1. Here y=sqrt(1+2sqrt(3+4sqrt(5+6x))), and the volunteers were Ms.^*Conner and Ms. O'Sullivan. I requested dy/dx.
   The complexity of this computation is that the function defined by the equation is three or four compositions "deep". But it is all the same (?) function, almost, sort of. So:
   dy/dx=(1/2)(1+2sqrt(3+4sqrt(5+6x)))^–1/2{0+2(1/2)(3+4sqrt(5+6x))^–1/2}{(1/2)(0+4sqrt(5+6x))^–1/2}{0+6}. I hope this is correct. If you are doing this "by hand" there are lots of ways of rewriting the result so that it looks ugly or pretty or both.
2. Here f(x)=e^7/(x2–5)sec(37x²), and the volunteers were Mr. Soni and Mr. Lin. I wanted f´(x).
   Logically notice that f is the product of two things. Then deal with the derivatives of the factors, which will need the Chain Rule. So here we go:
   f´(x)=(e^7/(x2–5)(–7/(x²–5)²(2x))sec(37x²)()+(e^7/(x2–5))( sec(37x²)tan(37x²)(37·2x)).
   There are several things to notice in this computation, but perhaps the most annoying (!) is the way the derivative of secant works. The Chain Rule declares that the derivative of Frog(Toad(x)) is Frog´(Toad(x))Toad´(x). But the derivative of secant is the product of the functions secant and tangent. Therefore the derivative of sec(Toad(x)) is sec´(Toad(x))Toad´(x), and sec´(Toad(x)) is sec(Toad(x))tan(Toad(x)). Sigh. Therefore the derivative of sec(Toad(x)) is sec(Toad(x))tan(Toad(x))Toad´(x) so that the derivative of sec(37x²) is sec(37x²)tan(37x²)(37·2x), as was written. I think this can be difficult to understand. Please try.
3. Here y=[sin(7x)–1]/[5–cos(4x)], and the volunteers were Mr. Hamden and Mr. ???. I wanted y´.
   The outermost (?) structure is a quotient. So the quotient rule gives:
   y´={(cos(7x)7)·[5–cos(4x)]–(sin(4x)4)[sin(7x)–1]}/[5–cos(4x)]² and perhaps the most interesting part of this is – on top, which is actually is – – –, where the first minus sign is from the quotient rule, the second minus sign is from the bottom's – before cos(4x), and the third minus sign is from the derivative of cosine.

Three chain rule examples with tabular information about a function

x	f(x)	f´(x)	f´´(x)
1	2	0	2
2	3	6	5
3	7	3	–4
4	2	5	7

This is a ludicrous exercise in how to "square" in weird ways. Also, I fixed the table values so now the "instantiations" (plugging in, darn it) can all be completed.

1. Here g(x)=(f(x))². I requested g(2), g´(2), and g´´(2). The volunteers were Mr. Kadriu and Mr. Palovic.
   Since g(x)=(f(x))², g(2)=(f(2))²=3²=9.
   Differentiation needs the Chain Rule, and we get g´(x)=2f(x)·f´(x), so that g´(2)=2f(2)·f´(2)=2·3·6=36.
   In order to compute the second derivative, we need a formula for the first derivative. Knowing that g´(2)=36 doesn't give us any information about g´´. So we start with g´(x)=2f(x)·f´(x) and recognize that the outermost structure is a product. So we get: g´´(x)=2f´(x)·f´(x)+2f(x)·f´´(x) so g´´(2)=2f´(2)·f´(2)+2f(2)·f´´(2)=2(6)(6)+2(3)(5)=102.
2. Here h(x)=f(x²). I requested h(2), h´(2), and h´´(2). The volunteers were Mr. Murphy and Mr. Rementer.
   Certainly h(2)=f(2²)=f(4)=2. But we need the formula h(x)=f(x²) to compute a derivative. Therefore (Chain Rule!) h´(x)=f´(x²)(2x), so that h´(2)=f´(2²)(2·2)=f´(4)(4)5·4=20.
   We need a formula for h´ to compute the second derivative. Just knowing a value of the first derivative is not enough. So start with h´(x)=f´(x²)(2x). This is a product. The second factor is 2x, which is sort of easy. The first factor is f´(x²), and this is a composition. The "outer" function is f´ and the "inner" function is x². Now we will get the second derivative, using the Product Rule and the Chain Rule.
   h´´(x)=f´´(x²)(2x)(2x)+f´(x)(2), so that when x=2 we get: h´´(2)=f´´(2²)(2·2)(2·2)+f´(2)(2)=f´´(4)(4)(4)+f´(4)(4)=7·16+5·4=132.
3. Here k(x)=f(f(x)). I requested k(2), k´(2), and k´´(2). The volunteers were Mr. Lee and Mr. Steinberg.
   Certainly k(2)=f(f(2))=f(3)=7. Now for the derivative. k is a composition of f with f. The Chain Rule gives: k´(x)=f´(f(x))f´(x). When x=2, this becomes k´(2)=f´(f(2))f´(2)=f´(3)·6=7·6=42.
   We should start with a formula for the first derivative in order to compute the second derivative. So k´(x)=f´(f(x))f´(x), and this is a product in which one factor is a composition (the composition has outer function f´ and inner function f). So k´´(x)=f´´(f(x))f´(x)f´(x)+f´(f(x))f´´(x). When x=2, this becomes k´´(2)=f´´(f(2))(f´(x))²+f´(f(x))f´´(x)=f´´(f´(2))(f´(2))²+f´(f(2))f´´(2)=f´´(3)·(6)²+f´(3)5=(–4)(36)+(3)(5)=–129.

Not a QotD but advice:
How to take my math exam

• Get familiar with the style of my exams. That's why I've given you links to past exams. You don't need to worry additionally about the format and perhaps weirdness of the questions. Get to know what to expect.
• Go slowly. Many of the dozen or so students from this class who have come to see me for help have repeatedly worked too fast, and when they try to work rapidly, they make very basic copying errors. Please don't do that. Please!
• Find friendly problems. You don't need to go from problem 1 to problem 2 to problem 3 to ... If you don't like a problem, skip it temporarily and find one that you can do with confidence. Then do another one, then ... you will get to all of the exam.
• Answer what's asked (don't invent questions). Certainly I write these exams so they can be done "by hand" and if you misread the question or change some of the numbers, you may not be able to answer the changed question, or you may make the question useless to the examiner.
• Do not simplify. So leave 3/6 alone, and don't change it to 1/2. And leave 4²–(13·3) alone and don't change it to –10. I will not reward you for doing arithmetic. And if asked to find the derivative of x²e^7x, please leave the answer as 2xe^7x+x²e^7x7 and don't change this unless there is another reason to work with this expression. Don't do unnecessary work. It takes time, and it also increases the chance to make mistakes.

