Chapter I: Introduction to The Derivative
(Draft Version 91405)
© 2005 M. Flashman
How do we come to understand Mathematics:
The discussion of how we can best arrive at mathematical understanding
has certainly been around for thousands of years. Current discussions of
pedagogy (the art of instruction) focus attention on the learner developing
understanding of concepts and techniques through a mixture of investigations,
guided experiences, and problem solving. When possible we will try to build
an understanding of calculus concepts on your prior physical and mathematical
experiences and familiarity with aspects of common phenomena. 
 Preface: In this chapter we begin exploring one
of the key concepts of the calculus, called the derivative. Though
the definition of the derivative in a mathematical setting is primarily
a matter of numbers and functions, the derivative is important
for its use in various applications that employ the numerical and
symbolic power of mathematics to study change. These applications have
become important interpretations of the derivative concept, while
providing powerful analytic tools for the construction of mathematical
models. The interpretations suggest and confirm the truth of mathematical
results when the mathematics fits well with the experience of the interpretation.
This symbiotic relation of mathematical abstractions with "real world"
interpretations may be a reason for the unusual success mathematics has
had in assisting scientific studies in so many disciplines. We start with
some key motivation for the derivative concept. 
In Section A we examine the geometric problem of describing lines
that are tangent to a curve. This problem should be familiar to
you already from Chapter 0.B3, though previously we considered it using
only algebra.
Next we'll turn our attention in Section B to the physics problem
of describing the velocity of an object moving along a straight
line. These two problems have been key models in the historical development
of the calculus.
In Section C, we'll
consider models from probability and economics (also considered
in Chapter 0) which in the 19^{ th} and 20 ^{th} centuries
have provided more applications for the calculus outside of the traditional
physical and natural sciences. Further examples and applications will be
found in the exercises illustrating the importance of the study of
change.With these motivating interpretations it should become apparent
that there is a common mathematical concept used to analyze each
of these models.
The mathematical abstraction
of that common concept leads to the definition of the derivative in
Section D. This concept has two key aspects, as a number and also as
a function . The definition, its initial application to "finding the
derivative," and other basic concepts in the study of the derivative will
round out this chapter. Procedures for finding the derivative of function
based primarily on the function's algebraic  symbolic character provide
the first example of a calculus, a calculus for finding derivatives.
What lies beyond
this chapter: More details of the calculus of derivatives will be covered
in Chapter II. We'll return to an examination of some traditional applications
of the derivative calculus in Chapter III. In Chapter IV we will reverse
our investigations and examine how information about a derivative controls
the possibilities for related "primitive" functions. Information about
the derivative is usually presented in the form of derivative (or differential)
equations. Trying to solve these equations provides a sensible transition
to the second key concept of the calculus, the integral, which,
like the derivative, is also both a number and a function. But for now
we can put this second concept aside and proceed to discuss motivating
interpretations for the derivative.
I. A. [Motivation] Finding the Slope of a Tangent Line
The Tangent Problem
Again: We are interested in finding the slope of a line that
is tangent to a given curve at a specific point on that curve. In
Chapter 0.B3 we considered the curve with equation Y = X^{
2} and used the geometric characterization that a tangent
line will meet the curve at only one point to investigate this problem.
Algebra for the equations of the curve and the tangent line led us to conclude
there was only one possible slope for the tangent line. our work showed
that the slope depended only on the first coordinate of the point. In fact,
for the point on the curve with coordinates (a,a^{2 })
we found that the slope of the tangent line had to be 2a. In this
section we will follow a different approach to find the slope of the tangent
line. Instead of finding the slope using algebra alone, we will begin by
estimating the value of this slope, combining algebra with our common
sense of what it means to be close.
X

