function f ◦ g given by f ◦ g(x) = f (g(x)) is differentiable
at c and that (f ◦ g)0 (c) = f 0 (g(c)) · g 0 (c). Conventional wisdom holds that students’
conception of the chain rule (as with other rules) is that of symbol manipulation.
This conception appears to be a straight-forward manipulation of symbols which can
easily be applied in problem situations. However such a situation carries a heavy
requirement for the function to be given by an expression, fostering students’ tendencies toward instrumental understanding. Even in such symbolic problem situations,
many students have difficulty in recognizing the need to apply the chain rule or in
applying it.
In discussing students’ understandings of the elements of the dihedral group,
Zazkis, Dubinsky, and Dautermann (1996) propose the VA model of mathematical
thinking in which visual and analytic modes of thought are mutually dependent in
problem-solving. The concept of derivative offers another example of a situation in
which a student might alternate between the visual (slope of a tangent to the curve)
and analytic (rate of change) interpretations in building her or his understanding
(Asiala, Cottrill, Dubinsky, & Schwingendorf, 1997).
From a visual perspective, the rule might be interpreted as the derivative of the
inner function compensating for, or scaling, the derivative of the outer. Such a conception lends itself to understanding the rule in relation to the applications mentioned
above. This was the understanding of both Newton and Leibniz as they developed
the calculus (Edwards, 1979). This visual perspective is not presented in current
pedagogical practices, perhaps due to the lack of a reasonable procedure for producing the graph of a composition given the graphs of the component functions. This
loss of a visual approach may explain the difficulty many students demonstrate when
working with the chain rule. The fundamental nature of the rule demands that the
beginning student must not only be capable of manipulating the symbols, but must
1
2
have a conceptual understanding of the chain rule. This project is an investigation
to describe how students come to understand the chain rule, and via this description,
propose activities which could bring students to the point of being better able to
build this understanding.
1.1 Purpose
The purpose of this study is to attempt to determine the correlation between
a student’s ability to deal with composition of functions and with using the chain
rule successfully. The data is also used to compare two instructional approaches to
the chain rule: a traditional lecture-recitation method and an innovative “reform”
method. Finally, an attempt is made to describe the understanding of the chain rule
concept on the part of the interviewed students.
It is supposed to answer the questions:
• How does a student come to understand the chain rule?
• In what way does the student’s understanding of (de)composition of functions
factor in the chain rule?
• With reference to the results of the RUMEC study (Clark et al., 1997), is
the chain rule concept described well via a schema, and do the current results
support the reported schema?
This study proposes to test the hypothesis: “The chain rule in and of itself is
not difficult for students to apply; rather it is the issue of composition of functions.”
Also, it will be the second attempt of the first cycle through the research paradigm,
which will result in a revised genetic decomposition.
Thus, the overarching purpose of this study is to test APOS Theory (described in
Chapter 2). The theory suggests that certain kinds of mental constructions should
be evident when students work and reflect on chain rule problems. It also states that
these constructions will be organized in a schema that should have certain characteristics at various levels of development. This study analyzes the data to determine if
the predictions of the theory hold up to scrutiny.
Finally, since APOS Theory is an evolving work, an opportunity to augment or
extend the theory may arise in this work, as it did in the RUMEC study.
1.2
Historical background
In attempting to gain an understanding of how students construct their concept of
the chain rule, it is beneficial to consider its historical development and its proof. The
3
chain rule arose quite naturally in the development of calculus by both Newton and
Leibniz as they worked with infinitesimals. The rule has evolved with the calculus
as Euler brought it to bear on functions and as Cauchy and Weierstrass developed
analysis. The proof has also evolved from a symbolic tautology to an argument based
on the limit of a difference quotient. The historical perspective presented here may
offer an explanation of students’ difficulty with the chain rule.
1.2.1
An overview of the development of calculus
Two types of problems from antiquity served to instigate the development of
calculus. One problem was to rectify a curve, that is, to construct a line with the
same length as that of a given curve. The other was the quadrature of a curve, or
the area beneath it. It was not the case that such problems were unsolved; actually
many specific curves had solutions for both types of problems. The ancient Greeks
used the method of exhaustion and the method of Archimedes (c. 200 b.c.) to find
many volumes and surfaces. The Greeks made use of indivisibles as they dealt with
concepts of infinite processes. What was lacking was a general method for any curve.
(Boyer, 1969)
The work of Oresme, Galilei, and Cavalieri (c. 1635) extended the use of indivisibles as the quadratures of families of curves were obtained. These mathematicians
approached the problems from a geometric point of view. During roughly the same
period, mathematicians such as Descartes, Wallis, and Fermat (c. 1655) were leading
an analytical revolution. The birth of coordinate geometry allowed the creation of
an arithmetic of infinitesimals (indivisibles). Wallis reset the work of Cavalieri in infinitesimal analysis. Fermat extended this and invented the method of differentiating
and integrating monomial expressions which are in use today. (Boyer, 1959)
Fermat did not recognize the inverse relation between these processes, nor did
those who followed his work. It was the genius insight of Newton and Leibniz to
deduce from this work (and the work on infinite sums of Gregory) the Fundamental
Theorem of Calculus. They saw and explicitly stated this inverse relation between
differentiating and integrating. Moreover, they showed that area, volume, and related
problems which had been solved by summation were reduced to anti-differentiation.
The distinction between these insights is indicated by the fact that the Fundamental
Theorem is stated in two parts. (Kline, 1972)
Newton (c. 1668) perceived a curve as the path of a moving point and spoke of
fluents x and y as the horizontal and vertical components of this motion. He defined
the rate of change of a fluent to be a fluxion, with the notation ẋ and ẏ. By interpreting
4
increasingly smaller increments in x and y, he found the derivative to be the ratio
of the fluxions which is also the slope of the tangent to the curve. It is interesting
to note that although Newton developed his calculus by working analytically, his
first published account of it (the Principia Mathematica) is presented in a classical
geometric setting. (Rickey, 1987)
The approach to calculus of Leibniz (c. 1673) was to view these infinitesimals
as mathematical objects, which he called differentials. In working on quadratures,
he noted the summation of the area under a curve is related to its anti-derivative.
Leibniz developed a careful notation for the calculus, much of which persists today.
He also derived many of the rules for computing derivatives and integrals, notably
the chain rule. (Kline, 1972)
Following this seminal work in the calculus, the next major advance was that of
Euler (c. 1748) and his incorporation of the concept of function. This was followed
by a long period in which the methods of the calculus were used to solve problems
and were refined. However, the notion of the infinitesimal had not been satisfactorily
defined by Newton or Leibniz. This lack incited Berkeley and Lagrange to call for a
rigorous formulation of calculus in the late 18th century. The issue was settled by
Cauchy and Weierstrass (c. 1821) with the development of analysis as a mathematical
field. (Edwards, 1979)
1.2.2
The Chain Rule
The current formulation of the chain rule, as stated above, is a result of this
development of real analysis. The concept itself—the derivative of a composition
of functions—was intuitively obvious to both Newton and Leibniz. As C. Edwards
(1979) points out discussing the work of Newton:
The tangent and area problems emphasize the importance of systematic
procedures for differentiation (the calculus of ẏ/ẋ, given f (x, y) = 0) and
anti-differentiation (the converse). Newton exploited the facility for differentiation and anti-differentiation by substitution methods—equivalent
to what we call the chain rule and integration by substitution —that is
essentially “built into” the calculus of fluxions. (p. 196)
Newton did not set the chain rule out as a theorem or procedure. It was a natural
algorithm employed in many of his examples. In a similar manner, the notation of
Leibniz led him to consider the chain rule as intuitive as simplifying a product of
5
fractions. If z = z(u) and u = u(x), then
dz
dz du
=
· .
dx
du dx
Again this process was not even given a name. In Leibniz’ union of geometric and
analytical thinking using infinitesimals, the chain rule is implicitly defined.
The work of Euler and Cauchy precipitated the chain rule in its present form.
By changing the objects of calculus from curves in space to arbitrary functions, the
intuitive nature of the chain rule was lost. It became a theorem to be proved. Cauchy’s
proof of the rule follows:
Now let z be a second function of x, bound to the first y = f (x) by the
formula
z = F (y).
z or F [f (x)] will be that which one calls a function of a function of a
variable x; and, if one designates the infinitely small and simultaneous
increments of x, y, and z by ∆x, ∆y, ∆z, one will find
F (y + ∆y) − F (y)
F (y + ∆y) − F (y) ∆y
∆z
=
=
·
,
∆x
∆x
∆y
∆x
then, on passing to the limits,
z 0 = y 0 F 0 (y) = f 0 (x)F 0 [f (x)].
(Cauchy, cited in Edwards, 1979, p. 313)
Edwards notes that Cauchy overlooks the possibility that ∆y = 0 in the work cited.
In fact, Cauchy does not treat the functional relationship of ∆y and ∆x, and so the
proof is false.
Often in mathematics, the proof of a theorem helps to convey the fundamental
concepts on which it is built. That is, by experiencing the proof, a student will
construct a relational understanding of the theorem. Unfortunately, the method of
proof for the chain rule is generally interpreted as an algebraic trick—multiply by 1
(∆y/∆y). Even though
F (y + ∆y) − F (y) ∆y
·
∆y
∆x
is simply Leibniz’s dz/du · du/dx written in terms of difference quotients, the geometric interpretation is lacking. This may be related to our inability to graphically
6
demonstrate the composition of two functions given by their graphs. The difficulties
that students have with the chain rule may be explained by its history, that is, the
function and composition ideas came later and clouded the intuitive idea of Newton
and Leibniz.
1.3
The RUMEC study
In a previous study (Clark et al., 1997), eight members (including this author) of
an academic community of researchers1 working with a common theoretical framework attempted to explore calculus students’ understanding of the chain rule and its
applications. Specifically, the students were asked, as part of a clinical, task-based
interview situation, to solve the following four problems and to explain what they
were doing. The students were prompted to use the chain rule explicitly in the first
problem if it was not chosen as the method of solution. Prompts were given in the
second problem relating to the chain rule if the student indicated difficulty. The
problems were:
2
1. Let f be the function given by f (x) = (1 − 4x3 ) . Compute f 0 .
2. Let F be the function given by F (x) =
you did.
Z
sin x
2
et dt. Compute F 0 . Explain what
0
3. Let A be a real number. Given that the following relation defines a function,
√
√
x y + y x = A, find its derivative.
4. A ladder A feet long is leaning against a wall, but sliding away from the wall at
the rate of 4 ft/sec. Find a formula for the rate at which the top of the ladder
is moving down the wall.
A preliminary analysis of the responses to these questions indicated that although
almost all of the participants were able to use the chain rule to correctly solve the
first problem, those who had difficulty with the other problems were unable to apply
the chain rule, even when prompted.
Clark et al. (1997)2 attempted to analyze the concept of chain rule based on
an Action-Process-Object-Schema (APOS) theoretical framework which is described
in the next chapter. Using the framework to analyze the mental constructions of
1
The group is known as the Research in Undergraduate Mathematics Education Community or
RUMEC.
2
For the rest of the paper, I will refer to this study as the RUMEC study.
7
the students, the RUMEC study set out to describe each of the 41 participants as
working with an action, process, or object conception of the chain rule (at the time
of the interview). In the negotiation of differences between researchers’ analyses, an
issue arose regarding whether a student’s general power rule (as used in Problem 1,
above) was her or his chain rule. The issue could not be resolved within the given
theoretical framework, and so the RUMEC study failed to meet its objectives.
An analysis of the problem situation concluded that the concept of the chain rule
is best described by a schema. Although the framework treated the mental construction of a schema, it was found to be insufficient to account for the data collected.
Specifically, this insufficiency was that a schema was defined but its development was
not described. The RUMEC study provided the description of schema development
in terms of the chain rule. In this way, it extended the previous APOS framework
(Clark et al., 1997).
The RUMEC study used the interview data to illustrate schema development and
attempted to return to the original objective of the project (describing each of the
41 participants). However, since the data were collected via interviews based on
the previous genetic decomposition, the “right” questions had not been asked. In
particular, no attempt was made to have students discuss the similarity/difference
of Problems 1 and 2. Also, these chain rule problems were at opposite ends of the
spectrum in terms of complexity. The current work is a follow up to the RUMEC
study. Incorporated in its design are a number of examples of problem situations
involving the chain rule, as well as opportunities to explore the students’ perceived
relationships between the situations.
1.4
An overview of this study
This study consists of two phases, one quantitative and one qualitative. Phase 1
involved the collection of data via questionnaires administered to 34 students. Phase 2
used in-depth, task-based interviews with six of the students from Phase 1.
The purpose of Phase 1 is an attempt to determine the correlation between a
student’s ability to deal with composition of functions and with using the chain rule
successfully. Data were collected to establish an individual’s baseline understanding
of function and composition, as well as her or his relative proficiency in mathematics.
Integrated in this phase was a comparison between two methods of teaching calculus
regarding the effectiveness of the instructional treatment of the chain rule.
Students were selected from second quarter calculus by requesting volunteers.
Data were collected on an instrument constructed for this study which assessed the
8
participant’s facility with (a) functions, (b) composition of functions, (c) differentiation, and (d) the chain rule. Each participant was asked to authorize access to her
or his academic record for the researcher to gather data, such as SAT scores and
predicted indices.
The questionnaire items were scored using a 5 point rubric based on general guidelines adapted from Carlson (1998). These codes along with the predictor data were
tested for correlational significance. A second analysis explored any differences between the two groups (traditional versus C4 L). Analysis of variance was used to
determine if there exists a significant difference in sample means of the chain rule
scores.
The second phase of the study involves interviews with participants from the first
phase to attempt to describe how these students construct the concept of the chain
rule. The six participants in the interviews were chosen based on their scores in
Phase 1, covering a range of chain rule scores and a range of overall scores. The
interview follows a guide designed to elicit the student’s understanding of the rule,
based on tasks from the previous instrument. This aspect of the study is based on
the following theoretical perspective and research paradigm.
Chapter 2: Theoretical Perspective and Research Paradigm
As Phase 2 of the study is of a qualitative nature, it is necessary to discuss our
theoretical perspective and research paradigm. This discussion is a brief overview; a
complete description can be found in Asiala et al. (1996).
2.1 Theoretical perspective
The APOS theory is based on the following statement:
An individual’s mathematical knowledge is her or his tendency to respond
to perceived mathematical problem situations by reflecting on problems
and their solutions in a social context and by constructing or reconstructing mathematical actions, processes and objects and organizing these in
schemas to use in dealing with the situations. (Asiala et al., 1996, p. 7)
Three basic types of knowledge—actions, processes, and objects—are observed,
which are organized into structures called schemas. An action is any repeatable physical or mental manipulation of objects to obtain other objects. It is a transformation
that is a reaction to stimuli that the individual perceives as external. In the case of
function, for example, a student indicates an action conception if he or she is limited
to requiring an explicit formula in order to interpret a situation as a function. A
student in such a case would be unable to do much with the function, apart from
evaluating it at a point. This student would have great difficulty dealing with situations involving composition of functions (Breidenbach, Dubinsky, Hawks, & Nichols,
1992).
As a student reflects on an action, it is interiorized and becomes a process. The
student perceives the action as part of her or him and has control over it. For functions, this means the student can think of a function as receiving input, performing
some operation, and returning output. This can be done mentally, without actually
performing the operations on the input (Breidenbach et al., 1992).
As a student realizes that an action can be brought to operate on a process,
the process is encapsulated to become an object. The object conception can be deencapsulated back to the process as needed. In the case of functions, this encapsulation/de-encapsulation arises as one considers manipulations of functions such as
adding, multiplying or forming sets of functions.