The derivative of secant
We know that the derivative of 1/f(x) is This can be applied to find the derivative of secant which is 1/cosine. If f(x)=cos(x) then –(1/f(x)²)f´(x) becomes (watch the minus signs!) –(1/(cos(x)²)(–sin(x)). The two minuses cancel, and we usually write
   –(1/(cos(x)²)(–sin(x))=(1/(cos(x)²)(sin(x))=(1/(cos(x))(sin(x)/(cos(x))=sec(x)tan(x).
The derivative of sec(x) is sec(x)tan(x). Recall the famous statement of John von Neumann, inventor of things like the atomic bomb and the digital computer: "In mathematics you don't understand things. You just get used to them."

The derivative of tangent
Well, tan(x)=sin(x)/cos(x) and these "top" and "bot[tom]" functions have known derivatives. This is a good candidate for the quotient rule, which describes the derivative of top/bot as (top´·bot–bot;´·top)/(bot)². Let's fill in with what we know. We will use these replacements:
   top → sin(x);   top´ → cos(x);   bot → cos(x);   bot´ → –sin(x).
Therefore:
   (top´·bot–bot;´·top)/(bot)² →  (cos(x)·cos(x)&ndash(–sin(x))·sin(x))/(cos(x))² 
         = ((cos(x))²+(sin(x))²)/(cos(x))² = 1/(cos(x))² = (sec(x))².

So the derivative of tan(x) is (sec(x))², slightly absurd. But Is this correct? Checking absurd things, even a little bit, might be a good idea. Well, to the right is part of the graph of tangent, which I hope you recognize. Look: in any interval in which tangent is defined, the curves tilt "up" (more properly, tan(x) is increasing in such intervals). Since tan´(x) is supposed to be (sec(x))² and sec(x) is never 0 and squares are always positive, this means that slopes of tangent lines to y=tan(x) will always tilt up. So we get some confirmation of the slightly absurd fact.

The parentheses are your pals
I tend to use lots and lots of parentheses when I compute derivatives because otherwise I get confused. I also tend to be a bit overcautious, perhaps. I always write "cosine squared" as (cos(x))² and not as cos²(x) because when I am a hurry and sloppy, cos²(x) sometimes changes (all by itself?) into cos(2x) or even cos(x²), both of which are, of course, very different.

Composition and domain and range
We then took a digression from CALCULUS and discussed what is probably officially a precalculus topic, and which I think can be much more difficult than learning how to write the derivatives of formulas: the behavior of domain and range when functions are composed. I wrote this as an independent web page but decided to take some class time since some of the ideas are relevant to the next topic. The next topic
we will learn about the Chain Rule, which is probably the most important of the differentiation "rules". The Chain Rule describes how to differentiate the composition of two functions each of whose derivative is already known.

The Chain Rule
Suppose that f and g are differentiable functions. The F(x)=fog(x)=f(g(x)) is differentiable, and F´(x)=f´(g(x))·g´(x).

Here o is supposed to be a little circle, and the little circle indicates composition. I will more frequently write F(x)=f(g(x)).

Suppose we are trying to analyze how a function, F, behaves under perturbation (change, or kicking) of the input variable. We could input x and get the output F(x). If we change the input variable by a small amount, h, to a new value, x+h, the new output is F(x+h). This might be complicated and difficult to understand. But if F is differentiable, there's a nice description of the output: it is F(x)+F´(x)h+(Err_F)h, where we have the old output, F(x), and a change which is "linear" or first order, that is, something (the derivative) multiplying the kick, and then "higher order terms" (H.O.T.) which may be complicated, but as h→0 much go to 0 faster than just constant multiples of h alone. Probably in real physical situations with small changes, the output changes would essentially be a multiplier of h because the H.O.T. would generally be difficult to observe.

Now consider just a "function box", F, and attempts to "experimentally" observe or evaluate the derivative. The situation could be diagrammed as shown above. But it could be possible to lift the lid (?) and look inside the F function box. It might be "wired" as shown, with the input to F first going to an internal function box, g, and then that output going to an internal function box, f. We will try to analyze how changes to the input propagate or are transmitted through this simple network. And maybe knowing things about f and g will help us learn about f.
Consider the following complicated diagram, please:

Function diagrams
I tried to make a supporting argument for the Chain Rule using function diagrams. You can read the text for a more standard discussion. If x is the input to a differentiable function, g, then the output is g(x). If we perturb or change the input to g a bit, the output, g(x+h), can be thought of in several parts: g(x)+g´(x)h+(Err_g)h. Here g(x) is the old output, g´(x)h is a change in output which is directly proportional to h, and there's other, higher order stuff, which→0 faster than h, so Err_g→0 as h→0.

Now we could take another differentiable function, f, with input w and input perturbation k. The output will be f(w+k), and this can be imagined it as f(w)+f´(w)k+(Err_f)k. If we wire up the output of g to the input of f, then we can try to think this way:

x+h→ g(x)+g´(x)h+(Errg)h
        w  \___________/
    THIS IS k WHEN CONSIDERING f's INPUT  
          w   +   k  →  f(w)   +  f´(w)k     +  (Errf)k 
                       f(g(x)) + f´(g(x))g´(x)h  +  (STUFF)h

This "multiplier" is exactly what characterizes F´(x). The amplification factors multiply, but you need to evaluate them at the correct values of their "arguments". This is the Chain Rule.

Basement example #1
If F(x)=(x²+7)³⁰⁰, what will F´(x) be? I don't need the Chain Rule, not really (?), to compute F´(x) because, after all, F(x) is "just" a polynomial (although F(x) is a polynomial of degree 600, and this polynomial is not presented in standard fashion). Success (rapid, accurate computation) here probably will result from recognizing that the Chain Rule applies.
If F(x)=f(g(x)), then g(x) is x²+7 so g´(x)=2x, and f(x) is x³⁰⁰ so f´(x)=300x²⁹⁹. Thus F´(x)=f´(g(x))g´(x)=f´(x²+7)(2x)=300(x²+7)²⁹⁹(2x). Whew!

How does a machine handle this?
Indeed, when asked to differentiate (x²+7)³⁰⁰, Maple does recognize the composition and uses the Chain Rule. This takes less than .001 seconds (timing is only good up to a thousandth of a second). It is possible to force (!) Maple to e-x-p-a-n-d this. That takes .004 seconds, and then differentiation of that formula takes .002 seconds. This is a lot more time!

What people do in practice
But now comes the realistic comment. Hardly ever does anyone bother writing down all of these intermediate steps. That is, in practice very few f's and g's are actually identified. What happens is that people see and differentiate the outside most function (f above), put in the inner function (g) in that derivative, and then multiply by g´. For example, consider sin(e^x+x²). What is its derivative? The outside function is sine, whose derivative is cosine. So I begin writing cos(what's inside)·(the derivative of what's inside). The result is cos(e^x+x²)·(e^x+2x). This expression is a formula for the derivative of sin(e^x+x²). Again, I urge you to consider the significance and necessity (!) of appropriate parentheses in these expressions. The "argument" of cosine is e^x+x² and the cosine expression is then multiplied by the expression (e^x+2x).

