Math 1106 – Elementary Applied Calculus
Hoffmann and Bradley’s Calculus for Business, Economics, and the Social and Life Sciences
Lecture Notes for Chapter 2, “Differentiation: Basic Concepts”
Notes after Class Meeting #3, Wednesday, August 22, 2007
After the foundation has been laid, with the concept of the definition of the limit of a function, we now move ahead into the goal of calculus: being able to measure the slope at any spot (i.e., for any value of its input variable) on a function. This is the promised “rate of change” that calculus concerns itself with – at least the first part of calculus, which is called “differential calculus.” It’s the meat in Section 2.1 of the text.
Recall how I reminded you on the first day of class, that you already knew something about “rate of change” through all your study of linear functions. The concept of slope for linear functions is easy to understand. And, for a linear function (unlike one whose graph snakes up and down sinuously), the slope is the same no matter where on the graph you choose (i.e., no matter what the value is for x). So we use that as our starting point for other forms of functions.
For any function, suppose that we start with the function f ( x )
=
16 x
2
. That’s clearly a parabola with vertex at the origin and opening up. Let’s choose a value for x: x = 1. Let’s develop a measure of how steep that function’s graph is at the point
( x , f ( x ) 1 , . Ideally, we want to imagine the tangent line at that point; that would be a line that just “kisses” the curve at that point.
(There is just one such line; any other line through that point cuts across the curve to the other side. “Secant” and “tangent” come from the Latin words for “to cut” and “to touch,” respectively.)
Now imagine a second point whose x-coordinate is a little bit more (or less, if you like) than 1. I won’t specify the exact value of what I mean by “a little bit more,” but I’ll just represent that amount by the letter h. That means I can refer to another point on the graph of the parabola whose coordinates are
(
1
+ h , 16 ( 1
+ h )
2
)
. The straight line that contains both of those points is called a secant line, because it’s easy to see that it’s not a tangent line at the first point we’re working with. But nonetheless, let’s measure the slope of that secant line. You remember how to do that: if you know the coordinates of two points on a line, you can determine it’s slope by developing the ratio we’ve come to call “rise over run.” But write the ratio this way and simplify: f ( 1
( 1
+
+ h ) h )
− f
−
1
( 1 )
= f ( 1
+ h h
)
− f ( 1 )
=
16 ( 1
+ h ) h
2 −
16
=
16 ( 1
+
2 h
+ h h
2
)
−
16
=
32 h
+ h
16 h
2
= h ( 32
+ h
16 h )
In this context, “simplify” means to fiddle with the algebra in order to find a way to eliminate the
h in the denominator, which I did in the last step above. What results is an expression that represents the slope of that secant line ... the one that “cuts” across the curve at (1,16). Its slope is evidently a little more or less than 32, depending on the exact value of h. Now, here’s the key point: If you could reduce the size of h to zero, you would in effect have the needed tangent line; but that would mean you had a tangent line whose tangent is (drum roll!) equal to 32 (since then
16h would be equal to zero because h is equal to zero).
All of that above was spelled out for a specific point on the graph of the function, together with the variable h. Now we can generalize it by simply letting the point of interest have the general coordinates of
( x , f ( x )
)
, which means the “helper” point that has a first coordinate h more than x
=
32
+
16 h