9
10
A schema is a coherent collection of actions, processes, objects and other schemas
that is invoked to deal with a new mathematical problem situation. A schema can be
thematized to become another kind of cognitive object to which actions and processes
can be applied. By consciously unpacking a schema it is possible to obtain the original
processes, objects and other schemas from which the schema was constructed. (Note
that this is the description of schema prior to the RUMEC study which extended
it as described in the next section.) A genetic decomposition is an attempt by the
researcher to describe the objects and processes in some set of students’ schemas.
While this description is not unique for any given concept, it is useful in guiding
instructional design and investigations (Asiala et al., 1996).
2.2
Schema development
The triad mechanism, introduced by Piaget and Garcia (1989), distinguishes three
stages in the development of a concept: Intra, Inter and Trans. The Intra stage is
characterized by the focus on a single object in isolation from any other actions,
processes, or objects. With the chain rule, for instance, the student may have a
collection of rules for finding derivatives including some special cases of the chain rule
such as the general power rule and perhaps even the general formula, but might not
recognize the relationships between them.
The Inter stage is characterized by recognizing relationships between different
actions, processes, objects and/or schemas. It is useful to call a collection at the
Inter stage of development a pre-schema. In the case of the chain rule, the student
may begin to collect these special cases and recognize that in some way they are
related.
Finally the Trans stage is characterized by the construction of a coherent structure
underlying some of the relationships discovered in the Inter stage of development. At
the Trans stage of the chain rule, the student would link function composition to
differentiation, and recognize various instantiations of the chain rule as linked in that
they follow from the same general rule through function composition.
It is worth noting that it is only when a schema reaches the Trans stage of development that it can properly be referred to as a schema. The reason is that at the
Trans stage the underlying structure is constructed through reflecting on the relationships between the various objects from the earlier stages. This structure provides
the necessary coherence in order to identify the collection as a schema, that is, it is
a coherent whole. This coherence is deciding what is in the scope of the schema and
what is not.
11
2.3
Paradigm for research
This research in mathematics education is based on Dubinsky’s method (Asiala
et al., 1996) consisting of three components in a cycle: theoretical analysis; design
and implementation of instruction; and data collection and analysis. The theoretical
analysis component influences the instruction component in that the activities and
tasks used in instruction are based on a genetic decomposition. In turn, the implementation of the instruction influences the empirical data component as those who
have been instructed are studied. Data are collected in as many forms as reasonable
in order to gain insight. Tests and homework, direct observation, questionnaires and
clinical interviews are possible sources for data collection. The data analysis leads
to a revision of the current genetic decomposition; thus the empirical data affect the
theoretical analysis.
On the other hand, the theory influences the empirical data since the data collection instruments are constructed based on the genetic decomposition. These instruments are designed to also get at issues involving the “goodness” of the genetic
decomposition. Typically, two genetic decompositions will be presented in a study;
the preliminary one used to design the instruction and the data collection, and the
revised genetic decomposition based on the observations and analysis.
2.4
Initial genetic decomposition
The results of the RUMEC study provide evidence that understanding the chain
rule involves the building of a schema. The schema consists of a function schema
coordinated with a differentiation schema. The student’s function schema would have
at least a process conception of function, function composition and decomposition
(breaking a composite function into two or more component functions). This is linked
to her or his differentiation schema which includes the rules of differentiation at least
at the process level (Clark et al., 1997). Note that a distinction is made between
a differentiation schema and the derivative schema of a student. The former, which
deals with taking derivatives, is a part of the latter which also contains notions of
interpretations and representations of the derivative (Asiala, Cottrill, Dubinsky, &
Schwingendorf, 1997).
The chain rule schema develops through the levels of the triad; Intra, Inter and
Trans. In the first level, the Intra- level, the student has a collection of rules for
finding derivatives in various situations, but has no recognition of the relationships
between them. This collection may include some special cases of the chain rule, and
perhaps even the general formula which is perceived as a separate rule rather than a
12
generalization of the others. The Inter- level, is characterized by the student’s ability
to begin to (mentally) collect all different cases and to recognize that these are related.
At this stage the collection of elements in the chain rule schema is being formed, and
the collection is called a pre-schema. At the Trans- level a student has constructed the
underlying structure of the chain rule. He or she must link function composition and
decomposition to differentiation, and recognize various instantiations of the chain rule
as linked in that they follow from the same general rule through function composition.
The elements in the schema must move from being described essentially by a list to
being described by a single rule (Clark et al., 1997).
Chapter 3: Literature Review
3.1
On functions
Functions are a key component in secondary and undergraduate curricula (see,
for instance, the foreword of Harel & Dubinsky, 1992) and much has been written
on the learning and teaching of the concept. Leinhardt, Zaslavsky and Stein (1990)
present an exhaustive review of the research papers which focus on functions, graphs
and graphing. The studies are organized by three issues: task, student learning,
and teaching techniques. The issue of task is discussed in terms of the action of
the learner, the situation in which the task is set, the variables involved and their
nature, and the focus of attention in the task. The issues of student learning are
presented as intuitions and as misconceptions which the students bring to a task.
Finally, the teaching techniques discussed are entry to the topic, sequencing the tasks,
explanations in the learning, and examples and representations.
Ferrini-Mundy and Graham (1991), Tall (1992), Even (1993), Thompson (1994b),
and Wilson and Krapfl (1994) offer more recent reviews of research on functions.
These note that function is a dichotomous concept—perceived by the student as a
process or as an object (see also Sfard, 1992). Harel and Dubinsky (1992) add that
students start with an action conception of function prior to moving to process and
object. These reviews agree that the concept of function is important to mathematics
and not well understood by our students. These studies and those about which they
report were searched for evidence of explorations of composition of functions and for
discussions of the chain rule. While many of the studies of students’ understanding
of function relate to this study, little was found regarding students’ understanding of
composition of functions explicitly, and no studies were found on their understanding
of the chain rule.
3.2
Students’ understanding of function
Many studies observed students’ understanding of function by presenting questionnaires of various function situations for the student to interpret (Even, 1988,
1993; Vinner & Dreyfus, 1989; Breidenbach et al., 1992; Dubinsky & Harel, 1992;
Schwingendorf, Hawks, & Beineke, 1992; Hitt, 1993; Ferrini-Mundy & Graham, 1994;
13
14
Lauten, Graham, & Ferrini-Mundy, 1994; Wilson, 1994; and Carlson, 1998). Typically these studies ask the student to give her or his definition of function, and then
they present many examples of situations to the student. The situations generally
range over graphs, tables, expressions, equations, sets of ordered pairs, parametric
equations, and finite sequences. In some cases, only verbal descriptions of the situation are provided. The student is then to decide in each case if a function is present
in the situation. Often, the student is asked to relate the answer back to the definition of function previously given, either at the time of the questionnaire or in a later
interview. These studies conclude that the notions students hold about function are
often different from what they give as a definition. Also demonstrated are the specific
types of functions which present difficulty to students.
Breidenbach et al. (1992), Dubinsky and Harel (1992), and Schwingendorf et
al. (1992) suggest that these difficulties indicate the lack of a process conception
of function. They offer evidence of an instructional approach which seems to help the
students construct this process. Even (1988, 1993), Hitt (1993), and Wilson (1994)
focus on the concept of function found in mathematics teachers. They offer rather
disturbing results as the teachers exhibited very limited conceptions of functions (see
also Vinner & Dreyfus, 1989). Ferrini-Mundy and Graham (1994) and Lauten et
al. (1994) provide descriptions of the students’ function concept and suggest methods
of employing graphing calculators as aids in learning. Carlson (1998) compares three
groups of students at different stages of their collegiate careers. She finds a very
narrow view of function is held by high-achieving second semester calculus students.
She also concludes that the concept of function develops slowly, and this development
is facilitated by reflection and constructive activities.
Other researchers investigated the concept from a problem-solving and graphical point of view. They describe methods of overcoming difficulties with functions
by using computer software and problem-solving techniques. These reports generally offer anecdotal evidence as support for their views. Confrey (Confrey & Smith,
1994; Confrey, Smith, Piliero, & Rizzuti, 1991) offers a computer environment called
Function Probe for student use in problem-solving situations. In a similar manner,
Schwarz and Bruckheimer (1990) use an environment Triple Representation Model,
Cuoco (1994) proposes the use of Logo and ISETL, and Goldenberg (1988) employs
the Function Analyzer. These environments allow the student to have control over a
function by switching between representations, exploring covariation, and changing
individual parameters.
15
3.3
Composition of functions
One of the very few papers that discuss the composition of functions is a teaching
methods paper (one which does not present any supporting observations of students)
in which Alson (1992) discussed a qualitative approach to graphing functions. He
defines two operators on the graphs which are essentially composition operators, one
for the domain and one for the range of the graph. His treatment offers insight into
the graphical interpretation of the composition of elementary functions. However, as
an instructional approach, the students must start with a rich concept of function.
Other methods may allow the student to develop this richness while constructing the
idea of composition.
Ayers, Davis, Dubinsky, and Lewin (1988) present evidence that computer experiences induce reflective abstraction as students learn about functions and composition.
Students worked in the microworld of the Unix operating system, using shell scripts
and pipes to represent functions. The authors present a genetic decomposition of
composition and probe the process conception of composition in their investigation.
They indicate that in order for a student to succeed in the task of de-composing a
function—given descriptions of two functions, H, G, find F such that H = F ◦ G—he
or she must have a process conception of function and of composition.
Dubinsky (1991) further discussed the issue of composition and reflective abstraction. Two aspects of reflective abstraction, coordination and reversal, are illustrated
by excerpts of interviews on composition. According to Dubinsky, an understanding
of functions as objects and as processes is necessary for understanding composition
(beyond evaluating each function at specific points with a formula). However, in
her work investigating students’ understanding of inverse functions, Vidaković (1996)
found students whose concept of composition of functions was based on a process
conception of function but not necessarily on an object conception.
3.4
Difficulties with differentiation
While the search for investigations on learning the chain rule gave no results, two
teaching methods papers on the chain rule are available. Mathews (1989) discusses
the use of a computer algebra system, muMATH, to verify the chain rule. Using
the computer to produce the symbols is expected to help the students understand
the truth value of the rule. Thoo (1995) proposes using arrow (or tree) diagrams as
a mnemonic device for the chain rule. He finds the diagrams especially useful with
functions of several variables. Thoo also believes that these will help internalize the
composition of functions.
16
Studies of students’ difficulties with calculus topics (Orton, 1983; Frid, 1992;
White & Mitchelmore, 1992; Thompson, 1994a; Thomas, 1995; Cottrill et al., 1996;
Vidaković, 1996, 1997; Asiala et al., 1997) offer insights to misconceptions and obstacles to understanding of calculus concepts. Specific concepts such as graphical
interpretation of the derivative, limit, rate of change, differentiation, inverse functions and the Fundamental Theorem of Calculus (FTC) have been explored. These
reports offer descriptions of the students’ possible mental constructions and pedagogical suggestions. Thomas (1995) also used Problem 2 of the RUMEC study
(Section 1.3) in her exploration of students’ understanding of the FTC. She found
that those students who were unsuccessful with the problem were unable to deal with
the area-accumulating function (a function defined through a definite integral with a
variable endpoint). This result is consistent with those of the RUMEC study on the
problem. Selden, Selden and Mason (1989, 1994) approached student understanding
with a problem-solving framework. They found students fail to use calculus strategies
when dealing with nonroutine problems.
Chapter 4: Phase 1
The purpose of this portion of the study is to attempt to determine the correlation
between a student’s understanding of composition of functions and understanding the
chain rule. The data will also be used to compare two instructional approaches to
the chain rule: a traditional lecture-recitation method and an innovative “reform”
method. The comparison is an attempt to establish whether there is a difference in
performance which might be attributed to a more conceptual approach to the topic
(the reform course) versus an instrumental approach. This study proposes to test the
hypothesis: The chain rule in and of itself is not difficult for students to apply; rather
it is the issue of composition of functions.
4.1 Methods: Subjects
The subjects for this study were students at a large, southeast, urban university
who had completed at least two quarters of calculus. A total of 34 students volunteered to participate in the study. They were solicited from second and third quarter
calculus classes near the end of the term. Some of the students had been through a
calculus course using cooperative learning, computers and alternatives to lecturing.
The reformed calculus course is known as C4 L (for Calculus, Concepts & Computers
and cooperative learning)1 . The others had completed a traditional, lecture-based
course. Seven students took the C4 L course both quarters, three took C4 L in the first
quarter only (then took the traditional course for the second quarter), seven took
a traditional first quarter then C4 L for the second quarter, and 17 students took a
traditionally taught course for both quarters. This data is recorded with the variable
C4L, defined in Appendix B.
All of the students were majoring in Arts & Sciences or Education programs; five
were graduate students in Education taking undergraduate mathematics courses for
certification. The students volunteered their time (up to 2 hours) to complete the
questionnaire. In order to make some sort of comparison between the volunteers, an
additional data point was sought. The University computes a predicted grade point
index (PGPA) at the time of admission. The PGPA was collected to compare the
academic strength of each student prior to her or his calculus instruction.
1
See Schwingendorf, Mathews, & Dubinsky (1996) for a detailed description of the project.
17
18
An alpha-numeric code assigned to each subject was used to identify all of the
written work collected as data. This code was also used on all audio and video tapes
made in the second phase of the study. This code is used in this report to present
data without compromising confidentiality.
4.2
Instrument
Data were collected on an instrument constructed for this study which tries to
assess the participant’s facility with (a) functions, (b) composition of functions, (c)
differentiation, and (d) the chain rule. The items on the written instrument (questionnaire) are given in the following section along with the coding scheme for each
item. The questionnaire is included as Appendix A.
The questionnaire was piloted with 4 students of known ability (having recently
completed the researcher’s final exam in second quarter calculus) to check for errors
and reliability. As there was no need for major revisions to the questionnaire, the
responses of the pilot study are included in this study.
The data was collected in a piece-wise fashion with individuals and small groups
over the course of a summer session. In each instance, the informed consent form was
read and explained. After the consent forms were signed, the following instructions
were given:
• These items are designed to explore your range of understanding on a number
of topics. Some will be very easy to do, some will be harder. Please do the best
you can.
• Please answer each question as completely as you can. Please do not leave any
answer blank. If you would like to write additional notes with an answer, feel
free to use the back of the page.
• Please do not write your name on any page of this questionnaire. The code
number at the top of this page will serve to keep your identity confidential.
The questionnaires were completed under ordinary test conditions with little interaction with the researcher.
4.3
Analysis
The 20 problems were coded (scored) using a 5 point rubric based on the following
general guidelines adapted from Carlson (1998):
5
Complete response to all aspects of the problem. Indicates complete
mathematical understanding of the problem’s concept. Includes only
minor computational errors, if any.
19
4
3
2
1
0
Responses falling between 5 and 3.
Demonstrates understanding of main idea of the problem. Not totally complete in response to all aspects. Shows some deficiencies in
understanding aspects of the problem. Incomplete reasoning.
Responses falling between 3 and 1.
Attempts, but fails to answer or complete problem. Very limited
or no understanding of problem. Contains words, examples, or diagrams that do not reflect the problem.
No answer. Written information made no attempt to respond to the
problem. Written information was insufficient to allow judgment.
These guidelines were used to construct specific rubrics for each item—a first estimation of each rubric appeared in the proposal for study. Each item was analyzed
using a constant-comparison method (Krathwohl, 1993). In this method, responses
were sorted based on the type of response found in the written work. The resulting
categories were then compared to the general rubric and the initial estimation, and
were given an initial code. More than one category resulting from the sorting operation would receive the same code occasionally, if the two types of responses both
indicated the same level of understanding. This scheme was repeated as new sets of
data became available. The specific responses were then used as descriptors in the
rubric for the item.