#2
I think I found the derivative of sin(x²), a function which occurs in the diffusion of light through things like plastic. Here the outside function is sine and the inside function is x². The derivatie is cos(x²)(2x). There is a great deal of copying when computing derivatives.

#3
I think I did another example, something like computing the derivative of F(x)=e^{5x2+7sqrt(sin(x))}. The outside most function is "e-to-the-" (?) whose derivative is "e-to-the-". Therefore the derivative, F´(x), will begin with e^{5x2+7sqrt(sin(x))} and continue by multiplying by the derivative of 5x²+7sqrt(sin(x)). That's a sum, so its derivative is 10x+7·(the derivative of sqrt(sin(x)). What's the derivative of sqrt(sin(x))? That itself is a composition, with the "outside" function sqrt or thing-raised-to-the-half-power. The outside derivative is (1/2)(thing-raised-to-the-minus-half-power). So the derivative of sqrt(sin(x)) is (1/2)(sin(x))^–1/2(cos(x)). Now put it all together:
if F(x)=e^{5x2+7sqrt(sin(x))} then
F´(x)=(e^{5x2+7sqrt(sin(x))})(10x+7((1/2)(sin(x))^–1/2)).

I strongly recommend that you use lots of parentheses when applying the Chain Rule.

Function	Derivative
f(x)	limh→0[f(x+h)-f(x)]/h
xn	nxn–1
CONSTANT	0
ex	ex
f(x)+g(x)	f´(x)+g´(x)
f(x)·g(x)	f´(x)·g(x)+f(x)·g´(x)
CONSTANT(f(x))	CONSTANT(f´(x))
1/f(x)	–f´(x)/[f(x)]2
f(x)/g(x)	[f´(x)g(x)–g´(x)f(x)]/[g(x)]2
sin(x)	cos(x)
cos(x)	–sin(x)
cos(x)	–sin(x)
tan(x)	(sec(x))2
sec(x)	sec(x)tan(x)
f(g(x))	f´(g(x))·g´(x)
In mathematics you don't understand things. You just get used to them.

There will be a few more minor entries to the table, but we are just about done.

QotD
I wrote an entirely absurd formula defining a function and asked people to write the derivative. To the right is a picture taken by Mr. Lin (thank you!) slightly after the event which contains the formula defining f(x). The person seen is certainly older and heavier than I imagine myself, and he just looks weird, as if he were preaching to a community of penguins. Or something.

It may help you to know that I wrote the answer on the board as "you" were working on it. Soon after, a student (whose name is unknown to me but I would like to thank him) came up and looked at what I wrote. He very calmly pointed out a mistake which I then fixed. Sigh: I never make mistakes except fairly often.

Here is the question, and here is its solution, as done by my silicon pal, Si: > time(); 0.003 > f:=(5*exp(4*x)-19*x^4)/(3*x^2*cos(5*x)); 4 5 exp(4 x) - 19 x f := 1/3 ------------------ 2 x cos(5 x) > time(); 0.005 > diff(f,x); 3 4 4 20 exp(4 x) - 76 x 5 exp(4 x) - 19 x (5 exp(4 x) - 19 x ) sin(5 x) 1/3 ------------------- - 2/3 ------------------ + 5/3 ----------------------------- 2 3 2 2 x cos(5 x) x cos(5 x) x cos(5 x) > time(); 0.005 The machine was first told what f was and learned it in about .002 seconds. This "learning" takes a relatively long time (.002 is lots of time) because the machine does not just remember the string of symbols. It actually stores the formula as a logical pattern implied by the various relationships, including arithmetic (addition, multiplication, etc.) and composition. This is so that when requests to do things (differentiation is only one "thing" that can be requested) are received, the known logical structure will be used in getting answers. The time required to differentiate took less than .001 seconds! (The time command does not show increments of less than one-thousandth of a second.) Different "invocations" of the program do take different amounts of time, however, even for identical computations. The program is very large, and when it is started, it may be stored in different chunks of memory in various ways, and this can increase the running time for some computations. But this is a straightforward machine computation. I do not know why the derivative is shown in what is, to me, a rather peculiar way. After some irritating algebra, what is shown above is exactly the same as my (corrected!) answer in the picture.

Here is the answer as done by "hand" calculation:

(5e4x(4)–19·4x3)(x2cos(5x))–(5e4x–19x4)(3·2x·cos(5x)+3x2(–sin(5x)(5)))
---------------------------------------------------------------------
                             (x2cos(5x))2

People had to realize that the derivative of e^4x is (e^4x)4, because this is also a use of the Chain Rule. The outside function is "e-to-the-" and the inside function is multiplication by 4.
Similarly, the derivative of cos(5x) is (–sin(5x))5: the outside function is cosine and the inside function is multiplication by 5.
The use of lots of parentheses is a good thing here, because then things become easier and less equivocal (uncertain) to read.

Student "commitment"
I remarked that this course is a 5 credit course. This is a big chunk of credit. I remarked further:

• I had said near the start of the course that this course would need a larger than usual commitment of time and energy from students. In recent conversations I found that this "qualitative" recommendation was misunderstood by people.
• More specifically and quantitatively, I think that YOU should study calculus about 2 hours every evening. You should start this today. Certainly you may consider this exaggerated, but Mr. Nakamura and myself are putting in more than usual effort in our instruction, and I feel both you need to do this also and you need to recognize the "weight" of 5 credits. You can recreate yourself -- you are not in high school now, and you can decide to be a different person. Please be a person who substantially pushes towards success. Yes, intelligence and talent (and luck!) matter, but persistence is essential to success. Please work with other students in the course.

What's the range?
Here's a function: f(x)=(1x+2)/(3x²+4). Since 3x²+4 is always at least 4, the bottom is never 0. The domain of this function is all real numbers. So the collection of inputs is all numbers. What can we say about the collection of all possible outputs, the range of this function? I claim this is not totally clear.

Certainly, if x is very very large positive, the fraction =(1x+2)/(3x²+4) will become small. (The limit as x→+∞ of f(x) is 0, we will later say.) A similar thing is true as x→–∞. Both of these limits or asymptotic behaviors or whatever you want to say occur because the bottom is a degree 2 polynomial, and the top is a degree 1 polynomial, and the high degree polynomial "dominates". I'll be more precise about this in the future. We learned a bit about f when I asked people to give me approximate graphs using their calculators. Also, notice that f(–2)=0 and that's the only x-intercept (crossing of the horizontal axis) for this function.

But what happens in between? What sorts of numbers can we get out of f? Well, f(1)=3/7, so 3/7 is in the range. And f(2)=4/16=1/4, so 1/4 (sigh!) is in the range. And ... and ... I think just listing numbers is not even simple fun, and is mostly pointless. One nice suggestion for an output, though, was 0. The only x which gives 0 is –2. (Why? Well, Top/Bottom is 0 exactly when Top=0 [think about it] and this is 1x+2=0.) This is good because now I know that the only sign change is at –2, and (Intermediate Value Theorem) this is where f's outputs change from positive to negative or negative to positive.