has the coordinates of
( x
+ h , f ( x
+ h )
)
. Then the string above in the previous paragraph starts off with this expression for the “rise over the run”: f ( x
+ h ) h
− f ( x )
=
16 ( x
+ h )
2 h
−
16 x
2
.
The expression on the left is call “the difference quotient” ... because it’s a quotient, and its numerator is the difference of two y-values (and its denominator is the difference of two xvalues).
I’ll let you “simplify” it the rest of the way (hint: look at Example 2.1.3 on page 102!). Don’t forget, the goal is to “get the h out of the denominator.” And the reason for that is because, ideally, we want to “take it to the limit.”
That is, we would dearly love to evaluate that with h = 0, but as it stands, that would make the denominator equal to zero. But if we algebraically twist things, we can find a way to “cancel” out that h in the denominator, and then it’s easy to substitute h = 0. If you do it right, the result comes out to be 32 x 16 h .
When we evaluate, therefore, the limit h lim
→
0 f ( x
+ h )
− h f ( x )
= h lim
→
0
16 ( x
+ h ) h
2 get the correct answer if we just evaluate the limit for the equivalent expression
−
16 x
2 lim h
→
0
32 x then we’ll
+
16 h .
But that’s easy: we just “take it to the limit” and substitute h = 0 into that to obtain 32x – we “got the h out” of the denominator, so making h equal 0 produces no division by zero.
This final result, the limit of the difference quotient for the specific function f ( x )
=
16 x
2
has yielded a new function called “the derivative of f.” We’ll normally name such a function after the function from which it was derived. Thus, this derivative function is f
′
( x )
=
32 x (pay heed to the use of the quote-mark to modify the name of the original function).
All of this uses a specific function. But when used generically, this shows the development of
“the Derivative of a Function,” as defined at the bottom of page 101 of the text. Since the original function is not a straight line (linear) function, its slope naturally varies as x changes.
Notice that the derived function also varies as x changes. It is the derivative that reports on the steepness (slope) of the original function.
The difference quotient by itself measures the average rate of change between two points on the given curve. When it’s taken to the limit, it produces what’s called the instantaneous rate of change at the particular point we specify. Pay heed to the definitions at the bottom of page 102 that communicate this. Also, whether the slope is positive or negative at a certain point is directly related to whether the function is increasing (going up) or decreasing (coming down) at that point, respectively.
I ended the meeting by listing the exercises at the end of Section 2.1 (beginning on page 108) that will correspond to the items in MathZone for this section: 3, 11, 13, 16, 19, 25, and 37.
See you in class next Monday!

Notes after Class Meeting #4, Monday, August 27, 2007
In today’s meeting, I used the limit of the difference quotient to derive the derivative function for a rational function (something like f ( x )
= x
4
2
). Then I went over the first few short-cut rules for differentiation involving certain types of functions: constant functions, power functions, constant multiples of functions, and the sum of functions. These are pretty easy to tuck away in the memory and use – a good deal easier than having to always haul out the difference quotient to do the work. Keep listening, more such rules are yet to come.
I also went over the various forms of notation for the derivative of a function. If the function is represented by the equation y
= f (x ) , then its derivative can be denoted by any of these forms of notation: y
′ = f
′
( x )
= dy dx
= d dx f ( x ) . No need to repeat those rules here; they’re clearly spelled out in Section 2.2 of the textbook.
A well-known application of the notion of the derivative is the relationship between distancetraveled, speed (velocity), and acceleration. If s (t ) represents how far an object moves for some given t units of time, then the derivative of that function, s
′
(t ) represents the velocity with respect to those same time units. And velocity is such an important concept that its function might be called v (t ) , but you know now that v ( t )
= s
′
( t ) . Similarly, acceleration is the rate at which velocity changes, so if we name the acceleration function a (t ) then that means a ( t )
= v
′
( t ) .
If you have any frustration in your use of MathZone for the homework assignments, and you decide to make use of the “Ask My Instructor” link, be sure to spell out the details of the exercise you’re writing about. MathZone does a poor job of presenting any details, and so it’s best for you to include screen snapshots, especially if you make a desperate plea that, “My answer was right, but MathZone said it wasn’t even though my answer looks just like the MathZone answer.”
Remember, I come from a long line of doubting Thomases, and I’ll need some visual evidence to be able to respond.
Notes after Class Meeting #5, Wednesday, August 29, 2007
Don’t forget: next Monday is Labor Day. I won’t be there in the classroom, so you don’t show up either!
Today two additional “rules for differentiation” were revealed from Section 2.3 of the text: the
Product Rule and the Quotient Rule, for any function that fits those descriptions.
It would be tempting to model those two rules along the simple, direct lines of the rule for sums of functions. (Remember, that one says, essentially, that “the derivative of a sum of functions is the sum of their individual derivatives.” This is particularly nice when differentiating polynomial-like functions.) Alas, the Produce and Quotient Rules are a little more bizarre in their makeup.
The Product Rule has a nice symmetry to it, and because of the “commutativity of addition” it’ll be hard for you to get it wrong: d dx
( f ( x ) g ( x )
)
= f ( x ) d dx g ( x )
+ g ( x ) d dx f ( x ) . (There’s that