Finally, the entire set of data was re-analyzed some weeks after the last questionnaire was completed. This last iteration checked the reliability of the generated
rubric. The refined codes were used in the statistical analysis reported in the next
chapter.
In the following discussion, each item from the questionnaire is given. When
applicable, notes on the item follow, stating such things as where it was originally
published, comments on why the item was chosen, and/or issues that arose in the
coding. The rubric for coding that item is described by giving a description of the
typical responses for each level that was observed in the data. If a code was not given
to any response for the item, it is not listed in the rubric. Likewise, no rubric has a
description for the code 0 since that description did not vary between items.
4.3.1
Functions
Item 1
Express the diameter of a circle as a function of its area and sketch its graph.
20
(Carlson, 1998) This item involved covariation in functions. Many students did
not give a graph as part of their answer. Perhaps this was due to misreading the
problem, rather than a misunderstanding.
5
4
3
2
1
The correct formula was computed and a graph was sketched which
was reasonably similar to that of the function.
The graph was missing; or a circle was sketched in addition to the
correct formula.
Used correct area formula in terms of d, but does not solve for d;
used an incorrect formula (such as circumference) and solves for d;
or, indicated thinking about a function by use of function notation
(with otherwise incorrect answer).
The graph was missing; or a circle was sketched in addition to a
response which was otherwise a 3.
Used some circle formulae without function notation.
Item 2
A student has marked the following as a non-function. State whether this student is
correct and why.
y
x
(Even, 1988, 1993) Items 2 and 3 raised the issue of univalence in functions.
5
4
3
2
1
Answered “No”. Used a definition of function in the explanation.
Answered “No”. Gave a statement about the vertical line test (without elaboration).
Answered “No”. Restated the situation; or stated that discontinuous
functions are allowed.
Answered “Yes”. Indicated a misunderstanding of the open circle on
the graph; or there was an indication of using an erroneous definition.
Answered “Yes”. Any other non-empty response.
21
Item 3
A student has marked the following as a non-function. State whether this student is
correct and why.
A correspondence that associates 1 with each positive number, −1 with
each negative number, and 3 with zero.
(Even, 1988, 1993) It was interesting to note that 8 out of the 34 students literally
read the sentence from left to right, that is, they saw the domain as {−1, 1, 3}.
5
4
3
1
Answered “No”. Used a definition of function in the explanation.
Answered “No”. Drew graph and gave a statement about the vertical
line test (without elaboration).
Answered “No”. Restated the situation; or stated that discontinuous
functions are allowed.
Answered “Yes” due to interpreting the statement literally left to
right.
Answered “Yes”. Any other non-empty response.
Item 4
Decide if it is possible to use one or more functions to describe this situation. If yes,
then describe your function(s) briefly. If no, then explain.


x = t3 + t
t is a real number.
y = 1 − 3t + 2t4 
(Breidenbach et al., 1992) Item 4 was intended to demonstrate whether the student
perceived a process in the situation.
5
3
2
1
Fairly clear description of either a function R → R × R or functions
x(t), y(t).
Indicated desire to solve for t.
Work for linear equation; substitutes expression for x into y; or
answered “Yes” with no reason given or with disconnected reason.
Answered “No” with reason stated.
Item 5
Tim and Donna live 1 km from their school. Usually, they walk to school together.
Yesterday, they both left their houses 10 minutes before school started. Tim started
to walk, but Donna was afraid to be late and started to run. After a while, Tim
22
time
time
distance
time
distance
distance
time
distance
distance
distance
realized that although he tried to walk faster and faster, he had to run if he did not
want to be late, and started to run. At about the same time, Donna became tired
and had to walk instead of run. They both reached school exactly on time. Which
of the following graphs is Tim’s and which one is Donna’s?
time
time
(Even, 1988, 1993) This item involves covariation in functions.
5
2
1
4.3.2
Correct choices.
Correct choice for Tim but chose parabola for Donna
Indicated need for an expression; or chose two parabolas.
Composition
Item 6
t
Given two functions v, w such that v(t) = 5t − 6 and w(t) = + 4, find (w ◦ v)(9).
3
Explain.
Item 6 was a composition problem to determine understanding of “◦” notation.
5
4
3
2
1
Answered 17.
Found (w ◦ v)(t) correctly, but did not evaluate.
Reversed composition; or incorrectly evaluated expression.
Subtracted functions then evaluated at 9.
Evaluated w(9), v(9) and stopped; multiplied functions; found w(9) ·
v(9); solved system of equations.
Item 7
√
Given that (f ◦ g)(x) = 5 2x + 3.
23
a. Find f and g that satisfy this condition.
b. Are there more than one answer to part (a)? Explain.
(Even, 1988, 1993) This item required decomposition of a composite function.
5
4
3
2
1
Two correctly labeled pairs.
√
Misunderstood 5 notation, but indicated understanding of composition.
“No” for part (b); reversed labels (process); or incorrect answer for
(a) and “Yes” for (b).
correct expressions unlabeled; or wrote identity as g = 1.
Left (a) blank; took a derivative; f = 0, g = 0; or g = 1/x.
Item 8
Find k so that g(x + 1) = g(x) + k, given that g(x) = 3x + 5. Explain.
(Carlson, 1998) This item invoked the student’s composition concept and compared it with a functional shift up.
5
4
3
2
1
Correctly finds k = 3 and explains answer in terms of slope.
Pointwise evaluation to find constant without explanation.
No explanation or work shown.
Error working with g(x + 1); used expression g(x + 1) + k; or divided
to solve for k.
Unable to evaluate g(x + 1); or did not evaluate g(x + 1).
In each of the following two questions, f, g, h are functions whose domains and ranges
are the set of all real numbers, and such that h = f ◦ g.
Item 9
If only the information in the following table were known, would it be possible to find
f (2)? If so, find it and if not explain why not.
x h(x) g(x)
−1
1
−3
4
π
1
π
0
2
(Breidenbach et al., 1992) The last two items of this section required students to
compose and decompose functions given without expressions.
5
4
Correct answer (f (2) = h(π) = 0) with work.
Correct answer with inconsistent or no work.
24
3
2
1
Indicated understanding of composition and table but no answer; or
wanted line for x = 2 in table.
Asked for function/equation/expression.
Interpreted as multiplication.
Item 10
If only the information in the following table were known, would it be possible to find
g(4)? If so, find it and if not explain why not.
x h(x) f (x)
−1
1
−2
2
3
1
4
−2
π
(Breidenbach et al., 1992) This problem cannot be solved as stated without assuming that f is one-to-one, since the domain of f is given as the set of real numbers,
rather than just the set of x-values in the table. However, this issue did not confound
the subjects in Breidenbach et al. (1992) as the students easily made what was to
them a natural assumption. Likewise, there were no indications that any of these
subjects saw the need for the one-to-one property.
5
4
3
2
1
4.3.3
“Correct” answer (g(4) = f −1 (−2) = −1) with work.
Work showed strong attempt but cannot see decomposition.
Indicated understanding of composition and table but no answer; or
wanted line for x = −2 in table.
Asked for function/equation/expression.
Interpreted as multiplication.
Differentiation
Compute the derivative of each of the following functions. Show all of your work.
Item 11
f (x) = 11x5 − 6x3 + 8
Item 12
3
g(x) = 2
x
Item 13
h(x) = x2 − 3 5x − x3
Item 14
y = 3ex − 4 tan(x)
Item 15
y = x2 sin(x)
25
5
4
3
2
1
4.3.4
Correct answer.
Minor error such as dropping (−) sign or arithmetic errors; error
with derivative of trig functions.
Errors involving elementary differentiation rules.
Did not use appropriate rule; or compound errors with rules.
Integrated; or guessed.
Chain rule
Compute the derivative of each of the following functions. Show all of your work.
Item 16
2
F (x) = 1 − 4x3
Item 17
4
4
G(x) = 2 5x2 + 1 − 4x 5x2 + 1
Item 18
H(x) = sin 5x4
Item 19
y = cos3 (t)
Item 20
2
y = e−t
5
4
3
2
1
4.4
Correct answer.
Minor error such as dropping (−) sign or arithmetic errors; or applied
chain rule but error with derivative rule.
Computed f 0 (g 0 (x)) or f 0 (g(x))
Applied chain rule indiscriminately; or attempted to avoid chain rule
by expanding/rearranging terms (Item 19).
No evidence of considering chain rule.
Data
The data for Phase 1 are presented in tabular format in Appendix B. The codes
are presented by item, by subject as ordered tuples, and by subject as combined
scores.
Chapter 5: Results: Phase 1
These data were analyzed in two ways. First, as one sample from a single population, a regression model was used to determine which rating (after PGPA) has the
highest correlation with the chain rule rating. It was expected that the composition
rating would have the most significant coefficient in the model after PGPA. This was
not the case however, as is detailed below.
In the second analysis, the data were split by course (C4 L vs. Traditional) and
compared in each category. The comparison between treatments was an ex post facto
study and, since the members of each sample were self-selected as volunteers for the
study, an attempt was made to confirm the similarity of the groups using PGPAs
(Schwingendorf, McCabe, & Kuhn, 1998).
5.1 Analysis 1
The predicted grade point average (PGPA) is computed as an attempt to state the
readiness for a student to attend the university. It was used here to give a measure
of the student’s ability prior to her or his instruction in calculus. The PGPA is
not computed for transfer students or for graduate students. In the case of transfer
students, the cumulative grade point average from the previous institution is used.
There was no PGPA score reported for the five graduate students in the study.
The codes for the five items in each category were summed to create a rating
in that category. Prior to summing, the subsets of codes were inspected as ordered
tuples (Table B.2) in order to decide if such a method was reasonable. This question
arose in observing that ratings in the center of the range for a category might come
about differently and thus tell different stories. For example, a rating of 15 could
come from [3, 3, 3, 3, 3] or from [5, 5, 5, 0, 0]. Should both ratings of 15 be treated
the same way? By returning to the written responses to compare such cases, it was
deemed reasonable to sum the codes. In the example stated, the rating of 15 gave the
same “global” information for the ability of each of the two students in that category.
Also at issue was how to deal with questionnaires with 20–25% of the answers
blank. This latter issue was more easily decided since it involved only three questionnaires. These were excluded from the analysis as being incomplete. Note that the
26
27
directions for the questionnaire specifically stated to leave no item blank. The IDs
for the deleted data sets are marked with an asterisk (*) in Table B.2 and Table B.3.
The descriptive statistics for the five variables used in the regression model are
given in Table 5.1 and the correlation data are in Table 5.2. The means indicate two
pairs of ratings. Function and Composition each have a representative code of a little
more than 3, which would be interpreted as demonstrating understanding of the main
idea of the problem, but showing some deficiencies in understanding aspects of the
situation. The representative rating for Differentiation and Chain Rule is slightly over
4, showing more complete ability in the situation. This difference appears to coincide
with the types of questions in the sections of the questionnaire. Items 1–10 varied
widely in difficulty and in coverage of topics within the headings of function and
composition of function. This variety is a by-product of the existing body of research
in these areas and correct responses might indicate relational understandings of the
situations. The remaining items are of a more instrumental nature, as indicated by
the directions “Compute the derivative of each of the following functions.”
In order to include PGPA in the regression model, only 27 observations are able
to be included. That model is presented in Table 5.3. We note that only Composition
and Differentiation have coefficients significantly different from zero. This result was
predicted by the theory—the genetic decomposition for the chain rule is composed
of the schemas for composition and differentiation. Table 5.4 gives the model using
31 observations without the PGPA score. The difference between models is slight,
although without the PGPA, the Composition rating is no longer significantly nonzero. Both models account for the variance with very low P-values.
The unexpected result is that the coefficient for Differentiation is larger than that
of Composition. It was hypothesized that the Composition rating would predominate.
Again, this could be explained by the different types of items in the two halves of the
questionnaire.
5.2
Analysis 2
The C4L variable in Tables B.2 and B.3 is a binary code for the first two quarters
of calculus. An “11” represents the student took the C4 L version of calculus both
terms. A “10” indicates taking the C4 L version first quarter and a traditional course
the second quarter, while a “01” indicates the opposite situation. A “00” indicates
being in a traditional course both terms. The means and standard deviations for
each variable in the regression model is given for each of the four values of the C4L
variable in Table 5.5.
28
Table 5.1
Descriptive statistics for variables
Mean
Median
Mode
Standard Deviation
PGPA
Function
2.73703
2.8
3.1
0.62212
14.6129
14
14
5.13599
Composition
15.9677
15
24
5.7531
Differentiation
22.6774
24
24
3.07014
Chain
Rule
21.0967
21
24
3.36010
Table 5.2
Correlation statistics for variables
PGPA
Function
Composition
Differentiation
Chain Rule
PGPA
Function
1
0.0089
0.20391
-0.022
-0.1579
1
0.33574
0.10597
0.13552
Composition
Differentiation
1
0.26548
0.40021
1
0.671989
Chain Rule
1
The data were compared by grouping those students with C4L values greater than
zero, that is, those who had at least one non-traditional course. This was done to
compare groups of almost equal size. Also, this grouping has historical precedence.
In previous studies using two groups, the influence of a non-traditional approach was
seen to show up when investigating topics taught traditionally (see Asiala et al., 1997;
Clark et al., 1997). This comparison is presented in Table 5.6.
Finally, a comparison was made based on the method of their first quarter course,
when the chain rule was introduced. These data are given in Table 5.7. There was
a significant difference for the Differentiation and Chain Rule ratings in favor of the
group who had a Traditional first quarter of calculus. However, these differences on
average are three points out of 25—which can be interpreted as making minor errors
on three problems. Since those items were of a more instrumental nature, this result
merely indicates that those students are less likely to make computational errors in
the test setting.
5.3
An alternate analysis
In order to address the difference found in the types of items on the questionnaire,
two new variables were computed. The Function rating was changed to include only
29
Table 5.3
Regression model with PGPA
Regression statistics
0.7978
Multiple R
0.6365
R2
0.5704
Adjusted R2
2.1882
Standard Error
27
Observations
Analysis of Variance
4
22
26
Sum of
Squares
184.502
105.3491
289.851
Mean
Square
46.125
4.7886
Coefficients
6.5755
-1.133
0.0154
0.1967
0.6163
Standard
Error
3.7905
0.7109
0.0965
0.0927
0.1426
t
Statistic
1.734
-1.594
0.1604
2.1217
4.320
df
Regression
Residual
Total
Intercept
PGPA
Function
Composition
Differentiation
F
Significance F
9.6323
0.00011
P-value
0.0946
0.122
0.8737
0.0435
0.000201
Lower
95%
-1.285
-2.608
-0.184
0.00444
0.3205
Upper
95%
14.43
0.3408
0.2158
0.3890
0.9121
Items 1 and 2, and a new Composition rating included Items 6–8. Those items were
selected as being the most like the last ten items. That is, they could be considered to
be straight-forward, computational situations dealing with functions and composition
of functions. We note that the original coding of these items was based on assessing
the depth of understanding in the situation, rather than on an instrumental, “right
or wrong” scheme. The items were not re-coded for this analysis.
The use of the new variables did not change the regression model or the results
between groups. This seemed to indicate that there would be no change in results if
one were to construct a new questionnaire using items of the same type for all four
categories to collect new data.