How can we systematically learn about f? Well, I do know (Quotient Rule) that f´(x)=[(1)(3x²+4)–(1x+2)(6x)]/(3x²+4)². When can this be 0? Only when the top is 0, so let me "simplify" the top:
(1)(3x²+4)–(1x+2)(6x)=3x²+4–6x²–12x=–3x²–12x+4.
So the top is 0 when –3x²–12x+4=0. We need to use the quadratic formula (hey, very few "random" degree 2 polynomials with integer coefficients actually can be written as a product of two degree 1 polynomials with integer coefficients so simple-minded factoring generally won't work!). The roots are –2–(4/3)sqrt(3) and –2+(4/3)sqrt(3). (I wrote something much more horrible in class, and this is what results after some arithmetic.) You can check (as I did in class) that these roots are at least approximately numerically correct and consistent with the graph as the x-coordinates of the "bumps".

What do I know about the SIGN of f´? This derivative is a quotient, and the bottom is (3x²+4)². The bottom is always positive. So the SIGN of f´ in this case is determined by the sign of the top. But the top is –3x²–12x+4. This is a parabola, and because the coefficient of x² is –3<0, the parabola opens down. From this and from knowing the two roots of the top, we see:

• If x<–2–(4/3)sqrt(3) then f´(x)<0.
• If –2–(4/3)sqrt(3)<x<–2+(4/3)sqrt(3) then f´(x)>0.
• If x>–2+(4/3)sqrt(3) then f´(x)<0.

So what will the range of f be, exactly? It seems "clear" looking even at the approximate graph shown to the right that the range will be all numbers between f's values at r_L and r_R. That is, the range is the interval [f(r_L),f(r_R)]. I can be more precise, of course. The range is [f(–2–(4/3)sqrt(3)),f(–2+(4/3)sqrt(3))], and this is (after some work which I would not inflict on people in class!) [(1/4)–(1/6)sqrt(3),(1/4)+(1/6)sqrt(3)].

What this is, and what this is not
I do not claim that the computations just done are wonderful. Maybe they are not even interesting. But if you absolutely need to know the answer to questions of that type, and you need to know with great accuracy, and great certainty, then the analysis we've done is probably the best way. It really allows us to get the correct numbers, and the procedure is understandable. It is also inherently more precise than using a graphing device, although I probably would try to get an approximate answer using a graph first.

By the way, to the right is, of course, a graph (in a weird window, check the axes!) of y=f(x). I hope you can "see" the range. Incidentally, [(1/4)–(1/6)sqrt(3),(1/4)+(1/6)sqrt(3)] is approximately [–0.039,0.539].

Building a BIG cylinder inside a sphere
One of the recent workshop problems asked students to analyze the problem of a cylinder "inscribed" inside a sphere. So the cylinder touches the sphere as mucl as possible. The task is to understand how to write a formula for the volume of the cylinder as a function of the cylinder height, and to also tell what the domain of that formula is, considering the origin of the problem. Here we will look at the formula obtained when the radius of the sphere is 3. The height of the cylinder is related to the cylinder's radius using Pythagoras, if we look at a cross-section of the cylinder. If x is the height of the cylinder, then V(x)=Π(9–x²/4)x. The domain of V(x), when considering this problem, is 0<x<6 or 0≤x≤6. It is almost one's own personal philosophy determines whether cylinders with dimensions and volume equal to 0 should be included in this analysis. I don't think a cylinder can have a negative height, and I don't think a cylinder insider a sphere of radius 3 can have a height bigger than the diameter of the sphere, which is 6.

We can graph V as a function of the height, x. When x is close to 0, the cylinder is short and wide, but the shortness (a factor of x) makes V(x) quite small. When x is near 6, the cylinder is tall, but the radius is very small (36/4 is 9!) so the V(x) is also quite small.
How can we find the cylinder of largest volume inside this sphere? That will be at the "top" of the graph, and I will locate the top by compute V´(x). So this is (Product Rule) Π(–2x/4)x+Π(9–x²/4). This will be 0 when (divide by Π, collect x²'s): 9–(3/4)x²=0, so x=2sqrt(3) (the negative root is not in the domain for this problem!). Indeed, if you look at the graph, the top of the graph seems to be at about 3.4. This specifies the cylinder. The radius and the volume can then be computed.

What this is, and what this is not
Again: this is not a profound problem! But it does show a systematic way to solving such problems. There may be computational details which can be irritating, but at least we have some method to work with. And I assure you that we will go into great detail about the general method later in this course.

We move on to lengthen the list of derivatives by a few more entries.

The derivative of sine: a guess
I began by (trying to) draw an accurate picture of sine and then discussing the slopes of the tangent lines to this curve. The derivative of sine, is, of course, exactly those slopes.

Where would the derivative of sine be 0? Well, where the tangent lines are horizontal. That should be at the tops and bottoms of the sine curve. These occur at Π/2, 3Π/2, –Π/2, etc.: lots of places because sine repeats every 2Π.

Let's look at sine between, say, x=–Π/2 and x=Π/2. The derivative, the slope of the tangent line, would start out at –Π/2 at 0, then it would increase (as the tangent line began to tilt up). Then it would tilt up more (the slope would be more positive) until it would begin to tilt "down": here the language gets complicated. I am not asserting that the derivative is negative, but I am merely asserting that the slope, which stays positive, begins to decrease. Eventually the slope becomes 0 again when x=Π/2.

What should happen between x=Π/2 and x=3Π/2? In that interval the sine curve is sort of a downwards reflection of the behavior in the interval [–Π/2,Π/2]. The derivative starts at 0, then becomes negative. It gets more negative, then gets less negative, and ends up at 0. The shape of the derivative exactly reflect the shape in the earlier interval, since the sine curve's shape is a flip of the earlier behavior.

The derivative repeats every 2Π, since y=sin(x) repeats every 2Π. I drew sort of what is shown below. The derivative of sine qualitatively looks like cosine, except maybe we don't know the scaling factor: how high the curve is. Things are wonderful: the scaling factor is 1.

The derivative of sine: via limits, algebra, etc.
In a calculus textbook, something like the following is done when f(x)=sin(x):

 sin(x+h)–sin(x)     sin(x)cos(h)+sin(h)cos(x)–sin(x)
----------------- = ---------------------------------- = PIECE #1 + PIECE #2
        h                          h

                   sin(h)
PIECE #1 = cos(x) --------
                     h

                   cos(h)–1
PIECE #2 = sin(x) ----------
                       h

                    [sin(h)]2             sin(h)               1
PIECE #2 = sin(x) ------------- = sin(x) -------- sin(h) ------------
                   h [cos(h)+1]             h             [cos(h)+1]

               sin(h)                     1
sin(x)→sin(x); ------ →1;  sin(h)→0; ----------- → 1/2.
Nothing happens!      h                    cos(h)+1

What's up with degrees? If you insist on using degrees in your calculus computations, well, things are different and really not so nice. I suggest, please, that you put your calculator in degree mode, and that you then graph sin(h)/h for h near 0. That is, try to see graphically what the limit of sin(h)/h as h→0 is. Well, if you do this (and I recommend it if you are "allergic" to radians) you will find out that limh→0sin(in degrees)(h)/h≈0.01745329252 (heh, heh, approximately). This means that, in degrees, the derivative of sine will be about .017453 multiplied by cosine. Well, no one in the world wants to work with such numbers floating around. They are ugly and complicate many computations. Yes, the Good Lord may have made the world, but human beings can fuss with things to have the details a bit more nice. One of the nice details is measuring angles in radians, so the derivatives of trig functions are simpler. (This is the same sort of reason that e is chosen as the base of the standard exponential function -- to make the derivative of "the" exponential function simpler.)