Y

Slope of Secant =

0

0

3

1

1

4

2

4

5

4

16

7

5

25

8

6

36

9

Estimating (the slope
of) the tangent line: Our estimates will depend on examining lines
that are close to the tangent line. One difficulty in our analysis
of the tangent problem is that we have not said precisely what a tangent
line is. The word "close" is also imprecise. We will continue our investigations
without resolving these issues here. The more experience one has investigating
a concept, the more clear its meaning can become. Articulation of
a careful definition for "tangent" will make more sense later, based on
greater experience .
Example I.A.1. Estimation
and analysis for Y=X^{2}: Let's reconsider the question treated
in Chapter 0.B.3 of finding a line tangent to the graph of Y=X^{ 2}
at the point (3,9). We call a line that passes through two (or more)
distinct points on a curve a secant line of the curve. As Figure
1 illustrates, there are many secant lines to the graph of Y = X^{
2} that pass through the point (3,9). Each of these secant lines is
determined by the first coordinate of a point on the graph of Y = X^{
2} other than (3,9). So using the first coordinates 0, 1, 2, 4, 5,
and 6 we have different secant lines determined by the points (0,0), (1,1),
(2,4), (4,16), (5,25), and (6,36). A glance again at Figure 1
should help in understanding that the lines determined by (2,4)
and (4,16) are better approximations for the tangent line than the
other lines in the figure.
Determining the slopes
of secant lines provides a numerical approach to finding the slope
of the tangent line (previously discovered to be 6 in Example 0.B.***).
In Table 1 we give these points and the slopes of the secant lines they
determine with the point (3,9) using the formula .
This table also helps in understanding that the slopes determined by
using (2,4) and (4,16) are better approximations for the slope of
the tangent line than the other slopes in the table.
It seems likely from
both the visual and numerical views that to obtain even better estimates
of the tangent line and its slope we should investigate secant lines determined
by points (x, x^{2}) with the first coordinate x
closer to 3. The graphing at this stage can become a little awkward because
of scaling. When x is very close to 3 the secant line on most
graphing devices is almost indistinguishable from the graph of the function.
For this reason we'll continue the analysis numerically. (Exercises pursuing
the graphical idea of closeness are better explored further individually
with some kind of graphing utility. See the exercises at the end of this
section.)
x

x^{2}


2.9

8.41

5.9

2.99

8.9401

5.99

2.999

8.994001

5.999

3.1

9.61

6.1

3.01

9.0601

6.01

3.001

9.006001

6.001

Notation: The
slope of the secant line determined by the number x will be denoted
here with the symbol .
Table *** displays more
slopes of secant lines for Example I.A.1, taken with first coordinates
much closer to 3. The pattern in this chart should be clear. The slope
of each secant line is exactly 3 more than the number x. This pattern
is not by chance. The slope of a secant line determined by the points (x,
x^{ 2}) and (3,9) where x is not equal to 3
is determined algebraically :
Thus, when x is
not equal to 3, the slope of the secant line determined by x
is x+3. See Figure 2. When x is close to 3, the
slope of the corresponding secant line, x + 3, will be close to
the slope of the tangent line. But in this situation when x is close
to 3, the number x + 3 is certainly close to 6. Therefore, we can
conclude that the slope of the tangent line at (3,9) must be 6.
Comment: At this
stage there are two ways to think of the use of the number represented
by "x." First, consider a number, x, as close to (but
not equal to) 3 as you can imagine. For this x, the slope of
the secant line is =
x+3, which is very close to 6. We'll describe this way of thinking
of the estimation as a static viewpoint, because the number x
is considered fixed.
A second way to think
of this situation is that we are progressively choosing values for x
closer and closer to (but never equal to) 3. For these x's,
the slopes of the corresponding secant lines, ,
which could be computed individually using the expression x + 3,
are closer and closer to the number 6. We'll describe this way of thinking
of the estimation as a dynamic viewpoint because the number x
is considered changing, varying close to 3.
Notation: There
are some important notation conventions and linguistic expressions that
have developed over hundreds of years to represent the kind of analysis
just made. We discuss now two ways that will be used throughout the text
and are common in mathematics today.
First, it is common
to write x .
3 to represent a statement that x is close (is approximately
equal) to 3. In using this notation we also will allow x to
equal 3 and will state explicitly when we exclude that possibility. With
this notation we can express the static viewpoint of our analysis. We supposed
x.3
with x…3
and found the slope of the corresponding secant line, .6.
A second very important
notation represents either the static or dynamic viewpoint of estimation
with an arrow. With this notation we write x 6
3 for the statement that x is close to 3, or that we
are considering x taken closer and closer to 3, or that
x approaches 3. It is usually presumed that x …
3, but at times it is convenient to allow this notation to include the
possibility that the number x may equal 3.
We can express either
the static or the dynamic viewpoint of our tangent line analysis with this
notation. We supposed that x6
3 with x…
3 and algebraically simplified the slope(s) of the corresponding secant
line(s), making it easier to see that the slope(s), 6
6 .
Include
a picture of the letter
in a car approaching a wall labelled 6, with the letter x on
a clock hand approaching the number 3.
 