30
Table 5.4
Regression model without PGPA
Regression statistics
0.7103
Multiple R
0.5045
R2
0.4495
Adjusted R2
2.4929
Standard Error
31
Observations
Analysis of Variance
3
27
30
Sum of
Squares
170.910
167.799
338.709
Mean
Square
56.970
6.2147
Coefficients
3.8272
-0.006
0.1413
0.6663
Standard
Error
3.52159
0.09409
0.08663
0.15379
t
Statistic
1.08679
-0.0712
1.63150
4.33257
df
Regression
Residual
Total
Intercept
Function
Composition
Differentiation
F
Significance F
9.16684
0.00023
P-value
0.28578
0.94363
0.11323
0.000152
Lower
95%
-3.3984
-0.1997
-0.0364
0.3507
Table 5.5
Comparison of courses
PGPA
Function
Composition
Differentiation
Chain Rule
Overall
mean
SD
mean
SD
mean
SD
mean
SD
mean
SD
mean
SD
00
2.62
0.70
15.31
6.13
15.31
5.16
23.63
1.26
22.00
2.48
76.25
10.80
01
2.91
0.35
13.86
3.98
19.43
7.35
23.00
2.83
21.71
2.69
78.00
12.74
10
2.67
0.40
16.33
2.08
13.00
3.46
21.67
4.93
20.00
2.65
72.20
6.24
11
2.82
0.88
12.40
4.45
15.00
5.57
19.80
5.07
18.00
5.52
65.20
15.80
Upper
95%
11.0529
0.18636
0.3191
0.9818
31
Table 5.6
Comparison of Traditional and any C4 L
PGPA
Function
Composition
Differentiation
Chain Rule
Overall
mean
SD
mean
SD
mean
SD
mean
SD
mean
SD
mean
SD
Trad (n = 16)
2.62
0.70
15.31
6.13
15.31
5.16
23.63
1.26
22.00
2.48
76.25
10.80
C4 L (n = 15)
2.83
0.56
13.87
3.89
16.67
6.43
21.67
4.05
20.13
3.96
72.33
13.46
t
-0.90
P -value
0.19
0.78
0.22
-0.65
0.26
1.84
0.04 *
1.58
0.06
0.90
1.70
t
-0.14
P -value
0.45
0.47
0.32
0.98
0.17
2.53
0.01 *
2.48
0.01 *
1.98
0.03 *
* significant difference between groups
Table 5.7
Comparison of first quarter course
PGPA
Function
Composition
Differentiation
Chain Rule
Overall
mean
SD
mean
SD
mean
SD
mean
SD
mean
SD
mean
SD
Trad (n = 23)
2.73
0.60
14.87
5.51
16.57
6.05
23.43
1.83
21.91
2.48
76.78
11.16
C4 L (n = 8)
2.76
0.71
13.88
4.09
14.25
4.71
20.50
4.75
18.75
4.53
67.38
12.76
* significant difference between groups
Chapter 6: Phase 2
6.1
Introduction
The purpose of this portion of the study is to attempt to describe the understanding of the chain rule concept on the part of the interviewed students. Also, it will
be the second attempt of the first cycle through the research paradigm, which may
result in a revised genetic decomposition.
The discussion and conclusions based on the Phase 1 results will be treated fully
in combination with the results from this interview data. The questions to be investigated here are:
• In spite of the lack of support for the hypothesis in the quantitative data,
is there evidence that students can be successful with the chain rule without
understanding composition of functions, and to what extent can this success
move toward understanding?
• With reference to the results of the RUMEC study, is the chain rule concept
well described via a schema, and do the current results support the reported
schema?
6.2
Subjects
Students who were interviewed were drawn from the 34 students who had participated in Phase 1. Each student was placed in a 3 × 3 matrix which was formed by
partitioning the ratings for the Chain Rule and the Overall ratings into three subintervals (high, middle, low). An attempt was made to interview one subject from each
of the nine subsets. However, only six interviews were conducted. Volunteers could
not be found to represent the cells left empty in Table 6.1. Of the two or three subjects in the three categories, either the student had requested to not be interviewed
on the informed consent form in Phase 1, or he or she declined the invitation during
this phase. Table 6.1 indicates the code number for the student interviewed as an
entry in the matrix.
32
33
Table 6.1
Matrix of interviews
Overall
Chain Rule
High (25–23)
High
Middle
Low
(96–81)
(78–66)
(63–45)
A113
Middle (22–20)
Low (19–1)
6.2.1
A134
A132
A125
A137
A123
Procedures
The interviews were audio and video taped with the camera positioned to record
the subject’s actions and written work. The audio tapes were transcribed to written
form and then coded for key events. These were then analyzed according to the
research paradigm described above. The video tapes were logged to ease reference to
key events found in the audio transcripts.
6.3
Instrument
The interview followed a guide designed to elicit the student’s understanding of
the rule, based on tasks from the quantitative instrument. New tasks were presented
to the student to probe the limits of understanding. The students were asked to reflect
on the questionnaire tasks; by asking them for definitions of function and derivative,
domain-range issues in composition, the statement of the chain rule, and a discussion
of any relationships between the 20 problems that the student perceived. The guide
was developed following the initial analysis of the quantitative data.
Then the interview addressed Leibniz’ rule as a way of gauging the robustness
of the student’s concept. A situation with no simple
anti-derivative was presented;
Z
sin x
2
et dt.” If the need to compute
“compute F 0 if F is the function given by F (x) =
0
an anti-derivative was too great, a situation where the integrand has an elementary
anti-derivative was presented to determine if the student can derive Leibniz’ rule from
the result. The text of the interview guide can be found in Appendix C.
6.4
Analysis
The interviews were transcribed from the audio tape recordings by the author and
an assistant. These transcripts were then proofread by the author, who had conducted
the interviews. Each transcript was then coded for key events in the interview and
those events were searched for common issues relating to the topic. Not surprisingly,
34
the events followed the interview questions rather closely, and so most of the issues
found were predicted by the interview guide.
Two new issues arose in the data. We found data that better assessed the interviewee’s understanding of the chain rule than was available in the questionnaire data.
Secondly, by interviewing two students whose data were excluded from the analysis
in Phase 1, we found limitations in employing only one means of data collection in a
study of mathematical knowledge and understanding.
6.5
Where the RUMEC study left us
Clark et al. (1997) introduced the notion of using the triad mechanism (Piaget
& Garcia, 1989) to describe the development of a schema in the RUMEC study
(see Section 2.2, page 10). This mechanism focuses on the relationships between
problem situations and/or between mental constructs that a student brings to bear
on a problem. The relationships may or may not be explicitly perceived by the student
as they are used. The authors found the triad useful, having exhausted themselves in
the attempt to describe the schema in terms of actions, processes and objects alone.
The RUMEC study was able to map the triad levels to the observed data in the
study and gave a genetic decomposition for the chain rule concept based on the triad
(see Section 2.4, page 11). However, it was unable to map the overall response of each
interviewed subject to the triad. The data collected did not indicate clearly if and
how each subject related various problem situations together. Simply put, the data
were collected prior to the researchers’ understanding of the complexity of schema
development. This present study was designed to fill that gap.
The following excerpts from Clark et al. (1997) describe how they interpreted
student responses with respect to the triad in slightly more specific terms than found
in the genetic decomposition.
Students who are at the Intra stage of schema development with respect
to the chain rule see the various rules for differentiation as unrelated. The
students are able to solve some of the problems by simply applying rules
which have been memorized and in some cases, not remembered correctly.
(p. 354)
Many of the students who participated in our study showed evidence of
being at the Inter level of development of the chain rule. They had collected some or all of the various derivative rules in a group and perhaps
35
could provide the general statement of the chain rule, but had not yet
constructed the underlying structure of the relationships. (p. 356)
Once a student’s collection of derivative rules and understanding of function composition attains a coherence as a schema the student has moved
to the Trans stage of development. That is, he or she is capable of operating on the mental constructions which make up her or his collection.
He or she is now able to reflect on the explicit structure of the chain rule
which these constructions were implicitly containing. (p. 358)
The following chapter reports the analysis of the six interviews based on these descriptions and on the initial genetic decomposition. The triad mechanism is seen in
the observations and the data for each subject is described by a level in the triad.
Chapter 7: Results: Phase 2
7.1
The triad description revisited
Three questions from the interview were written to evoke the students’ relationships between situations involving the chain rule. In Question 6, the student is to
sort the differentiation problems from the previous phase. Later, he or she is asked
to give her or his definition of the chain rule, and in Question 11, relate it back to
the list of problems just sorted. Finally, in Question 15 four problem situations are
presented to find if the student sees the involvement of the chain rule.
While these questions form the basis for the following discussion, the rest of the
last half of the interview is also taken into account when applicable. Certainly, the
success or lack of success on the Leibniz rule question indicates how robust a student’s
schema has become.
In the following discussion pseudonyms are used for ease of readability rather
than the alpha-numeric codes and to protect the identities of those interviewed. The
interviewer is indicated throughout by “I”. The code will be given—in the section
headings with the pseudonym—for cross reference with the data tables.
7.1.1
Eli [A125]
Eli may have been working at the Intra- level. In the grouping questions and in
his definition of the chain rule, he was working on the surface of the situation only.
Eli: Can I have like one question that is in two groups?
I: If you want.
Eli: Then probably I would, first I would group 4, 5, 8, 9 to one since they are
all trigonometric.
I: Uh-huh.
Eli: Then probably group #4 and #10 together because they have e in it, and
#2 just one group cause it has fraction, then #3, 6, & 7 that’s one group,
cause you have powers. . .
I: Uh-huh. What about the powers in 1?
Eli: Powers in 1? [Mumbling]
I: You said 3, 6, &7 grouped together because they have powers.
Eli: Yeah.
I: Right exponents. Do you have exponents in #1?
36
37
Eli: Oh, yeah.
I: You have 11x to the fifth. So would that go in there or were you thinking
that or was there a reason that doesn’t go in there?
Eli: Um, it’s because 3, 6, &7 all have [pause]
I: Expressions? Which things were you. . . ?
Eli: Well, I think [mumbling] they have [mumbling]
I: So you were pointing at #6 when you said multiply it out. . . what would you
be multiplying out?
Eli: Well, like 1 − 4x3 times 1 − 4x3 I think [mumbling]
I: Are you thinking of FOIL?
Eli: Yeah. [Mumbling]
I: OK.
Eli: Actually I might need 3 [mumbling] involved with the FOILs. So I would
probably put 1, 3, and 6 together.
I: Uh-huh
Eli: [mumbling]
I: OK, you asked me and I said sure you could. You have some of your groups
overlapped a little bit.
Eli: Uh-huh
I: What if I changed my mind and said, Um, no they can’t overlap. They have
to be disjoint. Can you still group them?
Eli: Um. . . yeah.
I: How would you do that?
Eli: I would put. . . I’d probably put #4 and #10 as one group and then 5, 8, and
9 as the other group. So a trigonometric group and the e group.
I: OK. What about the leftovers then? 1, 2, 3, 6, and 7? Would they be in a
group together or would they each be an individual group? How would they
be treated?
Eli: 1 and 3 as one group and 2 as another group. 6, 7 as another . . .
Eli gave a description of the chain rule but seemed unable to formalize it. This
was consistent with the trouble he had dealing with functions on the questionnaire,
where his combined code was 9/25 with 3 blank responses.
I: OK. Alright. Um the point of the questionnaire was to get you to think about
and use the chain rule in taking derivatives. OK. Uh, do you remember the
chain rule?
Eli: Uh, derivative of inside times derivative of outside. Not really.
I: It had to do with insides and outsides.
Eli: No, not really.
38
I: So for instance, #8 is a problem that involves the chain rule. You’d just take
the derivative. Here’s what you wrote down for, well it’s 18 on this paper.
The original function was sin(5x4 ). You wrote down cos(5x4 ) times 20x3 .
Eli: Uh-huh. Cause um the derivative outside times the derivative of inside.
I: OK. Why don’t you write that down? Um underneath the blue line that I’ve
drawn on that page. Write down, you can use words or symbols or whatever
you like as best as you can recall what the chain rule says. [pause] So what
did you write down?
Eli: Uh, I put symbol like A and B then I put um A inside, or B inside of A,
then I said derivative of A times derivative of B.
I: So the derivative of A, then what happens?
Eli: I just put derivative of A then write the B part down.
I: OK. Um in fact, the last five questions on this list were intended to be chain
rule questions and use the chain rule to complete the derivative. OK. Does
your rule there work for all five of those?
Eli: Uh, I don’t think so.
I: What’s wrong with your rule?
Eli: Because it doesn’t deal with the powers. Here it had A and B but, the ten
only had one power.
Finally, the four problem situations also were dealt with superficially. Eli decided
that the first two did not require the chain rule and the last two did. However, he
did not offer any explanation and ended by stating that he did not see it in the last
problem. The interviewer wrapped things up in an attempt to get some response
from him, but Eli was not able to add anything.
I: OK. Good. Um, here’s another sheet of paper. Read each of the following differentiation situations carefully. Without actually trying to solve the
problem, determine if the solution will involve the use of the chain rule in
some way. If so, describe how the chain rule is used.
Eli: [pause] Uh, number 1. Uh, no.
I: No. OK.
Eli: [long pause] Uh, I think that the first two, #1 and #2 are. . . have nothing
to do with the chain rule, 3 and 4 um is gonna have to use the chain rule.
I: Uh-huh, how does the chain rule get used in part c there, excuse me I meant
number 3?
Eli: Uh, 3? I have to take the derivative uh, 36πV 2 .
I: Um-huh
Eli: Then, um, . . . I don’t know. It just seem like I have to use the chain rule,
uh, I don’t know.
I: OK, and for #4 does that seem any more clear?
39
On #4 it says that you have formula of C and it says that F = −32t2 . So I
plug in F into the formula of C and I will have to use the chain rule there,
uh,. . . Well I can’t see the relationship of the inside and outside function.
I: Well, what variable are we using in C?
Eli: What variable? F .
I: I, right. Capital F . So then we’re saying OK, yeah but then you can even
find capital F in terms of time, in terms of little t. So that’s what is going
on there. So the inside function is all of that, −32t2 , that’s going to go in
for all of those F ’s.
Eli:
7.1.2
Tim [A134]
Another student working at the Intra- level was Tim. Like Eli, he grouped based
on the surface characteristics and did little in his second attempt.
I: Yeah I just want you to group these, like if you decide that number 3 and
number 6 go together, then they are in a group. Read through these and
decide . . .
Tim: Where are the groups?
I: Make up groups, and then put them in the different groups.
Tim: Oh, OK. This one [works for 2 minutes] So.
I: OK. So you put
Tim: 4, 5
I: 4, 5, 8, and 9 in one group. What do they have in common?
Tim: Sine, cosine and tangent.
I: Right, OK. And then we have 1, 3, 6, and 7.
Tim: This one is powers, power of 5, power of 6, so I don’t know how to put these
together.
I: OK, they are just the leftovers.
Tim: Yeah
I: Is there any other way that you could group these?
Tim: Ah, yeah. [pause] I could group the e functions together.
Tim did not recall the chain rule by name. In fact, it took much prompting using
an example for him to see what he had done.
I: OK, good. Let’s use the bottom of this page here, I will draw a line. Could
you write down the chain rule in whatever words or symbols you remember?
Tim: The chain. . . ?
I: The chain rule, for taking derivatives.
Tim: You mean y = x2 − 7x + 5 and you take the chain rule for this one?
I: No, the chain rule is just a rule we have for taking certain kinds of derivatives.
Do you remember using the chain rule?
40
Tim:
I:
Tim:
I:
Tim:
I:
Tim:
I:
Tim:
I:
Tim:
I:
Tim:
I:
Tim:
I:
Tim:
I:
Tim:
I:
Tim:
Um, y, it was something, um. . . Can I do an example out of the book? The
chain rule for this one? [writes out f (x) = y n , (y n )0 = ny (n−1) ]
That is actually called the power rule.
The power rule.
You have got something raised to a power so you bring the exponent down.
You would use that in a lot of these problems here, the power rule. The
chain rule is talking about something else. Let me show you an example
of a problem where you would use the chain rule, for instance #18 here.
The original problem was sin(5x4 ). [Student is shown written work from the
questionnaire, which has the correct solution to the problem.]