Leibnizian notation for the derivative. Using functional notation for the derivative, that is written like this:
( f ( x ) g ( x )
) ′
= f ( x ) g
′
( x )
+ g ( x ) f
′
( x ) . I call that a “mix and match” kind of rule, for you have to use both original functions as well as both of their derivatives in order to build the derivative of the product function. The derivative of a product ends up being the sum of two products. A very good example that you should read and strive to understand is Example 2.3.3 on page 127 of the textbook. It does a good job of first showing a natural occasion where a function is developed (the total revenue) that is the product of two other functions (the total production of items times the revenue per item).
The Quotient Rule is a different kind of animal, potentially a snake in the grass waiting to bite you if you don’t pay attention to the exact layout of the rule. This is because of the presence of the minus sign, which then restricts your ability to juggle the pieces since subtraction is not commutative. Of course, the Quotient Rule is used for the so-called “rational” functions, the ones that are seen to be the ratio of two other functions. Here’s the rule, expressed with functional notation, and using suggestive names for the top (Numerator) function and the bottom
(Denominator) function:


N ( x )
D ( x )


′
=
D ( x ) N
′
( x )
D (
− x )
N
[ ]
2
( x ) D
′
( x )
.
We spent quality time in class working through some MathZone questions. Please, any time you have some encounter with MathZone that you want to bring up in class, be sure to have specific information about which item to look at, so that we don’t have to waste precious time hunting around looking for your particular example. Always note the section and the exercise number and bring that with you to class in your notebook.
Next time (the Wednesday after Labor Day) we’ll continue with the material in Section 2.3
(second derivatives, and higher derivatives). Then, it’ll only be another week before the first test!
No fear! I bet you’ll ace it!
Notes after Class Meeting #6, Wednesday, September 5, 2007
Okay, everybody seems to have had a pleasant Labor Day holiday. Except for those students who were mysteriously absent today: I hope they’re not in the slammer for speeding on the Interstate or some other holiday-related misdeeds! ;-)
I finished up talking about Section 2.3, about the topic of second-derivatives (the derivative of a derivative) and higher derivatives. The classic example of a second derivative is acceleration, which is the first derivative of the velocity function, which itself is the first derivative of the position function; thus, acceleration is the second derivative of the position function. Later, in
Chapter 3, we’ll see how the second derivative plays a very important and illuminating role in understanding the shape of the graph of a function.
The rest of the class period was given over to answering questions arising, chiefly, from
MathZone. I want you to become fully comfortable with this tool, and the only way to do so is through using it. Many of you have gotten on board the MathZone train. But others have yet to even get to the station! That is very troubling to me, because the first set of homework (all the assignments up through Section 2.3) must be submitted by 8 a.m. next Wednesday, September