The derivative of cosine
Look again at the shape of the graphs drawn just above. If I move the coordinate axes Π/2 to the right, the graph of sine becomes the graph of cosine. The candidate graph for the derivative needs to be recognized. It is actually minus the graph of sine. This minus sign is slightly annoying, and sometimes I screw up and forget it or I put it where it shouldn't be. The derivative of cos(x) is &ndashsin(x). (There is a shift of Π/2 in both graphs.)

More trig functions? There are 6 trig functions. The most important are certainly sine and cosine, and then tangent. Sometimes useful is secant. I will find the derivatives of tangent and secant on Thursday. I will also then put in another line, the most important one, in the derivative table. So far we have what follows.

Function	Derivative
xn	nxn–1
CONSTANT	0
ex	ex
f(x)+g(x)	f´(x)+g´(x)
f(x)·g(x)	f´(x)·g(x)+f(x)·g´(x)
CONSTANT(f(x))	CONSTANT(f´(x))
1/f(x)	–f´(x)/[f(x)]2
f(x)/g(x)	[f´(x)g(x)–g´(x)f(x)]/[g(x)]2
sin(x)	cos(x)
cos(x)	–sin(x)
In mathematics you don't understand things. You just get used to them.

QotD
Here is the question, and here is its solution, as done by my silicon pal, Si:

> time();
                                                0.002

> f:=(3*x+exp(x)+7*sin(x))/(x^2+5*cos(x));
                                          3 x + exp(x) + 7 sin(x)
                                     f := -----------------------
                                                2
                                               x  + 5 cos(x)

> time();
                                                0.002

> diff(f,x);
                                   3 + exp(x) + 7 cos(x)   (3 x + exp(x) + 7 sin(x)) (2 x - 5 sin(x))
                                   --------------------- - ------------------------------------------
                                        2                                   2            2
                                       x  + 5 cos(x)                      (x  + 5 cos(x))
> time();
                                                0.002

A function f is differentiable at x if the limit lim_h→0[f(x+h)–f(x)]/h exists. When that limit exists, then its value is called the derivative of f at x, f´(x).

Then I discussed the economics of hiring an engineer. A few years ago I looked up generally available labor information. The annual cost of hiring a new mechanical engineer, with a first college degree (a bachelor's degree) with no previous professional experience, in the NY/NJ metropolitan area, was way above $100,000. Please note that this is the cost to the employing firm, so it is not just the salary. I wanted to contrast this with the fact that there are free computer programs which will take functions defined by formulas involving familiar functions and produce a formula for the derivative of that function. Therefore, computing derivatives of formulas can't be what people want since that's easy. What's wanted is the ability to model and investigate complicated situations, and understand how to apply appropriate technical tools. At the calc 1 level, that is what this course is about. Yes, you will need to learn to take a formula and produce a derivative, but that's only the beginning.

QotD
Find the derivative of f(x)=sqrt(x) from the definition of derivative.

My silicon friend and Γ
There is an important function called the Gamma function and its value at x is usually written Γ(x). I am not inventing this function. There are over 1,000,000 references to it on Google. I wanted to use this function to force people to think about derivatives. I chose this specific function because I thought that students were unlikely to be familiar with it, and also because their calculators would likely not be able to compute it easily. Γ is a function which students are unlikely to "meet" until junior and senior level courses. My silicon friend, my rather new laptop, can compute values of this function, and do arithmetic on these values. So all that was known is that Γ was a mysterious "box" (shown to the right!) which could be investigated with experiments. I told students I would be willing to use it to answer questions such as Γ(3)=2 and Γ(4.7)=15.43141. Students were bewildered, especially when I asked these questions:

Is Γ(x) differentiable at x=3? Give some evidence supporting your assertions. If the answer is yes, what is an approximate value of for Γ´(3)?

First I was asked to compute some values of Γ and I asked how this helped. I directed people's attention to the definition and asked how we could use that definition to possibly answer the questions. Well, fairly soon people asked me to compute some rather strange quotients, and with the computer I was able to respond:

• [Γ(3+.001)–Γ(3)]/(.001)=1.84682 (approximately!)
• [Γ(3+.0001)–Γ(3)]/(.0001)=1.84569 (approximately!)
• [Γ(3–.0001)–Γ(3)]/(–.0001)=1.84544 (approximately!)
  I liked this suggestion very much because it shows a slightly and appropriately suspicious nature: the limit is supposed to be two-sided, so let's check a little bit that things do seem to work correctly.
• [Γ(3+.00001)–Γ(3)]/(.00001)=1.84558 (approximately!)

Students then decided enough information was present to answer the questions. So: YES, Γ is differentiable at x=3. The computations shown above give evidence supporting this assertion, because for some "small" numbers, the difference quotient computed seems to get close to about 1.8455 or so. Therefore this would also be an approximate value for the derivative. I was sort of happy.

More about this silly question Since the darn function is actually quite useful, much more is known about it than I presented. A wikipedia article is here and a Gamma function calculator is here (on a web page created by a group interested [really!] in "Engineering Fundamentals"). A graph of y=Γ(x) for x between –5 and 5 is shown to the right. I would say that graph is VERY strange, very very strange. Here is maybe a little more support for the answers which were given to the previous questions. I do like pictures (probably too much!). My machine reports that Γ(3)=2. If the reported derivative is correct, then 1.8455 should be the slope of the line tangent to y=Γ(x) when x=3. This line should go through the point (3,Γ(3)) which is (3,2). An equation for this tangent line should be y–2=(1.8455)(x–3). The picture to the right shows (for x between 2 and 4) both the curve y=Γ(x) in red and the candidate for the tangent line in green. The picture looks good to me.