You mean give the rules of this one, how I worked out this one?
Yes, how did you find the derivative there?
Oh, [writes and crosses out f (x) = u(x), f 0 (x) = u0 (x)] [mumbles] [writes
f = (y · x), (y · x)0 = y 0 · x + x0 y] I don’t know, you know, if you give me an
example of how to do the chain rule, I know how to do products.
That is what I am saying, this solution you have right here. . .
Mm-hmm
you used the chain rule
yeah
to get that solution, which is correct. OK? Does this remind you of the chain
rule, then?
Uh-huh, so what you want me to . . .
We are starting with sin(5x4 ) and look at what you wrote down for your
answer, and try and remember how you came up with that idea.
Oh! So first I take derivative of outside, derivative of sine is cosine,
Right
so then I take derivative of inside, so inside is 5x4 , so I write down 20x3 .
OK
That is so easy, you know, I don’t know how to get the something that you
asked me to do.
You don’t. . . So the question was, what does the chain rule say?
If there is a function of u, something like that, and take the derivative of
that, right, first you take the derivative of outside first, then take derivative
of inside.
Tim was able to see the chain rule in the implicit situation, but could not see it
in any of the other three.
Tim: If I do. . . OK, OK. So if I pick one?
I: Uh-huh
Tim: This one not chain rule. Hold on, I use the chain rule, yes. This one use
chain rule.
I: How would you use the chain rule?
41
Tim:
I:
Tim:
I:
Tim:
I:
Tim:
I:
Tim:
I:
Tim:
I:
√
That one has y and when I take derivative, I write it as y 1/2 and when I take
derivative of that one, I get that and that and [writes (y 1/2 )0 = 21 y 1/2−1 =
1 −1/2
] So if y is something else, I have to take derivative of y.
2y
OK
Not really, y, there is something inside y. I have to use the chain rule for
that.
OK. Right.
I’m not sure about the next problem. I forgot.
The name of the next one is a related rate problem. Do you remember that
name?
That has to do with speed? What is the formula? I not clear on the question
you know, ‘find the formula for the rate.’ [pause] This one doesn’t have
anything to do with it. Do you think?
OK.
So I don’t know if it has derivative. The next one is . . . [reads problem to
self] . . . power rule. . . No, I not use chain rule.
OK
[pause] I don’t think I would use the chain rule for the last one.
OK. Good.
While it is the case that Tim correctly solved all of the chain rule problems on the
questionnaire, it seems clear that he is not working with a general formula. Rather
he is taking each problem as it comes to him and dealing with it individually.
7.1.3
Al [A137]
Al seemed to be at the Inter- level. While his grouping of the ten derivative
problems did not use the chain rule, as was the case for Eli and Tim, he was able to
connect his notion of the chain rule with the items 16–20 (which are numbered 6–10
on the interview sheet).
I: What I would like you to do is to write down the chain rule, using whatever
words or symbols that you like, and can remember.
Al: OK.
I: Feel free to talk out loud.
Al: The chain rule, I know that you have something, you have a function f (x)
[mumbling] , so you’re gonna have f (x) [mumbling —pause] that’s 2, you
have something to the power of whatever and you take the derivative of
that, [has written (x3 + 2x)2 ] that would be 2x cubed plus 2x that the chain
rule would be if you take the derivative of that again and multiply it by that
of the inside.
I: OK, so is there an example of that in those problems there?
Al: [silence] Yeah, #6.
42
I: #6. Good. So you described it, you didn’t write anything down.
Al: No.
I: That’s OK. That’s fine, we got it on tape which is all I’m worried about
really. Um, so here’s a point, the last five questions 6–10 were written with
the intention that you would have to use the chain rule to solve them. So
I’m wondering, um, if we compare what you have written down as the chain
rule with each of those last five problems, does your statement take care of
it?
Al: Takes care of #7.
I: Uh-huh.
Al: Then it takes cares of that because you have cosine of that and the entire
derivative of that; which I said is that you have something on the outside
take the derivative and times the derivative on the inside.
I: OK. That’s not quite what you said before though, right?
Al: Right, I said that each have to raised to powers.
I: Right, so now you’re talking about inside and . . .
Al: Yeah, I know that you’d have to use the chain rule for that and . . . I think
that I would definitely use the chain rule for that.
I: OK.
Al: And, it is a prerequisite condition and I haven’t done chain rule for like a
year now.
I: That’s fine, that’s fine. I was just trying to get whatever you thought of first.
And now what you are thinking of second. This is fine.
Al: OK.
He went on to work through the remaining three problems. The main distinction
between his work and that of Eli and Tim was that he seemed to be in control of
making the connections once the chain rule was pointed out.
In classifying the four situations, Al saw the chain rule in the implicit differentiation problem and in the last problem. He was unfamiliar with the related rate
problems and decided that it would not be used. It seemed as though he might have
answered differently had he worked through the solution to the problems.
I: Yeah, alright. What I have is a set of four situations, that are differentiation
problems that you probably dealt with in the first calculus course that you
took. I don’t want you to solve these problems, but if you would read through
each of these situations and decide if the solution would use the chain rule
in some way and if so, then how would it be used.
Al: OK. [silence] For one you would use it. You have x times the derivative of
y times the. . . that would be you use the product rule to get the derivative
of that, that is y to the 1/2 times the derivative of y. . . −1/2y to the −1/2
times the derivative of y.
43
I: And that’s the chain rule?
Al: Yeah. You have uh, y in itself is a function and y can be x cubed or you
know t cubed. And the square root of it, and to get to that you have uh,
1/2(t3 )−1/2 times the derivative of that which we’d write 3t2 .
I: OK.
Al: For one you’d use it. On to two now?
I: Uh-huh.
Al: [pause] I don’t know how to do #2.
I: Um, would it help if I told you that it was a related rate type of problem?
Al: Um, no.
I: OK. Alright. What you have is two things that are varying in this situation,
the height on the wall and then the distance from the wall on the bottom.
And, I’m given information about the rate at which the bottom is changing
and I’m asked about the rate at which the top is changing. So what happens
is you look for a relationship between those two original quantities. The
height and the distance from the wall and then you try and go from there to
an expression that talks about their rates.
Al: OK, but how would we do that?
I: Oh, you want to see the work?
Al: No,
I: Well, particularly if you draw a picture um you have a right triangle and so
you might label the height h and the bottom could be x. . .
Al: Then you used the triangle 1/2. You use that 1/2, you use that that triangle
thing?
I: Uh, I use a triangle thing. Not the area though, I use the Pythagorean
theorem to talk about the sides. We said that the ladder is L feet long so it
would be L2 = h2 + x2 .
Al: OK.
I: That’s our relationship. That relates these things together and then I would
need to find the rates by taking a derivative.
Al: OK. You want me to do three now?
I: Well, in that description, I didn’t say the word chain rule or use the phrase
chain rule. Do you think that the chain rule might pop up there or not?
Al: Um, no. [pause] No.
I: Um, yeah, so number three.
Al: [student reads to himself—long pause] You just take the derivative of that
and that would put that; you wouldn’t use it there either, I don’t think.
I: OK.
Al: I don’t think so. But, I’m not very good at word problems.
I: [laughs] Very few of us are.
Al: [chuckles] Yeah.
I: Do you recognize that problem as something that you have done before?
44
Al:
I:
Al:
I:
Al:
I:
Al:
I:
Al:
I’ve probably done it countless times. More than likely.
OK, but it’s not familiar to you.
Yeah. I’ve done that like probably ten times but. . .
The second one. . .
Yeah.
Right.
Cause almost every calculus course has that.
It’s a classic. OK, how about four?
[pause—reading] This, I don’t think that I’ve ever seen before. I would use
it here.
I: You would.
Al: Uh-huh. Yeah, because you have −42t2 and put that into F and that’s
squared cubed and then that times that.
So Al had made some connections between situations involving the chain rule, but
is still working toward a cohesive whole.
7.1.4
Ray [A123]
Ray is the most interesting case (he is the topic of section 8.4 below) as he had a
rather highly developed sense of the chain rule but no facility with it generally. This
is due to his making the decision to not memorize the derivation formulas. In the
following excerpt, Ray was attempting to do item 17 from the questionnaire, which
he had left blank.
Ray: This one is messier. I would like to approach this one as a composition.
I: 17?
Ray: 17, yeah, I would like to approach this one as a composition of two functions
as two different things, but I don’t remember those rules either. As far as
like the chain rule and except for this one is like 2x4 is f (x) and g(x) should
be [mumbles] I would have to use the chain rule to find out what that was
the derivative was. Oh, boy, foggy memory of the chain rule, I really did rely
on the, that, that. I actually had memorized the page where the derivative
rules were and then I didn’t have to memorize the derivative rules. Except
right before a test I would memorize them temporarily, I would cram them
into my RAM space.
I: So, um, the situation right now is that you know that there is a chain rule,
and it would apply here for a composition, um, try to dig out as much as
you can.
Ray: It’s like f (g(x)) prime, is g(x) prime. Something like x has derivatives inside
there, different things. . . something along these lines. How ever did you grade
my tests? [sighs] I know that you take the derivative of each one separately
and compose them somehow, either added or multiply them, I think that they
45
are added and beyond that that’s all the fog that covers the skyscrapers of
my mind.
I: That’s called pollution. Um, what other rules do you remember by name?
Ray: Let’s see here, the chain rule, the um sum rule is pretty simple, you can take
them separately. Um, the product rule I remember, by name, um that being
the special case, once again. I don’t exactly know how to do it. There’s a
rule for division, the quotient rule. Those are the important ones—chain,
product and quotient. Then we have like power rule—the easy one it’s like
x2 ; 2x. I remember that one and that really took care of a lot of them. The
power rule. I used that for the first four of them.
I: Right. Right.
As the interview progressed Ray was able to describe which rules he would use to
solve the problems, provided he could see the formulas. With much interaction on
the part of the interviewer, he recalled the formula for the product rule and the chain
rule. He demonstrated the use of the chain rule on item 18.
Likewise, with the four situations, he was able to see the chain rule in most of the
problems. In both the second and third problems (related rates) he did not see the
need for a derivative; hence no need for the chain rule. Apparently, he had associated
acceleration with rate and with integration.
Ray:
OK. [pause] OK, I’m trying to remember . . . got a rate . . . this one doesn’t
even scream for derivation to me. Because you have one rate . . . L is going
to be a constant. Um, looking for a second. . . this doesn’t even really scream
for a derivation except there’s too many variables and to get rid of L that
we’d. . . but no I don’t think that I would use the chain rule on number two I
would set up some kind of trigonometric relationship to two different sides,
time being a constant thing, but distance, the trigonometric relationship
between the distances would dictate everything out.
I: How would they dictate them?
Ray: Well, in the same amount of time that it sits here, [mumbles to self] cosine. . . goes to. . . opposite [mumbles to self] opposite over adjacent is the
tangent. . .
I: Right.
Ray: um I would probably use the tangent, but L doesn’t help a lot and um, x
velocity over 4 feet per second, and I solve for x to get um, I’m not sure. . . I’d
try to solve for x through some large algebra and that would be the answer.
I: OK. OK.
Ray: As the rate, there’s not acceleration there. Which would scream for . . . no.
That would scream for integration, acceleration, but actually. Yeah, yeah.
There’s no change here, it is constant velocity so this doesn’t even ask for
derivation. Therefore I wouldn’t probably use the chain rule to solve this
problem. That was number 2. [Reads to self]. . . oh,oh, S cubed. . . solve for
46
V . . . it gives. . . I don’t think that I would use the chain rule here because I
don’t really compose V , actually yeah they are, S has V in it. [mumbles to
self] Yeah, I’d use the chain rule for finding d(S) and because I would have
to solve for V and S would move into the problem I would have to use the
chain rule for finding d(V ) as well, I would use the chain rule twice here.
That’s number three.
Um [reads #4 to self]. . . hey I remember this problem. . . [mumbles to self]
. . . find the expression . . . uh, that’s a composition—I’m talking about four
now—you would substitute −42t2 in for everywhere that you would have F
and that would be an expression for the rate of change of C. Um, [mumbles to self] well, that’s obviously the composition of functions. [mumbles
to self] As I’m trying to find the rate of change, it’s the equivalent of acceleration. . . [mumbles to self] yeah, I would I don’t think I would even have
to derive here. . . this is question number four. . . but both of these. . . I would
think that I would have to use the chain rule because I’m putting one function
into another.
Ray seems to be at the Inter- level. He made the connections between the various
situations, but he was doing so with what might be considered semi-objects. That
is, one might consider a complete idea of a rule as being the totality of knowing the
formula, how to apply it and when to apply it. Many students seem to focus on and
excel at the first two to the exclusion of recognizing the need in context. Ray, on
the other hand, had focused on the latter two, choosing to look up the formula when
needed.
7.1.5
Peg [A132]
Peg indicated a rather strong Inter- level understanding as she grouped differentiation problems. She used the chain rule as an initial criteria, selecting two of the five
chain rule problems. The remaining problems were classified as being trigonometric
or exponential. When asked for a second way to sort, she kept the chain rule as a
main criterion.
I:
Peg:
I:
Peg:
So is that five different groups?
[pause] Yes. Five different groups.
OK, what was the discriminating features?
OK, One is like the most straight forward, where it is just strictly using the
uh, what’s the name of that rule? The power rule.
I: Uh-huh.
Peg: It’s just straight out the power rule. Uh, two and three are pretty much
exactly the same thing except for two, the expression needs to be rewritten
to use that rule. 6 and 7 is also just the power rule except for you have uh,
uh, you have a power rule and then it’s a chain rule.Uh, 4, 5, 8, & 9 all have
47
trigonometric uh terms in them to where then you also have to know their
trig functions and then 10 is the e function which also has it’s separate rule.
I: OK. When you went through that. . .
Peg: And that’s pretty much from my point of view which was from the easiest
to the most difficult. Although, I guess that the e function is not really that
difficult; it’s just knowing that it, it’s just not as, it’s not difficult it’s just
different because it just doesn’t seem to follow the format as all of the others.
I: Uh-huh. Um, was there any other way that you considered grouping these?
Peg: [pause] Uh. . . in the time frame, I just went with the first thought that popped
into my mind. I’m sure that if I was to sit here longer, I could think of other
ways to do it.
I: OK. What might be other criteria?
Peg: [pause] Uh, the other ones would have been, may be, the ones that just had
single terms as opposed to having two terms that have to be differentiated.
Or anything that that has a chain, or pulling out anything that has the chain
rule and knowing that also have to add more terms to the final expression.
Peg gave a very clear statement of the chain rule, using function notation. She
augmented this with a comment about inside and outside things. But her notion of
chain rule and composition were apparently driven by parentheses.
I: So the last couple of questions that I have for you, write down the chain
rule using whatever words or symbols that you like. It’s as best as you can
remember it. You have here like examples of it. Number 6 and 7 which is
16 and 17 on our papers here. You used the chain rule in your answer. So if
you want to use those to help spark some ideas. . .
Peg: [pause] [mumbles to self] [pause] OK.
I: Read that for the tape.
Peg: OK. Uh, when you have the derivative of f (g(x)) it’s f 0 g(x) times g 0 (x). It’s
basically you take a derivative of the outer term and I’m using outer term
because it’s just the way that I look at it, composed with the inner term and
then multiply it by the derivative of the inner term.
I: OK. Great. Um, actually the last five statements here on the list, 6 through
10,
Peg: Uh-huh.
I: were intended to be, to use the chain rule and you named 6, 7, & 8 as using
the chain rule. Do you see the chain rule being used in 9 and 10?
Peg: Yes, I do, but I had pulled them out separately because they had the trigonometric functions also.
I: OK, no that’s fine. Does your rule that you have written there at the bottom,
does it apply or does it take care of all six of those or uh, all five of those
cases?