Review of Analysis:
We have found a method for approximating the slope of the desired tangent
line. We analyzed the slopes of secant lines, ,
which were controlled by the choice of numbers for x, the first
coordinate of another point on the graph of Y = X^{ 2}. When x 6
3, with x…
3, by simplifying algebraically we found that 6
6. Since when x6
3 the slopes of secant lines are better estimates for the slope of the
tangent line, we concluded the slope of the tangent line must be 6.
Example I.A.2: Generalization
for Y = X^{ 2}. We continue our estimation analysis to find
the slope of the line tangent to the graph of Y = X^{ 2} at a generic
point with coordinates (a, a^{ 2}). We consider the
slope of a secant line formed by this point and the point (x,x^{
2}) which we again denote .
See Figure 5.
Generalization:
Frequently a single example has enough of the basic idea of a solution
to a problem that it can be used as a template to organize a more general
analysis, either as a computation or an argument. When the example has
dealt with a specific number, say 3, and the more general analysis replaces
that specific number with a general number represented symbolically,
for instance using the letter a, we say we have generalized the
example from the specific number, 3, to any real number, a.

Using algebra we simplify
(since x is not equal to a) to find .
When x6
a, x…
a, it should make sense now that x + a 6
2aand thus 6
2a.
Therefore, since when
x6
a the slopes of the secant lines improve as estimates for the slope
of the tangent line, the slope of the tangent line at the point with coordinates
(a, a^{ 2}) must be 2a.
Comment: The
technique of estimation has become one of the most effective tools for
organizing the study of the tangent problem. It is not clear when the precise
exposition of the method was first made, but it is certain that the concepts
used in this technique were being developed by many individuals prior to
the work of Newton and Leibniz. Perhaps it was the blending of these ideas
with algebra that began to yield results in a more systematic fashion,
raising expectations in those days of achieving a more systematic solution
to many other problems as well.
The next few examples
illustrate some of the power and ease that the interaction of algebra with
the analysis of estimation brings to solving tangent problems.
Example I.A.3.
Find an equation for the line tangent to the graph of the equation
Y = 3X^{ 2}  X + 1 at the point (1,3).
Solution: We
find the slope of the required line using the estimation method of the
last examples. Let
denote the slope of a secant line for the curve that passes through the
point (1, 3) and the point on the graph with coordinates (x, 3x^{
2}  x + 1). Simple algebra now shows that when x …
1, Thus when x61
with x…1, ,
so the slope of the tangent line must be 5. Using the point slope
formula we conclude by finding an equation for the tangent line: Y =
5 (X  1) + 3.
Example I.A.4.
Find the slope of the line tangent to the graph of Y = X^{ 3} at
the point (a,a^{ 3}).
Solution: We
find the slope of the required line again using the estimation method.
Let denote
the slope of a secant line for the curve that passes through the point
(a,a^{ 3}) and the point (x, x^{
3}). [See Figure 1.7.] Simple algebra now shows that when x…a,
Thus when x6a
with x…a, ,
so the slope of the tangent line must be 3a^{ 2}.
Comment: Notice
that the slope in this last example is a nonnegative number for any a.
Based on the sketch of the graph of Y = X^{ 3}, can you explain
why this is so?
Example I.A.5. Find
the slope of the line tangent to the graph of Y=1/X at the point (a,1/a)
where a…0.
Solution: Again
we use the method of estimation of the last examples. Let
denote the slope of a secant line for the curve that passes through the
point (a,1/a) and the point (x,1/x). [See Figure
8.] Simple algebra now shows when x …
a and x and a…
0 that
.
Thus when x6
a with x…
a and x…0, ,
so the slope of the tangent line must be .
Comment: Notice
that the slope in this last example is always a negative number
for any a. Based on the sketch of the graph of Y = 1/X, can you
explain why this is so?
Compare the last two
examples. Describe the relationship of the sign of the slope of the tangent
line and the graph of the equation?
Exercises I.A.
For the equations in problems 1 through 7, use estimation analysis to find
the slope and an equation of the line tangent to the graph of the equation
at the indicated point.
1.Y = X^{ 2}  X ; a) (1,0)
b) (1,2) c) (a,
a^{ 2} a).
2.Y = 2X^{ 2}  3X + 1 ; a) (1,0)
b) (1, 6) c) (s, 2s^{
2}  3s +1).
3. Y = X^{ 3}  X;