12! No late homework will be accepted! This will comprise approximately 1/3 of your homework grade in the course, so you don’t want to miss out on that.
By the time you read this, the practice test will have been uploaded to my website at http://science.kennesaw.edu/~bthomas . Before next Monday’s class (which will be devoted exclusively to reviewing and preparing for Wednesday’s test), be sure to take that practice test in order to get familiar with what you’ll be facing on Wednesday. And if you have ANY questions, bring them to the review on Monday, for your benefit as well the benefit of everyone else!
Do you feel a need for further tutoring or help with your algebra skills. Some students wait until the end of the semester to start lamenting their failure to get that squared away. How sad! And even sadder, they go and hire some tutor at an exhorbitant rate and complain about how it didn’t seem to help. Well, if that sort of describes what you’ve been thinking about, let me direct your attention to Aleks. Look in the front of your book, and there you’ll find a page that has information about Aleks, which is an online tutoring facility that has won rave reviews for the way it can assist people to reinforce their algebra skills. Kennesaw has worked a deal that, for anyone who chooses to go with Aleks, gives them 6 weeks of free use of it. Further, we have tailored the Aleks course so that it will only address the important aspects of Algebra that we feel are of most use in Math 1106. So, if you want to avail yourself of Aleks, then contact me via email ( bthomas@kennesaw.edu
) and I’ll share with you the special course-code for the KSUtailored version of Aleks.
And while I’m talking about free tutoring, have you tried out the free facilities within MathZone that allow you to address a question to a real human being and get some excellent feedback? At the top of the MathZone screen, you’ll see a button marked “Online Tutor” – try it and let me know about your experiences.
See you Monday at the review.
Notes after Class Meeting #7, Monday, September 10, 2007
Notes after Class Meeting #8, Wednesday, September 12, 2007
Notes after Class Meeting #9, Monday, September 17, 2007
Well, when I wrote that cheery “See you Monday at the review” I knew it was probably false. I just didn’t know quite how false it would be. As of now, I hope to be back with you on Monday,
September 24, if a week’s recovery back home will get my body in sufficient shape after quadruple coronary bypass surgery!
I hope you had a good test review that Monday – it depended on your presence and participation, and your having worked on the practice test ahead of time. Did you? If you didn’t do those things, I hope by now you’re kicking yourself, for the actual test was not markedly different from the practice test. I have not heard of the test results yet, but I hope that the average was in the high 80’s. [Mr. Kathiresan has now provided me the test scores: average score was 78.
Hmmmmmmmm. Someone must not have believed me when I said the practice test was a good study guide!]
Don’t go slitting your wrist if you got a bad score (remember from the syllabus how you can compensate for your lowest test grade). But don’t slack off if you got a really good score. We’re just beginning the semester’s work, and challenges await you down the line.

Today Mr. Kathiresan will have handed back the tests to you. Then he should have covered the most important rule for differentiation of all: the Chain Rule. Implicit in understanding this rule is the need to recognize when it’s needed: for composite functions. I sure wish they’d name those “chain functions” instead so that there would be an ironclad link to the name of the chain rule.
If you’re not keeping pace with your MathZone homework, then this stuff just gets to feeling harder and harder. But if you keep up your exercises, piece of cake! For instance, I’m spending this week concentrating on my physical and breathing exercises after my heart surgery so that I can hope to be back in the classroom with you next week. But if I didn’t do that, imagine how hard it would be for me to just walk across campus to get to the Burruss Building, much less put up with a bunch of slackers for half of an afternoon! ;-) Just kiddin’.
Folks, this stuff just gets harder the more you procrastinate and don’t do your homework in a timely manner. Don’t say I didn’t warn you when the end of the semester arrives and ... well, you get the idea.
The Chain Rule is the most important of the rules of differentiation. It gives a way to derive the slope function from functions that the other rules don’t adequately address by themselves. But to use it, you must be able to discern when a given function is composed of (or “built up from”) a chain of other functions.
For example: the function f ( x )
=
( x
2 −
3 x
+
7 )
14
is composed of the chaining together of two functions. We could actually get its derivative by “expanding” it out to the 14 th
power, but that would be cruel and tedious work. Rather, notice that function f(x) is composed of (i.e., is the
“chaining together” of) the two functions together is represented by writing f ( x )
= g ( x )
= x
14
and g h ( x )
= g ( h ( h ( x x
))
)
= x
2 −
3 x
+
7 . This chaining
. This means that the input variable (represented by x) is first acted upon by the function named h to produce the intermediate result h(x); then this result is used as input to the function g, which raises the intermediate result to the 14th power. In other words, f ( x )
= g h ( x )
= g ( h ( x ))
=
( h ( x ))
14 =
( x
2 −
3 x
+
7 )
14
.
Think of the symbol " " as a single link in a chain, indicating the linking of the two functions g and h. (Be sure that you notice that the order in which the linking is done is critically important.
In most cases, and indeed in this case, the function obtained through the linking g h is not the same as the one obtained by linking h g . Try it, and see what the opposite chaining of h and g produces! In other words, function chaining, a.k.a. “function composition”, is not commutative.).
Once you’re adept at decomposing a function into its constituent parts (the building-block functions that link together to obtain the given function), then the chain rule gives a very quick way to get the derivative, provided you can differentiate the two linking functions.
The chain rule says when f ( x )
= g h ( x )
= g ( h ( x )) , then f
′
( x )
= g
′
( h ( x )) * h
′
( x ) . Another way to think of this chaining is to regard the function h as being the “inside” function and the function g as the “outside” function. In this way, we “peel the onion” differentiating as we go: get the derivative of the outer function (and keep the inside function “inside”), peel off the outer function, and then multiply what you’ve got so far by the derivative of the remaining inside function.