Peg: [pause] Uh. . . in the way that I would look at it, the way that I look at the
problem, it does. I look at the outer term and the inner term which would,
48
the outer term that I’m looking at is being either how you solve strictly for
the power rule or for the trigonometric functions and then going inside and
actually you know. . .
She hesitated to include cos3 (t) in the chain rule group since t was in the parentheses
2
alone (or e−t due to lack of parentheses). She had attempted to solve it by rewriting
it as a triple product. This is consistent with the codes she received in Phase 1 on
composition and her lack of success with the final interview problem (discussed in the
next section).
I: OK. [pause] You did a very interesting thing when you solved 19.
Peg: I don’t think that’s correct.
I: I don’t think that it’s very far wrong actually. But what you did is rather
than use the chain rule, you expanded the expression. The expression was
cosine cubed of t. And so you wrote down three products of cosine t which
is cosine cubed.
Peg: Uh-huh.
I: And then, you took a derivative from there.
Peg: Even with that, I didn’t finish it did I?
I: No, you needed to do the other product.
Peg: Yeah, but, yeah, I made it to that.
I: But, my question is, if you can recall, this was a long time ago, um, was
there something that didn’t say chain rule to you in that problem?
Peg: That would be the one that I’m still the least clear on still today because it’s
not as straight as the chain rule because the uh, the variable t doesn’t have
anything associated with it like all of the other ones, that there’s something
inside the parenthesis to differentiate. Where this one. . . so would it be 3
cosine squared t uh times minus sin? Is that the correct answer.
I: Uh-huh.
Peg: Oh.
I: That’s exactly it.
Peg: But I, it’s just not as obvious of the chain rule as all of the others.
Peg was no longer familiar with implicit differentiation or related rate problems.
After being reminded, she does not see the chain rule since there is nothing inside the
other. She did see the chain rule in the last problem although the interviewer had to
point out the expression for F in the paragraph.
Overall we see Peg making connections based on a structure, the use of parentheses. She gave a general rule (again written with parentheses) but had some difficulty
recognizing problem situations which fit her rule. Peg may be working at the Trans-
49
level, that has been built on a weak structure. Her understanding of the chain rule—
discussed in the next section—is rather strong, but the data is equivocal regarding
the triad level.
7.1.6
Jack [A113]
The one example of the Trans- level is Jack, who not surprisingly had the highest
overall score of those interviewed. Jack grouped (and then partitioned) the ten differentiation problems based on levels of difficulty. He clearly saw the chain rule as one
of the criteria along with the power rule, product rule and trigonometric rules. He
stated that the trig rules are the most difficult, indicating that the other rules have
reasons behind them, but the trig rules must be memorized.
Jack:
10, I would probably put, I don’t know, I would be tempted to put 10 in a
group by itself just because with it just being e it’s basically the chain rule,
and the chain rule and the power rule together. That’s it. That one is done.
Um, 7 and, let’s see 7 requires it, 7 requires product rule and chain rule, and
the all powerful power rule. . .
8 just requires product rule. I know, no I probably would just go ahead and
group 4, 5, 8, & 9 all together because they have the trig function and because
I mean, the trig functions are the only derivatives that can throw you off real
easy if you don’t know them, because chain rule, if you understand chain rule
and you understand the use of things like exponentials and logarithms then
you aren’t gonna get messed up bad on chain rule. You aren’t gonna get
messed up bad on the power rule if you know simple mathematics. You’re
not gonna get messed up too bad on product rule as long as you remember
to keep everything straight. But, the trig functions, you know, you got, I
can’t even think of them at the moment. I haven’t used them in a while, but
it’s like you know one it’s the other and it’s just negative and one it’s just
the other, period. And it’s like if you forget that sine, the derivative of sine
just by whether or not it’s got a sign change in it, you just messed yourself
up big time and you’re gonna get a wrong answer because they cycle; and if
you start off on, off with the wrong derivative of it then you have messed up
the cycle already. Then, no matter, if you know all the others you are gonna
be messed up anyway. So. . .
In the next questions, Jack gave the textbook answer for the chain rule. He
elaborates on his answer describing inside and outside functions, as opposed to Al
and Peg who spoke of inside and outside terms. And finally he outlined a proof of
the rule.
Jack:
Um, huh, [pause] let’s see how would I write that down? OK. Let’s see.
[pause—writing] Um, that’s just the way that it runs through in my head.
[had written f (g(x))0 = f 0 (g(x)) · (g 0 (x)) · x0 ] Without any words.
50
I:
Jack:
I:
Jack:
OK.
That’s just the way that I think of it.
OK. Read that for the tape.
Um, you take the derivative, essentially, whenever you use the chain rule you
are essentially looking at a function that has got another function within it
um, I don’t know, it’s sort of like doing a, handling a composite, um and
in order to take the derivative of that composite you have to first take the
derivative of the outside function and not even do anything with what’s the
inside of it, the function that’s on the inside of it, you take the outside, it’s
derivative first and leave the inside function alone and then multiply that by
the derivative of the inside function, and then multiply it by the derivative
of the variable, or however many times you have to break it down. Because
you can have a huge function that’s got a lot of stuff inside of it and you’d
have to do the chain rule several times to get the x variable.
I: OK.
Jack: So, I mean you could have, that’s just like a simple composite f of g, but
you could have, if you have like h(f (g)) then you’d have to do the derivative
h with f of g inside of it and then the derivative of f with g inside it and
then the derivative of g and then derivative of x.
I: OK.
In the four problem situations, Jack saw the chain rule being used in all of them.
His explanation for each indicated thinking about the functions in the situation.
Overall, he had as well established an understanding of and facility with the chain
rule as one can hope for in a first year calculus student. The Trans- level is not so
evident by the preceding quotes, but is more clear in his work on the Leibniz rule
problem which is discussed in the next section.
7.2
Using versus understanding
A working hypothesis of this study is the need to understand composition of
functions as a condition for understanding the chain rule. While the first phase results
did not add evidence to this claim, the claim was not refuted since the questionnaire
items did not assess understanding of the chain rule, rather they could only indicate
success (or lack of success) with the rule.
7.2.1
Coding the aspects of understanding
The interview data provide us with sufficient information to assess the student’s
understanding of the rule. In order to discuss understanding, we have noted three
aspects from the interview:
1. Did the student mention the chain rule prior to question 10 of the interview?
51
2. Did the student arrive at a coherent statement of the chain rule? With or
without a prompt?
3. Was the student successful with the Leibniz rule problem? With or without a
prompt?
These questions were coded as an ordered triple with codes Yes or No for the first
aspect and codes Unsuccessful, Prompted, or Successful for the second and third
aspects. The following excerpts illustrate the codes for each subject which are summarized in Table 7.1 below.
Eli received the code [N, U, P]. As we observed in the excerpts given in the previous
section, Eli’s statement of the chain rule never indicated that the derivative of the
outside function was then composed with the inside function. When working on the
Leibniz problem, he insisted on finding an antiderivative. The interviewer provided
him with the simpler Leibniz problem (H(x)) and had done most of the talking up
to this point, getting Eli to work out the closed form of H:
I: Alright. OK, so now that’s H(x). That’s just computing it all out. Right?
You just did this side, it’s still H(x). So I could put an x in and take a sin
of it and do that other stuff. What’s H 0 ?
Eli: [pause] So you take derivative of this?
I: Uh-huh
Eli: So this is the chain rule again.
I: Always. [pause] OK, so you have. . . Oh, you didn’t label it H 0 . But H 0 (x) is
4 times sin(x) squared times the cos(x). Right? Now we started off with one
of these deals where it’s defined in terms of its definite integral and you were
able to take the antiderivative and evaluate it at the endpoints and come up
with that, and then take the derivative and you get this. Now back to the
original expression up here, do you see a way to jump straight to that answer
without finding an antiderivative?
Eli: Yeah.
I: What is it?
Eli: You just um, replace t by sin(x) then just multiply the derivative of sin(x).
I: OK, then let’s go back to the first one that I asked you. F (x) is the definite
2
integral from zero to the sin(x), et dt. What’s the derivative of F ?
Eli: The derivative? The derivative of. . .
I: The derivative of F at the very bottom here like you found H 0 here.
2
Eli: [mumbling] [writes F 0 (x) = e(sin x) · (cos x)] OK.
Tim’s code is [N, P, U] where the first two are seen in the excerpts above. In the
last problem, he recognized the fundamental theorem of calculus (FTC) version of
the situation (G(x)), but could not make a connection back to the Leibniz version.
52
I: I am asking you to find F 0 where F is the function given by F (x) = the
2
integral from 0 to the sin(x), et dt.
2
Tim: It is the inside, the anti-derivative [writes F 0 = et ].
I: Let me show you one other one, here is a function G(x) which is defined
2
almost the same way, zero to x of et dt. What’s G0 ?
Tim: Can I do this? [works and mumbles]
I: Whoa, whoa. You have taken a derivative there, not an anti-derivative.
Right?
Tim: Oh! Sorry, I’m sorry.
2
I: Let me just help you out here. There is no anti-derivative of et .
Tim: Oh? It is a trick.
I: There is no simple anti-derivative that we can just write down on paper.
So your method there, while it would probably work, if you could take the
anti-derivative of that, it won’t work here.
Tim: [pause—working] [attempts to find an anti-derivative] Yeah, it does not work.
I: But can you answer the question? Can you figure out what G0 is if you know
this is how G is defined?
Tim: [1.5 minute pause] I’m lost!
I: Really?
2
Tim: Yeah, I think G0 (e) = et .
I: OK. G0 of . . .
2
Tim: G0 (e) = et
2
I: G0 (e)? is et .
Tim: Uh-huh
I: OK
Tim: Is that right?
I: Let me ask you this, when you answered this first one, you jumped right to
it and said ‘oh, F 0 is the insides’.
Tim: Same thing here.
I: They are the same. So it doesn’t matter that there is a sin(x) up here?
Tim: Oh?
I: This one goes 0 to x, this one goes 0 to sin(x). So I am wondering if that
makes a difference in the answer?
Tim: So, yeah, yeah, hold on. Sine . . . [works and mumbles] [writes −1 ≤ sin x ≤ 1,
0 ≤ sin x ≤ 1] Yeah, I think it doesn’t matter for that.
I: It doesn’t matter.
Tim: No.
Al got [N, P, U]. In his solution to the Leibniz problem, he quickly stated that the
derivative is found by substituting sin(x) in for t. Then when working through the
53
FTC problem, he wanted to use the chain rule and multiply by 2x. After the prompt
to correct G0 , he did not go back to modify his original answer.
I: OK. Great. What I’m asking you for here is compute F 0 if F is the function
2
given by F of x = the definite integral from zero to the sin(x), et dt.
2
Al: I’d take et and just put the sin(x) squared in there, and then you’d pretty
much have zero and it would be this right here.
I: That’s the derivative? That’s F 0 ?
Al: Yeah.
I: OK, just clarifying.
Al: OK.
I: Let me ask you a similar question. This one is called G, it’s basically the
same except um, the definite integral goes from zero to x. What would um. . .
Al: Just change the t to the. . . x squared.
I: OK, so G0 .
Al: times 2x, and over here you’d take derivative too. I don’t know . . . what I’m
saying, you’d take the derivative, uh . . .
[end of side one]
I: So you’re wondering do you take the derivative. . .
Al: Take the derivative of x2 or not.
I: [talking together] Take the derivative of x2 or not.
Al: Yeah.
I: Alright. Do you remember seeing situations like this where you have the
functions defined through definite integrals?
Al: Uh-huh.
I: OK, it was part of the fundamental theorem of calculus.
Al: Oh, [mumbling] I never. OK, I know this part was. . . right and this part was
right.
2
I: OK, ex you’re sure about, and you’re wondering about, um, the 2x part.
OK. [pause] So, suppose I tell you that you don’t do 2x in this problem. The
2
answer is just ex . That comes from the fundamental theorem of calculus,
which essentially says that if you have a function defined in this manner the
derivative of that function in this case G0 is just the inside written in terms
of x.
Al: Uh-huh. Yeah. And that would be the answer.
I: Without the 2x?
Al: Yeah.
I: Right. Now what I want you to do is to go back and look at this other
problem.
Al: Uh-huh.
I: And see, you have e to the sin(x) squared, and the question is, are you still
comfortable with that?
54
Al: [pause] Yeah.
I: OK.
Al: Yeah.
Ray and Peg received codes of [Y, S, P]. Ray’s first thought on the Leibniz problem
was, since a definite integral is a number, then the answer is 0. It only took a small
prompt for him to see the function involved. When he saw the FTC version of the
question, he recalled the rule and quickly gave the answer. He then saw that the
original question involved a composition and struggles a bit before coming to the
answer for F 0 .
Peg mentioned the chain rule when confronted with a composition problem (that
is not a derivative problem). So the connection between the two is strong. Peg
recognized that the situation involved a theorem that she once knew, but denied
the FTC. The interviewer prompted her through the FTC and also pointed out the
composition.
Jack got a [Y, S, S]. Jack indicated a cohesive schema for the chain rule as he
attempted to deal with the Leibniz problem, which he did not recognize. To deal
with the unfamiliar, he wanted to compute something, so the interviewer provided
him with a simpler Leibniz rule problem with a monomial integrand. Jack worked
through the integration, then needed a little prompt to label his answer (the function
H). When asked to go back to the original problem, he was able to generalize the
use of the chain rule and to arrive at the correct solution.
7.2.2
Comparing the codes
The results of the codes for understanding are presented in Table 7.1. It is sorted
according to a lexicon order on the codes. Also presented are the combined codes
from Phase 1.
Two patterns in this table are of particular note. First is the support for our
working hypothesis. With the exception of Peg, the scores increase for the Composition codes as one is more successful at understanding. We note that Peg left items
9 and 10 blank on the questionnaire because she did not want to deal with the tables. If we exclude the codes for items 9 and 10, the Composition column becomes
[6, 9, 9, 9, 15, 15]T which also lends support to our hypothesis.
The second pattern is the third aspect code and the C4L variable. Those who were
able to deal with the Leibniz rule problem were all in a C4 L section at some point.
Both of these points seem to say something, but must be understood in context.
These patterns exist in these six students’ results only and do not generalize to any
55
Table 7.1
Aspects of understanding
Eli
Tim
Al
Peg
Ray
Jack
Code
C4L
Function
Composition
Differentiation
Chain Rule
[N, U, P]
[N, P, U]
[N, P, U]
[Y, S, P]
[Y, S, P]
[Y, S, S]
11
00
00
10
11
01
9
4
19
14
18
15
7
11
12
9
21
24
16
22
24
25
19
25
20
25
19
21
1
24
other population, not even the Phase 1 group. It would be unwise to attempt to draw
any strong conclusions from this data.
Chapter 8: Discussion
The analysis of the data is consistent with the genetic decomposition for the
concept of the chain rule based on the triad, which was presented at the start of
this work. Some evidence is presented to support the notion that understanding of
composition of functions is key to understanding the chain rule.
The type of instruction was a factor in how a student performed on these tasks.
The differences between the types seem to be related to the difference between using the chain rule and understanding the chain rule, and an explanation of these
differences is offered.
The need for collecting differing types of data is established. Two students scored
high quantitatively but did not perform well in the interview, while another student
failed the quantitative work but demonstrated a high level in the interview. Finally,
the use of writing as a pedagogical tool is recommended based on the results.
8.1 Revised genetic decomposition
The overarching purpose of this study was to test APOS Theory, specifically
the proposed use of the triad mechanism to describe the development of a student’s
schema. For the six participants in the interviews, we claim that it does. The interview data show students working at all three levels by discussing how they appear
to be making connections as they approach problem situations which involve the
chain rule in the solution. After presenting evidence of perceiving relationships in the
situation, a working definition of understanding the chain rule as it appears in the
data was constructed. This construct presents a parallel story as the six subjects are
ranked in understanding along the same lines as the triad levels they demonstrate.