Our task beginning this week is building a table of derivatives. It will turn out that for functions defined by familiar formulas, the derivatives can be written "easily". The quotes around the word mean that I hope you will be able to write something which looks like the derivative for functions defined by formulas of reasonable size, and that most of the time (97%?) you will be correct. I hope that you'll also be able to look at the output of computer programs that are supposed to find derivatives, and you'll have some feeling for what the outputs should be. For example, if you ask a program for the derivative of x¹⁷ and the output is arctan(x), maybe you'll go, "Huh?" Here's what we have so far:

Function	Derivative
xn	nxn–1

I actually proved this in the last lecture when n is a positive integer. It is, in fact, true for any constant n. So examples would be:

• x¹⁷ whose derivative is 17x¹⁶.
• sqrt(x), which you might see as x^1/2, and its derivative is (1/2)x^–1/2, which can also be written as 1/(2sqrt(x)).
• 1/x⁹, which is x^–9, and its derivative is –9x^–10, which can also be written as –9/x¹⁰.

Constants
I want f´(x)=lim_h→0[f(x+h)–f(x)]/h. What if f is a CONSTANT function, so its values are all the same? Well then the top of the difference quotient, f(x+h)–f(x), will be CONSTANT–CONSTANT, and it will be 0. So the derivative will be 0.

Function	Derivative
xn	nxn–1
CONSTANT	0

e^x
Let's consider the derivative of an exponential function, say a^x, where a is a constant. Then the difference quotient, [f(x+h)–f(x)]/h becomes [a^x+h–a^x]/h. As I mentioned in class, just plugging in h=0 yields 0/0, and this doesn't help. We can try the algebra that's available:
[a^x+h–a^x]/h=[a^xa^h–a^x]/h=a^x((a^h–1)/h).
So we need to consider (a^h–1)/h as h→0.

We actually analyzed this limit graphically and numerically in Lecture #3 for several values of a. When a=2, it seems that the limit exists and equals .693, while if a=3, 1.109 was the approximate value of the limit. Since f´(x) will be equal to a^x multiplied by whatever the value of lim_h→0(a^h–1)/h is, I want to choose a value of a so that the formulas are as simple as possible. So maybe we can get a value of a so the limit of (a^h–1)/h as h→0 is 1. This can be done. In fact, here is a fairly irritating way of thinking about this special number, which people call e. If h is small, we would like (a^h–1)/h to be approximately 1 when h is small, well, look at the following sequence of ideas:

n	(1+{1/n})n
1	2.000000000
2	2.250000000
3	2.370370369
4	2.441406250
5	2.488320000
10	2.593742460
100	2.704813829
1,000	2.716923932
10,000	2.718145926
100,000	2.718268237
1,000,000	2.718280469

• Since (e^1/100–1)/{1/100} should be approximately 1 we could multiply by 1/100 and see that (e^1/100–1) is approximately 1/100.
• Now add 1 to both sides and see that e^1/100 is approximately 1+{1/100}.
• If we took the 100^th power (raised things to the 100) then the 1/100 and 100 in the exponent of e cancel (repeated exponentiations multiply) so that e is about (1+{1/100})¹⁰⁰.

Well, the numerical value of (1+{1/100})¹⁰⁰ about 2.704813829. To the right is a table of values of (1+{1/n})ⁿ. This method of "computing" or approximating e is actually very very slow. ere are several million digits of e, if you need them. As I mentioned in class, it is certainly possible to prove that the formula converges as n gets large. This can be done with "only" high school algebra. It is quite tedious, and, to me at least, has little redeeming social value. (I may have the wrong attitude here, so please forgive me.)

Function	Derivative
xn	nxn–1
CONSTANT	0
ex	ex

What does that limit statement mean?
The definition f´(x)=lim_h→0[f(x+h)–f(x)]/h is sometimes a bit difficult to understand. What if I just throw out "lim_h→0"? Well, certainly f´(x) is NOT the same as [f(x+h)–f(x)]/h (hey: one of them has an h and the other doesn't even mention h!). So it really means f´(x)=[f(x+h)–f(x)]/h+ERR, where ERR (stands for "ERROR", of course) is some "mess", and all I know about it (and care about it, at this time!) is that it goes to 0 as h goes to 0. I don't like division, so let me multiply by h. Here's the result:
f´(x)h=f(x+h)–f(x)+ERR·h.
I don't like subtraction, so let me add f(x). f(x)+f´(x)h=f(x+h)+ERR·h.
People usually put the "ERR" term on the other side, so let me do that. It is irritating, but I won't change the sign on the ERR term, because right now I am interested in the qualitative aspect. So what I have is:

f(x+h)=f(x)+f´(x)h+Err·h

So what the heck do we have? If we think of the function as taking an input variable, x, and smooshing (?!) it around to get an output value, f(x), then f(x+h) is what results if we "kick" the input value by a little bit, h. If f is differentiable, then the output seems to split up into several pieces.

1. The old value, f(x). If h is very very very small, we should expect to see only something that looks like f(x). This is continuity -- there will be no big jumps if the input is changed just a little.
2. Suppose h is small, but not (somehow!) very very very small. Then we should see some change in the output. If f is differentiable, the major change will be a CONSTANT multiplying the input change. The multiplier is the value of the derivative.
3. So the two terms before aren't all. There's also "ERR·h". This is stuff that is much smaller than h. In reality, I would not expect to observe this for small h. It is "higher order" than h (people think of h³, for example: when h is small, higher order stuff is much smaller).

Sums
Suppose f and g are differentiable functions. Then I know:
     f(x+h)=f(x)+f´(x)h+Err_f·h
     g(x+h)=g(x)+g´(x)h+Err_g·h
I used different notation for the error terms for f and g to keep track of stuff better than I did in class. I can add these equations. Here's the result:
    f(x+h)+g(x+h)=f(x)+f´(x)h+Err_f·h+g(x)+g´(x)h+Err_g·h
but this is an awfully silly way to write the result. I should write it in such a way that the "structure" of the equation is shown. Here:
    (f(x+h)+g(x+h))=(f(x)+g(x))+(f´(x)+g´(x))h+([Err_f+Err_g]h)
The left-hand side is the function f+g at the input value x+h. The first piece, (f(x)+g(x)), is the unperturbed value of the function f+g at x. The second piece, (f´(x)+g´(x))h, is a multiplier not involving h, multiplied by h. The third piece is h multiplied by some stuff: [Err_f+Err_g], and all this stuff→0 as h→0. Hey, this is the higher order vanishing. So I know the next line in the table.

Function	Derivative
xn	nxn–1
CONSTANT	0
ex	ex
f(x)+g(x)	f´(x)+g´(x)

Products
This is harder and I will be a bit more detailed than what I did in a hurry in class. Suppose I know
     f(x+h)=f(x)+f´(x)h+Err_f·h
     g(x+h)=g(x)+g´(x)h+Err_g·h
and now I multiply the equations. Well, the left-hand side isn't bad, but there are three terms on the right-hand side of each equation, so there will be NINE terms if I distribute out the product/sums. I am really considering the product function of f and g. Here is how to organize the result. In class we thought a bit about how to organize this. In "print" such discussion is more difficult to write.