a) (1,0)
b) (1,0) c) (a,
a^{ 3}  a).
4. Y = X^{ 3}  X^{
2}; a) (1,0)
b) (1, 2) c) (t, t^{ 3}
 t^{ 2}).
5. Y = X^{ 4};
a)
(1,1)
b) (2,16) c) (a,a^{
4}).
6. Y = 1/(X^{ 2})
X…0 ;
a) (2,.25) b) (a, 1/a^{
2}).
7.
X>0 ; a) (9,3)
b) (a,)
8. For each of the following, use
your calculator to estimate the slope of the line tangent to the graph
of this equation at (0,0) with secant lines when x = .2,.1,.05, .2, .1,
and .05. Discuss briefly what you think the slope of the tangent might
be.
1. Y = sin(X).
2. Y = tan(X).
3. Y =
9. Use a graphing utility to sketch graphs
of Y = X^{ 2} with the secant lines determined by the point (3,9)
with x = 2.9, 2.99, 2.999, 3.1, 3.01, and 3.001. Each of these should use
a different viewing window to allow the two points to be visible easily.
Comment on the relation of the curve to the secant lines.
10. For each of the following equations
use your calculator to estimate the slope of the tangent line at (0,1)
using a secant line when x=.001.
a) Y = 2^{ x}
b) Y = 3^{ x} c) Y =
(.5)^{x} d) Y = (1/3)^{x}.
Tangents constructed
from given points not on a curve. One of the traditional problems for
tangents and circles is to construct tangent lines to a circle through
a point given outside of the circle. See Figure 7. The construction from
the point P to the circle with center O finds the midpoint M of the segment
and then constructs the circle with radius
and center M. The points where the two circles meet are T and S, and the
lines TM and SM are both tangent to the circle.
The next four problems
ask you to solve comparable problems for some other curves.
11. Use the result of Example
I.A.2 to find equations for any lines that are tangent to the graph of
Y = X^{ 2} and pass through the point (3,5).
12. Use the result of Example
I.A.4 find equations for any lines that are tangent to the graph of Y =
X^{ 3} and pass through the point (2,1).
13. Use the result of Example
I.A.5 to find equations for any lines that are tangent to the graph
of Y = 1/X and pass through a) (1,3/4) ;
b) (1,1/2).
14. Find the Xintercept of the
line that is tangent to the graph of Y = X^{ 2}  2 and passes
through the point (2,2).
Normals to Curves: Exercises
that explore some properties of normal lines.
15. Normal to the
parabola Y=x^2 at (a a^2) meets normal at (x, x^2) at (0,?) and distance
from this point to (a,a^2) is ?.
16. Normal to hyperbola Y=1/x
17. Tangents and the vertex of
a parabola. Show(?) that the slope of the tangent at the vertex of a parabola
deteremined by a quadratic function is 0. Use this to find the vertex of
a couple particular parabolas, then show the vertex of Y= Ax^2 + Bx +C
has first coordinate x=B/2A.
By the time of Archimedes
and Apollonius (dates), the Greeks geometers were familiar with the properties
of parabolas and their tangents. In the following exercises we investigate
one of these relations known to the Greeks and discussed in our preliminary
work on parabolas and tangents in section ***.
18. Use Y=x^2.
1. Show slope of tangent
is m at the point x= m/2.
2. Find slope of secant line between
(a,a^2) and (b,b^2) [aºb].
3. Show that the the slope of the
tangent to Y=x^2 is parallel to the secant line between (a,a^2) and (b,b^2)
[ aºb] at the
midpoint between a and b.
19. Use Y=x^2 + 6x.
1. Show slope of tangent
is m at the point x= (m6)/2.
2. Find
slope of secant line between (a,a^2 +6a) and (b,b^2+6b) [aºb].
3. Show
that the the slope of the tangent to Y=x^2 +6x is parallel to the secant
line between (a,a^2 +6a) and (b,b^2+6b) [ aºb]
at the midpoint between a and b.
20. Use
Y=Ax^2+Bx+C
1. Show
slope of tangent is m at the point x= (mB)/2A.
2. Find
slope of secant line between (a,Aa^2+Ba+C) and (b,Ab^2+Bb+C) [aºb].
3. Show
that the the slope of the tangent to Y=Ax^2+Bx+C is parallel to the secant
line between (a,Aa^2+Ba+C) and (b,Ab^2+Bb+C) [ aºb]
at the midpoint between a and b.
21. Investigating
the relation of secants to tangents with Y=1/X; Y= x^3.
22. Using
symmetrically placed points to estimate the slope of a tangent line:with
Y=1/x and y=x^3. How about [sin(a+h) sin(ah)]/2h=sin(h)cos(a)/h . Estimate
when h >0?
23. On
the graph below (Figure 9), estimate the slope of the tangent lines to
the given curve at the indicated points.
24. Suppose
x63, for each of the
following expressions, give a number that the expression approaches. Give
some numerical evidence for your conclusion.
Here are some samples
from sections of the graph of some functions with the size of the viewing
window. Based on this information, estimate the slope of the tangent line
for each of these functions at the indicated point on the graph.