The students who have moved into the Inter- level demonstrate some ability to extend their understanding of the chain rule, as well as the ability to discuss the rule
coherently.
The interview data did not provide evidence which required the modification of
the genetic decomposition given in Section 2.4. So the description of the schema for
the concept of chain rule developed by Clark et al. (1997), including this author, has
the support of this study with the six interviewed students. Clark et al. concluded
that the chain rule schema
56
57
must contain a function schema which includes at least a process conception of function, function composition and decomposition. The function
schema is linked to a differentiation schema which includes the rules of
differentiation at least at the process level. (p. 359)
Clark et al. (1997) further stated that this schema is developed through the three
levels of the triad. In the Intra- level, the student is able to deal with many types of
problems as distinct cases. Some students received high codes for their work on the
chain rule section of the questionnaire. But their concept of the rule is insufficient to
discuss the chain rule in any detail or to extend it beyond their familiar collection of
rules.
At the Inter- level, the student begins to reflect on these various cases and builds
connections relating them. Students in this study not only used phrases such as
“inside function/term/stuff” and “outside function/term/stuff”, but also used them
as the criteria when presented with new situations. In other words, they were using
their perceived relationship (or more general idea) for the chain rule when asked if
the chain rule would appear in the solution to a problem situation.
A student working at the Trans- level has acted on the relationships from the previous level to construct a coherent whole, a system by which he or she can determine
what situations are part of the schema and what are not. This investigation used the
Leibniz rule problem as a way to test the robustness of the subjects’ schema (or preschema, if at the Inter- level). This problem features a composition with a function
defined through an integral, which pushes most students’ function schema to the limit
(Thomas, 1995). The problem was presented with simpler cases on hand in order to
help the students progress to the point of needing the chain rule. The simpler cases
were intended to help the student deal with a function defined through an integral
or to give the student a chance to compute something if the need to do so was great.
The analysis of the responses to this problem was focused on how the chain rule was
incorporated into the problem, once the composition in F was established.
This genetic decomposition is not “revised” in the sense that it has changed as a
result of the data analysis. It has however been revisited with new data and found
to be consistent with that data. The triad mechanism is useful in describing the
development of this schema, and will likely be as useful with other schemas. This
mechanism extends APOS Theory by augmenting our understanding of how schemas
might come to be built in the minds of our students. In particular, this genetic
decomposition will be used as a basis for the design of future studies involving the
58
chain rule and for the development of instructional strategies, such as those presented
below.
8.2
Understanding of composition and understanding of the chain rule
Can we conclude that understanding of composition of functions is key to understanding the chain rule? The results from the first part of this study were largely
inconclusive. The students performed well on the differentiation problems and the
chain rule problems. There was not the same degree of variation as in the function
and composition sections, and this did not allow significant correlations to be found.
This is a problem of assessment; the difficulty in measuring (depth of) understanding
in a written instrument. Since the questionnaire data did not measure understanding
with respect to the chain rule, the evidence does not contradict the hypothesis.
There is a small amount of supporting evidence in the Phase 2 data. As seen
in Section 7.2, for the six subjects who were interviewed, there exists a correlation
between understanding the chain rule and the codes for (three of the five) composition
items. A new study is needed to address this hypothesis. That study might employ
interviews based on the interview guide for this study with a larger sized sample. Since
the focus would be on understanding, collecting data with a questionnaire alone is not
advised (see Section 8.4). A questionnaire could be useful for screening participants
for interviews and for setting the stage for the interview tasks, as it was for this study.
8.3
Comparing instructional methods
The comparison between the traditional and reform methods is an account of using
versus understanding, of instrumental versus relational. In the interview setting,
the students who had experienced some reform methods were more successful in
extending and discussing the chain rule. In the quantitative data, the traditional
group performed better in the differentiation and chain rule sections, in the sense of
statistical significance.
8.3.1
Understanding the chain rule
In Table 7.1 some evidence is found that the students who had experienced C4 L
instruction demonstrated better understanding than the others. Eli is the exception
to this statement, but it fits with his role as an outlier in the quantitative phase and
with his lack of cooperation in the qualitative one. This comparison can only be made
with these six (or five) students and so must be seen as an example, rather than as
a trend. On the other hand, a current review of a series of studies in which student
performance data is available indicates that, on average, students who had instruction
59
based on APOS Theory performed at the same level or above those students who had
traditional methods of instruction. (Clark, Dubinsky, Loch, McDonald, Merkovsky,
& Weller, 1999). The interview evidence of this study is also consistent with that of
Meel (1998).
8.3.2
Using the chain rule
The questionnaire data did give us information on how well the students were able
to use the chain rule. Tables 5.6 and 5.7 both indicate the students in the traditional
course scored higher in the chain rule problems. In the first table, the C4 L group
was taken as C4L > 0, which cut the participants into almost equal sized samples.
In that case, the mean Chain Rule score for the Traditional group was 22.00, a 9.3%
increase over the C4 L group. In the second table, the groups were formed based on
the type of instruction for the first semester. In that case, the mean of the Traditional
group (21.91) was statistically significantly higher than that of the C4 L group (18.75),
16.9% higher.
These results are also consistent with previous studies. Heid (1988) found that
students from traditional calculus courses performed slightly better on skills based
problem situations than students from reformed courses. Dubinsky and Schwingendorf (1991) found the slight differences between groups favored the reform group.
Palmiter (1991) found that the reform students performed better than the control
on both conceptual as well as computational tests. However, her results may have
been skewed since the treatment group was aware of the study and the control group
was not. For our results, there are two global factors and one local factor which may
explain the difference. Globally, a traditional course by its nature emphasizes skill
acquisition. Typically, a student in a traditional course will complete 60–75 problems
in a week, with most of the problems being computational over a small range of topics. Another global factor concerns C4 L in particular and reformed calculus programs
in general. In order to utilize technology as a tool for calculus, time has to be spent
getting familiar with it. C4 L pays a price in its methodology, and part of that price
is less time for practicing skills.
In addition, the implementation of the C4 L course that these students experienced
was new to the institution, as the instructors has just joined the faculty. This local
factor involved changing student expectations for the mechanics of a course, dealing
with a commuter student body (having come from a residential situation), and networking problems inherent in a new computer laboratory. There is no practical way
60
to measure the effect of this factor on the difference in the Chain Rule scores between
courses1 .
8.3.3
Finding a balance
These results are not surprising but do present the question of our teaching objectives. If a conceptual understanding is the goal (and how can it not be?), then
this evidence indicates a focus on developing a schema is necessary, as was done in
the reform method seen here. Of course, Ray is an example of going too far away
from the instrumental knowledge to the conceptual. However, it is certainly easier
to fix this than the reverse. Once a conceptual basis has been established—once a
schema has been constructed—it is merely a matter of practice and drill to augment
the understanding with skills. The use of quizzes on computational items or the use
of gateway examinations might help balance any over-emphasis on the conceptual
understanding by a particular methodology.
8.4
Quantitative versus qualitative data
In the quantitative phase, three sets of data from the questionnaires were excluded
from the analysis because there were too many zeros in the code data—which was
interpreted as incomplete questionnaires. Two of these students were interviewed in
the second phase, Eli and Ray.
In the previous chapter, the case was made that Eli should have been excluded
from the quantitative analysis for non-participation. In fact, he was fairly nonparticipatory in the interview, too. The decision to exclude Ray from the quantitative
analysis was justified. However, it was not the case that he was not participating—he
was just caught off guard. Had he been told what to expect when he came to fill out
the questionnaire, he may have prepared and done very well as he had in his calculus
classes.
In this set of interview participants, two individuals with high scores in the Chain
Rule codes are found: Tim got 25 and Al got 19. Based on this data alone, we might
have concluded that these two “know” the chain rule or even “understand” it. Yet
both have some difficulty extending the concept. Tim even had difficulty expressing
the rule with prompting. At the same time, Ray got a 1 for the Chain Rule in the
written work. This would have to be interpreted as a failure to understand the rule
in a test situation.
1
Over the intervening couple of years, this factor has diminished—especially in the perspective of
the students.
61
The qualitative data paint almost the opposite picture for these three students.
Which is the accurate representation of reality? One conclusion of this study is the
necessity for collecting multiple data types when attempting to describe a student’s
understanding of a concept. The question raised is, how can we as teachers assign
grades based on written exams and homework only?
8.5
Limitations
The results of the study may be limited in their potential to be generalized in two
ways. First, since the subjects were volunteers, there was no control over the number
of subjects for the study nor over the range of abilities of students. The issue might
be addressed by trying to get n ≥ 50, a goal which was unattainable in this study.
With such a number, the spectrum of abilities should be filled out. The statistical
methods used in Phase 1 might have been extended by deleting variables from the
model or throwing more powerful models at the situation. That this was not done
came from a certain sense of purity; the bad data is just accepted as is. The model
was based on the genetic decomposition of understanding a concept. However, the
data which was collected reported a measure of using the concept. It also came from
the fact that only this set of students was being described, rather than an attempt
to describe the entire population of first year calculus students.
The second issue is also about self-selection. Since the comparison of the two
methods took place after the term, no pairing of students across treatments was
possible. Students at the host institution are free to choose whatever section of a
course they desire. Thus other factors (or student characteristics) may be involved
in whether the student chose to take a Traditional or a C4 L course. Schwingendorf,
McCabe and Kuhn (1998) discuss statistical methods that have been developed to
treat this issue, which arises often in health and industrial studies. One such method
is factoring in the students’ PGPA scores, which seems to have accounted for any
natural differences between the classes in this study.
8.6
Instructional strategies
A student’s concept of the chain rule appears to be a schema which is developed through the triad mechanism. The components of this schema are function
de-composition and differentiation. While the students of this study seem to have
62
the differentiation well in hand, they appear to need to move from a focus on “inside/outside” structure based on parentheses to a perception of composition of functions. To aid in the progression through the triad, we should give students opportunities to reflect on a variety of situations in order to build some connections and then
to reflect on those connections.
Short writing assignments may be the tool to take care of both of these needs.
By requesting an explanation which identifies the functions involved in a chain rule
problem, along with the answer, the student will have to approach the situation functionally, rather than symbolically. Likewise, short essays that compare and contrast
banks of problem situations will foster the reflections necessary.
Regular writing assignments will also address the assessment issue raised above.
The written discussion can allow the instructor to see the student extending the
concept to new situations. Wahlberg (1998) found evidence of both learning on the
part of the student and enriched assessment on the part of the teacher using writing
assignments on the topic of limits. While writing assignments take time to grade,
they are a reasonable alternative to scheduling, conducting and transcribing personal
interviews with your students.
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63
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concept of function: Aspects of epistemology and pedagogy, MAA Notes 25 (pp. 59–
84). Washington, DC: Mathematical Association of America.
Tall, D. (1992). The transition to advanced mathematical thinking: Functions,
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Thomas, K. (1995). The fundamental theorem of calculus: An investigation into
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Lafayette, IN.
Thompson, P. W. (1994a). Images of rate and operational understanding of the
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PRIMUS, 5, 291–295.
Vidaković, D. (1996). Learning the concept of inverse function. Journal of
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68
Vinner, S., & Dreyfus, T. (1989). Images and definitions for the concept of
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APPENDICES
Appendix A: Questionnaire from Phase 1
Code:
Questionnaire for Calculus Topics Study
Jim Cottrill
Directions:
• These items are designed to explore your range of understanding on a number
of topics. Some will be very easy to do, some will be harder. Please do the best
you can.
• Please answer each question as completely as you can. Please do not leave any
answer blank. If you would like to write additional notes with an answer, feel
free to use the back of the page.
• Please do not write your name on any page of this questionnaire. The code
number at the top of this page will serve to keep your identity confidential.
69
70
1. Express the diameter of a circle as a function of its area and sketch its graph.
2. A student has marked the following as a non-function. State whether this
student is correct and why.
y
x
71
3. A student has marked the following as a non-function. State whether this
student is correct and why.
A correspondence that associates 1 with each positive number, −1
with each negative number, and 3 with zero.
4. Decide if it is possible to use one or more functions to describe this situation.
If yes, then describe your function(s) briefly. If no, then explain.


x = t3 + t
t is a real number.
y = 1 − 3t + 2t4 
72
time
time
distance
time
distance
distance
time
distance
distance
distance
5. Tim and Donna live 1 km from their school. Usually, they walk to school
together. Yesterday, they both left their houses 10 minutes before school started.
Tim started to walk, but Donna was afraid to be late and started to run. After
a while, Tim realized that although he tried to walk faster and faster, he had
to run if he did not want to be late, and started to run. At about the same
time, Donna became tired and had to walk instead of run. They both reached
school exactly on time. Which of the following graphs is Tim’s and which one
is Donna’s?
time
time
t
6. Given two functions v, w such that v(t) = 5t−6 and w(t) = +4, find (w◦v)(9).
3
Explain.
73
7. Given that (f ◦ g)(x) =
√
5
2x + 3.
(a) Find f and g that satisfy this condition.
(b) Are there more than one answer to part (a)? Explain.
8. Find k so that g(x + 1) = g(x) + k, given that g(x) = 3x + 5. Explain.
74
In each of the following two questions, f, g, h are functions whose domains and ranges
are the set of all real numbers, and such that h = f ◦ g.
9. If only the information in the following table were known, would it be possible
to find f (2)? If so, find it and if not explain why not.
x
−1
4
π
h(x) g(x)
1
π
0
−3
1
2
10. If only the information in the following table were known, would it be possible
to find g(4)? If so, find it and if not explain why not.
x
−1
2
4
h(x) f (x)
1
3
−2
−2
1
π
75
Compute the derivative of each of the following functions. Show all of your work.
11. f (x) = 11x5 − 6x3 + 8
12. g(x) =
3
x2
76
Compute the derivative of each of the following functions. Show all of your work.
13. h(x) = x2 − 3
5x − x3
14. y = 3ex − 4 tan(x)
77
Compute the derivative of each of the following functions. Show all of your work.
15. y = x2 sin(x)
16. F (x) = 1 − 4x3
2
78
Compute the derivative of each of the following functions. Show all of your work.
4
17. G(x) = 2 5x2 + 1
18. H(x) = sin 5x4
4
− 4x 5x2 + 1
79
Compute the derivative of each of the following functions. Show all of your work.
19. y = cos3 (t)
2
20. y = e−t
Appendix B: Phase 1 Data
The following tables summarize the distribution of codes for each item in the
questionnaire. In Table B.1, the data is presented in the form n(a, b) where n is the
number of responses receiving that code, a is the number in the C4 L group (C4L codes
11, 10, 01) and b the number of traditional students (C4L code 00). The maximum
value for a or b is then 17. The other tables give the codes by each student’s ID,
which was used to protect the identities during the analysis. The (*) next to an ID
indicates a questionnaire which was considered to be incomplete.
The C4L variable used in Tables B.2 and B.3 (and in Table 5.5) is a binary code for
the first two quarters of calculus. An “11” represents the student took the C4 L version
of calculus both terms. A “10” indicates taking the C4 L version first quarter and a
traditional course the second quarter, while a “01” indicates the opposite situation.
A “00” indicates being in a traditional course both terms.