Please: I don't want all this stuff to be memorized. I haven't memorized it. But I do know the general idea, and that's what I'd like you to get used to. Here is the left-hand side:
     (f(x+h)·g(x+h)). This is the function f·g's value at x+h.
Here are the pieces of the product of the two right-hand sides:
     (f(x)·g(x)). This is the function f·g's value at x: the "old", unperturbed value of f·g.
    (f´(x)·g(x)+f(x)·g´(x))h. This is the first-order term, stuff (no h's!) multiplied by one h.
Here is all the rest of the stuff. There are (good grief!) six different terms. But I can "pull out" one h from all of these terms, and what is left in all six of these terms are things that →0 as h→0. Wow!
     (f(x)·Err_g+f´(x)·g´(x)h+f´(x)·Err_gh+f(x)·Err_f+Err_f·g´(x)+Err_f·Err_gh)h.
If you see this, sort of, then you can see the next line of the table.

Function	Derivative
xn	nxn–1
CONSTANT	0
ex	ex
f(x)+g(x)	f´(x)+g´(x)
f(x)·g(x)	f´(x)·g(x)+f(x)·g´(x)

A quote from von Neumann John von Neumann (1903-1957) was a mathematician who was raised in Hungary and spent most of his career in the United States. He worked in many areas of pure and applied mathematics. His ideas were influential in quantum mechanics, the development of nuclear weapons, game theory, and the theory and construction of digital computers. In mathematics you don't understand things. You just get used to them. The last entry in the derivative table is frequently called "the Product Rule". I think the Product Rule is one of those things you need to "just get used to". Another way of looking at products Well, first let's reconsider addition. We could imagine f and g being functions that somehow model intervals which are growing in length. At a certain time, the intervals have some length (the lengths labeled just f and g to the right). If we increase time ("+h") then each interval grows. I used the Greek capital Delta () to indicate how much each would grow. If we combine the functions with addition, then the growth just adds. To me this is sort of straightforward. I hope it is to you. Since the growths add, the average growths add, and the "instantaneous growths" (the derivatives!) also just add. A simple physical (?) model of multiplication in this setting would be to just make the intervals into sides of a rectangle. Then the area of the rectangle will be the product function, f·g. When we allow the intervals to grow, the area grows in a more complicated manner. Look at the picture: the increase of the area has three "chunks". One of them is an increase in f multiplied by g. Another is f, multiplied by the increase in g. Then there's the corner piece, which is increase in f multiplied by increase in g. When the increments get small, the corner piece's decrease is much faster than the two other pieces' decreases. (Sorry: my use of language is not very good here.) So I think that the instantaneous increase in f·g will be given by f´(x)g(x)+f(x)g´(x): these are the first order terms of the increase. Of course, this is the product rule. Examples The derivative of 37x88: well, this is a product. It is 37·x88. Here f(x)=37 and g(x)=x88. So f´(x)g(x)+f(x)g´(x) becomes 0·x88+37·88x87. Most people think that multiplication by a constant deserves an entry of its own. So back to the table: Function Derivative xn nxn–1 CONSTANT 0 ex ex f(x)+g(x) f´(x)+g´(x) f(x)·g(x) f´(x)·g(x)+f(x)·g´(x) CONSTANT(f(x)) CONSTANT(f´(x)) 47x32+sqrt(Pi)x12–.007x2+99 has derivative 47·32x31+sqrt(Pi)·12x11–.007·2x+0. So now we can differentiable polynomials. People who aren't as neat as I am might leave out the +0. :-) 8ex–9x1/3 has derivative 8ex–9·(1/3)x–2/3. Consider (5ex–9x)(x12–12/x3). This is a product with f(x)=5ex–9x and g(x)=x12–12/x3 (but also recognize that –12/x3 is –12x–3). This has derivative (5ex–9)(x12–12/x3)+(5ex–9x)(12x11–12(–3)/x4) Look at, say, x12. This can also be written x7·x5. If we use the Product Rule, we will get 7x6·x5+x7·5x4. If we "clean up" a bit, we seem to get 7x6+5+5x7+4 which is 7x11+5x11 which is 12x11. That is the correct and expected answer, so the weird Product Rule does reinforce what we already know.

The picture now encourages me to believe the assertion of the problem. The text supplies this hint:
     Show that f(x)=x–cos(x) has a zero in [0,1].
The phrase "has a zero in [0,1]" means that there is some root of f(x)=0 in the interval [0,1]. Well, f(0)=0–cos(0)=–1<0 and f(1)=1–cos(1). Since cos(1) is between 0 and 1, 1–cos(1) will be between 1 and 0. So we know that f(1)>0. Since the values at the endpoints are both positive and negative, I know that 0 is between the endpoint values. The Intermediate Value Theorem then applies to show that f(x)=0 has a solution between the endpoints, which are 0 and 1.

Can we locate this root more precisely?
So far what we know is that there is a root of cos(x)=x in [0,1]. Such equations occur quite frequently in applications. It is rare that solutions to such equations can be written exactly in terms of simple operations (roots, logs, etc.) and classical constants, such as e and Π. But it may be (usually is!) important to know them accurately. Can we get better information?

If f(x)=x–cos(x), we know f(0)<0 and f(1)>0. I've indicated this with the + and – labels on the ends of the unit interval to the right.
f(.5)=–.3775..., so we now know there is a root in the interval [.5,1].
f(.75)=+.0183..., so we now know there is a root in the interval [.5,.75].
f(.625)=–.1859..., so we now know there is a root in the interval [.625,.75}
Etc. By this I mean we can continue chopping the interval, looking for the sign of f's value at the center, making the length of the interval where a root is located as small as we like. This is the key idea of the Bisection Algorithm. The weird entry condition below, f(a)·f(b)<0, means that f has different signs at the two ends of the interval.

The Bisection Algorithm
Entry conditions
A continuous function f(x) defined on an interval [a,b], with f(a)·f(b)<0;
a positive tolerance E for the error.

Output
An interval [c,d] so that d–c<E and f(c)·f(d)<0. This identifies an interval of length less than E which must contain a root of f(x)=0.

"Loop" structure
Given [p,q] with f(p)·f(q)<0: let m=(p+q)/2. Compute f(m). If f(m)·f(p)<=0, then replace q by m else replace p by m.
Exit check If q–p<E then return p and q as c and d in the output else go to loop.