So, let’s get the derivative of the function given above ( f ( x )
=
( x
2 −
3 x
+
7 )
14
), where g ( x )
= x
14
is the outside function and derivatives of the two functions g and h: h ( x )
= x g
′
( x )
=
2 −
3
14 x
13 x
+
7 is the inside function. First, get the
and h
′
( x )
=
2 x
−
3 . Then f
′
( x )
= g
′
( h ( x )) * h
′
( x ) which is equal to g
′
( x
2 −
3 x
+
7 ) * ( 2 x
−
3 ) and then (by the definition of g
′
( x )
=
14 x
13
) this becomes 14 ( x
2 −
3 x
+
7 )
13
( 2 x
−
3 ) .
Warning! You can skip the next few paragraphs if you don’t like “shaggy dog” stories.
Now, this chain rule’s powerful and orderly application can be exploited with more complicated functions, functions that are built up by chaining together 3 or more functions that are its “links.”
Here’s where the analogy to peeling an onion works, as we see the various layers getting differentiated and peeled off. Unfortunately, concrete examples of this are hard to come by yet, and will have to wait until later. But working symbolically, suppose we have a function that is the chaining together of 4 functions, like this: f ( x )
= g h m r ( x )
= g ( h ( m ( r ( x )))) .
Then the derivative of the function f is gotten by successively applying the chain rule, and the
“peeling of the onion” can be seen here: f
′
( x )
= g
′
( h ( m ( r ( x ))))
⋅ h
′
( m ( r ( x )))
⋅ m
′
( r ( x ))
⋅ r
′
( x ).
Are your eyes watering yet? ;-) A “real” example would perhaps be more enlightening. How about this little 3-link function: f ( x )
= 3
9 x
−
( x
2 + 5
7 x
2 +
3 ) . Now, you don’t have to get all dogmatic about it, and do a thorough job delineating the specific functions that are chained together, but I trust you can see that buried at the lowest level is 7 x
2 +
3 . That’s the innermost level, but the chain rule wants us to start at the outermost level: the cube root function.
I’ll start by restating the function in terms of fractional powers, since that facilitates the use of the
Power Rule for differentiation: f ( x )
=


9 x
− x
(
2 +
(
7 x
2 +
3
)
1
5
)
1
2


1
3
. Differentiating that is like “peeling the onion.” f
′
( x )
=
1
3



9 x
− x
(
2 +
(
7 x
2 +
3
)
1
5
1
2
)



−
2
3
*

 9
−
1
2 x
(
2 +
(
7 x
2 +
3
)
1
5
)
−
1
2
*



2 x
+
1
5
(
7 x
2 +
3
−
)
4
5 * 14 x





The “evil nun” would probably make me “simplify” that so there are no fractional exponents, etc., but as far as I’m concerned, it’s a legitimate representation of the derivative. Notice how the very last thing that was developed, and multiplied with, was the derivative (14x) of that lowest level thing ( 7 x
2 +
3 ) I’d earlier identified. Good thing I wasn’t asked to get the second derivative of f though: because then I’d have to be using the product rule with three factors!
Okay, here’s the answer for the person asking that “obvious” question (“Do we have to know how to do this?”): “No. It’s all about appreciating what lies beneath the surface of calculus. I