80
81
Table B.1
Distribution of codes by item
5
4
3
2
1
0
1
4 (1,3)
8 (3,5)
1 (0,1)
10 (7,3)
4 (3,1)
7 (3,4)
2
12 (4,8)
3 (3,0)
8 (5,3)
3 (1,2)
8 (4,4)
3
7 (3,4)
3 (2,1)
15 (8,7)
4
2 (0,2)
5
27 (14,13)
6
20 (9,11)
3 (2,1)
7
17 (11,6)
8
3 (1,2)
6 (3,3)
9 (4,5)
9 (5,4)
6 (4,2)
4 (2,2)
2 (1,1)
1 (0,1)
2 (2,0)
1 (0,1)
7 (4,3)
1 (0,1)
3 (1,2)
7 (2,5)
1 (1,0)
4 (2,2)
2 (0,2)
15 (8,7)
3 (1,2)
3 (1,2)
3 (1,2)
8 (5,3)
2 (1,1)
9
8 (6,2)
3 (1,2)
4 (3,1)
13 (3,10)
3 (1,2)
3 (2,1)
10
2 (2,0)
6 (3,3)
5 (4,1)
11 (4,7)
3 (0,3)
7 (4,3)
11
34 (17,17)
12
23 (12,11)
5 (2,3)
3 (1,2)
2 (1,1)
1 (1,0)
13
22 (11,11)
6 (2,4)
3 (1,2)
1 (1,0)
1 (1,0)
14
21 (7,14)
8 (6,2)
4 (3,1)
15
25 (9,16)
2 (1,1)
2 (2,0)
16
28 (12,16)
4 (3,1)
17
12 (5,7)
14 (7,7)
18
23 (10,13)
19
20
8 (4,4)
1 (1,0)
1 (1,0)
3 (3,0)
2 (2,0)
1 (1,0)
1 (1,0)
3 (2,1)
2 (1,1)
2 (1,1)
1 (1,0)
6 (3,3)
2 (1,1)
1 (1,0)
1 (1,0)
1 (1,0)
14 (5,9)
4 (3,1)
1 (1,0)
13 (6,7)
21 (9,12)
5 (3,2)
3 (2,1)
1 (1,0)
2 (2,0)
3 (1,2)
1 (1,0)
82
Table B.2
Student codes in tuples
ID
C4L
[Items 1–5]
[Items 6–10]
[Items 11–15]
[Items 16–20]
Overall
A111
01
[4, 5, 5, 2, 5]
[5, 5, 1, 5, 4]
[5, 5, 5, 1, 2]
[5, 3, 5, 5, 5]
82
A112
01
[4, 3, 3, 1, 5]
[4, 5, 5, 5, 5]
[5, 5, 5, 5, 5]
[5, 5, 5, 5, 5]
90
A113
01
[1, 5, 4, 0, 5]
[5, 5, 5, 5, 4]
[5, 5, 5, 5, 5]
[4, 5, 5, 5, 5]
88
A115
01
[2, 1, 1, 1, 5]
[5, 5, 5, 5, 5]
[5, 5, 4, 5, 5]
[5, 4, 5, 4, 5]
82
A121
11
[2, 4, 0, 2, 5]
[5, 5, 5, 5, 4]
[5, 5, 5, 5, 5]
[5, 5, 4, 5, 5]
86
A122
01
[5, 1, 3, 3, 2]
[5, 5, 5, 5, 2]
[5, 5, 5, 5, 5]
[5, 5, 5, 0, 5]
81
A123*
11
[4, 3, 3, 3, 5]
[5, 5, 5, 3, 3]
[5, 5, 5, 3, 1]
[1, 0, 0, 0, 0]
59
A124
11
[1, 1, 0, 0, 5]
[3, 1, 1, 3, 3]
[5, 5, 5, 4, 4]
[5, 4, 5, 4, 3]
62
A125*
11
[0, 4, 0, 0, 5]
[1, 1, 4, 1, 0]
[5, 3, 2, 4, 2]
[5, 4, 5, 3, 3]
52
A126
10
[2, 5, 3, 3, 5]
[5, 5, 5, 0, 0]
[5, 4, 1, 3, 3]
[5, 4, 2, 2, 4]
66
A127
00
[5, 5, 3, 5, 5]
[5, 5, 5, 5, 4]
[5, 5, 5, 5, 5]
[5, 5, 4, 5, 5]
96
A128
00
[5, 5, 5, 3, 5]
[5, 3, 1, 2, 2]
[5, 4, 5, 5, 5]
[5, 5, 5, 5, 5]
85
A129
10
[2, 4, 4, 2, 5]
[5, 3, 1, 3, 3]
[5, 5, 5, 4, 5]
[5, 5, 5, 2, 5]
78
A130*
00
[0, 5, 5, 0, 0]
[0, 1, 0, 2, 2]
[5, 5, 4, 5, 5]
[5, 1, 5, 2, 4]
56
A131
00
[0, 3, 0, 1, 1]
[5, 5, 1, 2, 2]
[5, 5, 5, 5, 5]
[5, 5, 5, 4, 5]
69
A132
10
[2, 3, 3, 1, 5]
[1, 3, 5, 0, 0]
[5, 5, 5, 5, 5]
[5, 4, 5, 2, 5]
69
A133
00
[4, 1, 3, 1, 5]
[1, 3, 4, 2, 2]
[5, 5, 4, 3, 5]
[5, 3, 5, 5, 3]
69
A134
00
[0, 1, 0, 1, 2]
[1, 3, 5, 1, 1]
[5, 3, 5, 4, 5]
[5, 5, 5, 5, 5]
62
A135
00
[4, 2, 1, 3, 5]
[2, 1, 3, 2, 0]
[5, 5, 3, 5, 5]
[5, 4, 5, 2, 1]
63
A136
00
[1, 3, 5, 0, 5]
[1, 0, 2, 4, 0]
[5, 5, 5, 5, 5]
[5, 4, 4, 5, 4]
68
A137
00
[4, 5, 3, 2, 5]
[5, 0, 4, 2, 1]
[5, 4, 5, 5, 5]
[5, 2, 5, 2, 5]
74
A138
00
[4, 1, 1, 3, 5]
[4, 3, 2, 1, 1]
[5, 4, 5, 5, 5]
[5, 4, 5, 2, 5]
70
A139
00
[2, 5, 3, 2, 5]
[5, 5, 5, 2, 2]
[5, 5, 5, 4, 5]
[5, 4, 5, 5, 5]
84
A140
01
[0, 1, 3, 1, 5]
[1, 2, 1, 0, 0]
[5, 5, 4, 4, 2]
[5, 2, 3, 5, 4]
53
A141
01
[2, 3, 3, 1, 2]
[5, 5, 0, 4, 3]
[5, 5, 5, 4, 5]
[5, 4, 4, 4, 1]
70
A142
00
[5, 5, 5, 5, 5]
[5, 5, 5, 4, 3]
[5, 5, 5, 5, 4]
[5, 4, 4, 2, 5]
91
A143
11
[0, 5, 5, 3, 5]
[4, 5, 3, 2, 2]
[5, 1, 5, 5, 5]
[4, 4, 5, 2, 5]
75
A144
00
[2, 5, 3, 2, 5]
[5, 5, 3, 3, 4]
[5, 5, 4, 5, 5]
[4, 5, 5, 2, 5]
82
A145
11
[2, 3, 3, 0, 1]
[3, 5, 2, 2, 2]
[5, 2, 0, 4, 1]
[2, 1, 1, 2, 4]
45
A146
00
[2, 3, 3, 1, 5]
[5, 4, 5, 2, 2]
[5, 3, 3, 5, 5]
[5, 5, 5, 5, 5]
78
A147
00
[3, 1, 0, 2, 5]
[5, 5, 5, 2, 0]
[5, 2, 5, 5, 5]
[5, 5, 3, 2, 5]
70
A155
00
[4, 5, 4, 3, 5]
[5, 3, 5, 5, 4]
[5, 5, 5, 5, 5]
[5, 4, 5, 5, 5]
92
A156
00
[0, 2, 3, 2, 2]
[5, 4, 1, 2, 2]
[5, 5, 4, 5, 5]
[5, 4, 5, 5, 1]
67
A157
11
[1, 2, 5, 2, 5]
[1, 4, 1, 2, 2]
[5, 4, 3, 3, 3]
[4, 3, 4, 2, 2]
58
83
Table B.3
Combined code data
ID
C4L
PGPA
Function
Composition
Differentiation
Chain
Rule
Overall
A111
01
2.70
21
20
18
23
82
A112
01
2.30
16
24
25
25
90
A113
01
3.40
15
24
25
24
88
A115
01
3.10
10
25
24
23
82
A121
11
2.40
13
24
25
24
86
A122
01
3.00
14
22
25
20
81
A123*
11
3.00
18
21
19
1
59
A124
11
2.40
7
11
23
21
62
A125*
11
2.80
9
7
16
20
52
A126
10
2.60
18
15
16
17
66
A127
00
3.90
23
24
25
24
96
A128
00
23
13
24
25
85
A129
10
17
15
24
22
78
A130*
00
10
5
24
17
56
A131
00
2.90
5
15
25
24
69
A132
10
3.10
14
9
25
21
69
A133
00
1.20
14
12
22
21
69
A134
00
4
11
22
25
62
A135
00
15
8
23
17
63
A136
00
14
7
25
22
68
A137
00
3.00
19
12
24
19
74
A138
00
2.50
14
11
24
21
70
A139
00
2.80
17
19
24
24
84
A140
01
2.80
10
4
20
19
53
A141
01
3.10
11
17
24
18
70
A142
00
2.30
25
22
24
20
91
A143
11
3.80
18
16
21
20
75
A144
00
3.20
17
20
24
21
82
A145
11
3.70
9
14
12
10
45
A146
00
1.80
14
18
21
25
78
A147
00
3.10
11
17
22
20
70
A155
00
2.30
21
22
25
24
92
A156
00
2.40
9
14
24
20
67
A157
11
1.80
15
10
18
15
58
2.30
Appendix C: Interview Guide for Phase 2
C.1
Overview
The interview will follow a guide designed to elicit the student’s understanding of
the rule, based on tasks from the previous instrument. New tasks will be presented to
the student to probe the limits of understanding. The students will be asked to reflect
on the questionnaire tasks, possibly by asking them for definitions of function and
derivative, domain-range issues in composition, the statement of the chain rule, and
a discussion of any relationships between the 20 problems that the student perceives.
The guide will be written following the piloting of the written part. This will also
then be piloted with the pilot subjects.
The interview will also address Leibniz’ rule as a way of gauging the robustness of
the student’s concept. A situation with no simple anti-derivative
will be presented,
Z
sin x
2
such as “compute F 0 if F is the function given by F (x) =
et dt.” If the need
0
to compute an anti-derivative is too great, a situation where the integrand has an
elementary anti-derivative will be presented to determine if the student can derive
Leibniz’ rule from the result. The data from these situations will be analyzed and
compared to that of Thomas (1995).
The interviews will be structured with four basic components, although each component will be modified to fit the student.
C.2
Review of function and composition
The student will be presented with a copy of her or his questionnaire from Phase
I and will be asked to discuss two function situations; one which he or she dealt with
successfully and one less successfully. The same will be done for the composition
questions. The goal will be to get a definition of function from the student. The
student will be allowed to ask about any of the first ten problems.
1. Here is a copy of your work from the questionnaire for question ??. Look it
over and describe how you solved the problem.
2. Would you change anything in that answer?
3. Repeat (1) and (2) for another function and two composition problems.
84
85
4. Questions 2–3 dealt with whether a situation can be a function. What is your
definition of function? Omit if the definition has been established.
5. Do you have any questions about the first 10 questions?
C.3
Classifying the 10 derivative problems
The student will be asked to group in any way that makes sense the ten derivative
problems. It is expected that he or she will use the elementary rules as a guide. This
may help determine if the student sees all five of the chain rule problems as a single
class.
6. Here is a copy of the last 10 problems on one sheet. The directions for all of
these were to compute the derivative of each of the following functions and to
show all of your work. I would like you to group these 10 problems in whatever
manner makes sense to you. You can have as many of them in a group as you
wish and you can form as many groups as you want.
7. Describe how each of the groups were determined.
8. Is there any other way that you considered grouping these? If so, describe it as
well.
9. If no answer given has disjoint sets: What if I required that your groups are
disjoint? Can you still group them?
C.4
Establishing the notion of chain rule
The student will be asked to state the chain rule. The questionnaire problems
will be used to get as general a statement as the student can agree with. The student
will be asked to discuss the proof (or need for proof) of the chain rule as well as any
intuitive notions of the rule that they have.
10. Write down the chain rule using whatever words or symbols that you like.
11. The last five questions were intended to be solved using the chain rule. Compare
each of those situations with what you have written. Does your statement deal
with each or can you modify it so that it does?
12. What is the key idea in the chain rule?
13. Do you recall how you learned the chain rule for yourself? Did you use mnemonic
devices or any sayings or tricks to keep the rule straight?
86
14. In mathematics, theorems need to be proved using more elementary definitions
and theorems. Does the chain rule need a proof? If so, what can you say about
the proof; if not, why not?
C.5
Extending the notion
The student will be asked to solve the Leibniz Rule problem from the RUMEC
study. A simplified version of the problem will be offered if needed to find a solution.
Finally, a set of word problems from a first semester calculus course will be presented.
The set will include a related rate problem and an implicitly defined function problem.
The task will be to identify any situations that require the use of the chain rule in
the solution, discussing in detail how the rule might be used. The student will not
be expected to solve the problems; however if a solution is required to complete the
task, one will be provided to the student.
15. Read each of the following differentiation situations carefully. Without actually
solving the problem, determine if the solution will involve the use of the chain
rule in some way. Describe how the chain rule is used.
(a) Given that the following relation defines y as a function of x,
√
√
x y + y x = 42
find its derivative.
(b) A ladder L feet long is leaning against a wall, but sliding away from the
wall at the rate of 4 ft/sec. Find a formula for the rate at which the top
of the ladder is moving down the wall.
(c) The volume V and the surface area of a sphere are related by the equation
36πV 2 = S 3 . Find dS/dV , the rate of change of the surface area with
√
respect to the volume when the volume is π 6 ft3 .
(d) The following formula give a reasonable estimate of the Heat Capacity C
of a certain element at very low Fahrenheit temperatures F :
C = 0.171F 3 + 236F 2 + 10, 800F + 16, 600, 000
Suppose that the temperature F is found at time t according to the equation F = −42t2 . Find an expression for rate of change of the Heat Capacity
with respect to time.
87
Z
sin x
2
16. Compute F if F is the function given by F (x) =
et dt. Freely offer the
0
fact that there is no Zelementary anti-derivative of the
integrand. One simpler
Z
x 2
sin x
situation is G(x) =
et dt. Another is H(x) =
4t2 dt. Have all three
0
0
on separate pages.
0
17. Finally, do you have any questions?
VITA
88
VITA
Jim Cottrill was born in Akron, Ohio on 1 February 1963, the son of Wayne and
Paulette. He was married to Julie Ann Vetter, daughter of Ron and Judy, on 15 June
1991 in Chillicothe, Ohio. They have a daughter Cecelia, born in 1996, and a son
Stephen, born in 1998.
Jim attended the University of Akron where he earned a Bachelor of Arts degree
in Secondary Education, in the fields of Mathematics and Chemistry, in 1986. He
taught mathematics, computer science and chemistry for the Akron Public Schools.
In 1992, he was accepted into the doctoral program at Purdue University. After
teaching at Georgia State University, Jim received his Doctor of Philosophy degree
in Mathematics Education from Purdue University in 1999.
He has co-authored three articles in the Journal of Mathematical Behavior : “Understanding the Limit Concept: Beginning with a Coordinated Process Schema”
(1996); “The Development of Students’ Graphical Understanding of the Derivative”
(1997); and “Constructing a Schema: The Case of the Chain Rule” (1997). His research interests are in undergraduate mathematics education and in the training of
pre-service teachers.
Jim has accepted a position as Assistant Professor in the Department of Mathematics at Illinois State University in Normal, Illinois beginning August 1999.