don’t expect you to be able to compose like Beethoven, but I do expect you to be able to appreciate his music – maybe even be able to identify his style when you hear it in an elevator.”
End of “shaggy dog” story.
Okay, on Wednesday Mr. Kathiresan will be covering the topic of “marginal analysis” and you business and accounting majors should be in hog heaven! This is important stuff, and you can be sure it’ll be on the next test, so read up on it before going to class on Wednesday.
Notes after Class Meeting #11, Monday, September 24, 2007
Okay, I’m back. Sorry I missed a few of you who chose to not come to class today.
I’ll be handing back your completed tests on Wednesday, so everybody may want to be in class.
There were some questions concerning MathZone work which we went over. I pointed out how a couple of the exercises for Section 2.4 were flawed, and that I would protect the score of anyone who worked on them and was scored incorrectly. The MathZone people have since fixed the problem (although it meant the disappearance of the actual work of the few people who had submitted the 2.4 homework ... well, at least their scores are preserved.)
It’s important to understand how the language of business equates to the language of calculus with regard to derivatives. In business, they’re called “marginals.”
Business is always interested in keeping track of marginal costs, revenue, and profit … that is, the cost (or revenue or profit) for just one more of the things being produced. This is usually very close to the value of the derivative function using the current production number. In fact, it is so close to being right, that many people just go ahead and call the derivative and marginal cost (or revenue, or profit) one and the same.
Recall that the derivative is defined as the limit of the difference quotient as the denominator approaches the value of zero: for small values of instead of
∆ x ):
∆ x (some authors will use the letter “h” g
′
( x )
= lim
∆ x
→
0 g ( x
+ ∆ x )
∆ x
− g ( x )
means g
′
( x )
≈ g ( x
+ ∆ x )
∆ x
− g ( x )
.
Remember those important words: for small values of
∆ x . That wavy-looking equals-sign means “almost equal to.”
From this, we solve and get the result that sizes of that “almost equation” by g ( x )
+ ∆ x
∗ g
′
( x )
≈ g ( x
+ ∆ x ).
(Just multiply both
∆ x and then add g (x ) to both sides and you get that result.)
This says, in other words, that an output value of the function just a little bit (
∆ x ) in the future will be almost equal to the previous output value, plus the amount gotten by multiplying the derivative at the previous value by the small additional amount being added to “x”.
In a business situation, “profit” for instance, the additional amount is set to the value of 1, so that the derivative at the previous value is a reasonable estimate of the additional amount to be added

to the function’s previous output value to get the next output value; so, for instance, the “marginal profit” (the additional profit attributable to one more unit being produced) is nicely approximated by the value of the derivative of the profit function for the current production level. In short, the marginal profit (the profit for “one more” unit) at the current production level is equal to the derivative at that production level.
Put another way
∆ y
≈ ∆ x
⋅ y
′
. The exercises assigned for this section all are intended to show just that: the actual difference in the output values (the
∆ y ) is almost exactly duplicated by multiplying the derivative by
∆ x .
Here's an example. Suppose you were desperate to have a good approximation for the value of the cube-root of 127:
3 127 .
You know it must be just a bit more than the cube-root of 125, which is exactly equal to 5 (since 53 =125). But, how much more than 5 is it?
Notice that in this case the function is the cube-root function: while
∆ x
=
2 . So, that would mean x
+ ∆ x
=
125
+
2
=
127 .
3 127
= f ( x
+ ∆ x )
≅ f ( x )
+ ∆ x
⋅ f
′
( x )
=
3 125
+
2
⋅ f f ( x )
= x
1
3 , and that x=125,
And hence,
′
( 125 ).
To complete that calculation, you need to have determined that in this case, since f ( x )
= x
1
3 , then f
′
( x )
=
( ) x
−
2
3
=
3 3
1
( )
2
3 and so
127
= f f
′
( 125 )
( x
+
=
∆ x )
3
1
3
( )
2
≅ f ( x )
+
=
∆ x
3
⋅
⋅
1
5
2 f
′
(
= x )
1
75
=
3
. Therefore,
125
+
2
⋅
1
75
=
5
+
2
75
≅
5 .
026666667 .
How does this compare with the actual value of
3 127 ?
My TI-83 says it's equal to 5.026525695. Rounding both figures to 4 decimal places, they're pretty close: the approximation of 5.0267 is close to
5.0265, after all.