A simple program implementing the bisection method Here's a bisection program program written in the Maple programming language. I wouldn't use this for serious, real applications. Such numerical programs must have much more careful analysis of the kinds of errors which can occur and should be carefully tested. I hope the logic is clear. bisection := proc (f, a, b, E) local p, q, m; p := a; q := b; while E <= q–p do print(p, q); m := (1/2)*p+(1/2)*q; if f(p)*f(m) < 0 then q := m else p := m end if end do end proc; The function f has been defined by the formula f(x)=x–cos(x) in another statement: f:=x->x–cos(x);. This bisection program prints out the intermediate stages, so you can see the program focusing on the interval in which the root sits. > bisection(f, 0., 1., 0.001); 0., 1. 0.5000000000, 1. 0.5000000000, 0.7500000000 0.6250000000, 0.7500000000 0.6875000000, 0.7500000000 0.7187500000, 0.7500000000 0.7343750000, 0.7500000000 0.7343750000, 0.7421875000 0.7382812500, 0.7421875000 0.7382812500, 0.7402343750 0.7392578125 When the program ends, it reports the last variable's value, the middle of the subinterval. We will get more sophisticated root-finding methods, but this one is simple to understand and to use. Algorithm? I mentioned the word "algorithm". Let me give some further information about this word in the form of quotes from The Art of Computer Programming by D. E. Knuth: The modern meaning for algorithm is quite similar to that of recipe, process, method, technique, procedure, routine, except that the word "algorithm" connotes something just a little different. Besides merely being a finite set of rules which gives a sequence of operations for solving a specific type of problem, an algorithm has five important features: Finiteness An algorithm must always terminate after a finite number of steps. Definiteness Each step of an algorithm must be precisely defined; the actions to be carried out must be rigorously and unambiguously specified for each case. Input An algorithm has zero or more inputs, i.e., quantities which are given to it initially before the algorithm begins. These inputs are taken from specified sets of objects. Output An algorithm has one or more outputs, i.e., quantities which have a specified relation to the inputs. Effectiveness An algorithm is also generally expected to be effective. This means that all of the operations to be performed in the algorithm must be sufficiently basic that they can in principle be done exactly and in a finite length of time. Knuth continues on the same page to contrast his definition of algorithm with what could be found in a cookbook: Let us try to compare the concept of an algorithm with that of a cookbook recipe: A recipe presumably has the qualities of finiteness (although it is said that a watched pot never boils), input (eggs, flour, etc.) and output (TV dinner, etc.) but notoriously lacks definiteness. There are frequently cases in which the definiteness is missing, e.g., "Add a dash of salt." A "dash" is defined as "less than 1/8 teaspoon"; salt is perhaps well enough defined; but where should the salt be added (on top, side, etc.)? ... a computer programmer can learn much by studying a good recipe book

Average and Instantaneous Rates of Change
A few lectures ago I tried to analyze a number of real phenomena. I hope that background should help you accept the following definitions:
The average rate of change of f in the interval [x₀,x₁] is (f(x₁)–f(x₀)/(x₁–x₀). Geometrically, this is the slope of a secant line connecting the two points (x₀,f(x₀)) and (x₁,f(x₁)) on the graph of y=f(x). The instantaneous rate of change of f at x₂ is a stranger thing, that probably can't be physically measured in most cases. It is the slope of the tangent line at (x₂,f(x₂)). The instantaneous rate of change of f at x₂ is better and better approximated by the average rate of change if the numbers x₀ and x₁ are close to f.

If (x,f(x)) is a point on the curve, then people usually rewrite the average rate of change using with the secant line joining the point (x,f(x)) and (x+h,f(x+h)). Below are two views of the resulting picture, one with h<0 and one with h>0. The official definition uses this idea of approximating the slope of the tangent line with the slope of the resulting secant line. In the right-hand picture, the secant line nearly "overlays" the curve since the curve I drew is fairly flat there.
   

The definition
Consider lim_h→0(f(x+h)–f(x))/h. If this limit exists, then the value of the limit is f´(x), the derivative of f at x, and we say that f is differentiable at x.

Most of this course will be devoted to studying the derivative of a function and its uses. Actually, the next few lectures will show that, for familiar functions defined by formulas, the derivative can be computed fairly easily. So computation of derivatives, while both necessary and useful, is not the ultimate aim of the course (hey, such computation can be described carefully enough so that [capable] programmers can create differentiation programs!). We will spend most of the time investigating how to use derivatives.

Example
A traditional first example is f(x)=x². Then we need to consider (f(x+h)–f(x))/h. Let's look at the top of the fraction.
     f(x+h)–f(x)=(x+h)²–x²=x²+2xh+h²–x²=2xh+h²=h(2x+h).
In the last step, I factored out an h because I was thinking ahead:
     (f(x+h)–f(x))/h=(h(2x+h))/h=2x+h.
Now lim_h→02x+h=2x. We're done.
Conclusion The function f(x)=x² is differentiable at all x's, and its derivative is given by f´(x)=2x.

A tangent line
What is an equation of a line tangent to y=x² at x=3? Here we need a POINT and a SLOPE.
     (3,9). If x=3, y=3³=9.
     f´(3)=2·3=6. The derivative's value at x=3 is the slope of the tangent line at x=3.
An equation for the tangent line at x=3 is therefore y–9=6(x–3). I'd probably leave the equation this way unless there was a reason to change it (if I were requested to provide the answer in a different form, or if I needed to compute with it more).

Is it correct?
Here I wanted to consider the formula for the derivative, f´(x)=2x and consider if the answer were reasonable. The graph of y=x² to the left of the y-axis is decreasing, and the tangent lines should be tilted "down". Their slopes should be negative. And the algebraic candidate we have for the slopes of tangent lines, 2x, is negative when x<0. On the other side of the y-axis, the tangent lines tilt "up", and their slopes seem to be positive. Of course, for x>0, 2x>0.
In this case, considering the answer and seeing that it is reasonable and consistent with other information is easy. Certainly in more complicated situations, such checks are more difficult. But if at all possible, within the limits of time and effort, please try to make such a check. Everyone makes mistakes: humans, computers, humans using computers, etc. A few seconds "thought" can catch errors that can be very irritating later.

And now for xⁿ
Everyone knows the answer. O.k., but why? (Why is "the answer" actually the answer, not why does everyone know it.) If f(x)=xⁿ where n is a positive integer, then:
     f(x+h)–f(x)=(x+h)ⁿ–xⁿ.
We need to consider (x+h)ⁿ. It is possible, using the Binomial Theorem, to write an explicit exact expanded form of this object. I don't need that. I need much less information. So:
     (x+h)ⁿ=(x+h)(x+h)···(n times)···(x+h).
There are lots of ways to multiply things out here: you need to choose in each factor either the left (x) term or the right (h) term. But how may ways are there which would get only x's? For this, we would need to make only the x choice each time. There is exactly one way to get all x's, so xⁿ would only come out one time. How many ways would result in one h and all the rest (n–1 of them) x's? Well, we could choose the h from the first term and choose all x's from the other terms. Or we could choose an h from the second term and all the others (the first term and the terms after the second term) x's. Etc. Here by "Etc." I mean that we could take an h from exactly one of n factors, with all the other choices being x's. So since there are n factors, there are n ways to get a product with one h and all the rest x's. So in the result, there's nhx^n–1. What about the rest? We took care of the "no h" term and the "one h" terms. So all the rest has at least two factors of h. So, actually, we now know:
     (x+h)ⁿ=xⁿ+nhx^n–1+h²JUNK.
In this expression "JUNK" is not something bad. It represents terms I don't need to care about at this time. In fact, later in the course, we will try to understand some aspects of JUNK and how they can be useful. Anyway, let's continue:
     (x+h)ⁿ–xⁿ=xⁿ+nhx^n–1+h²JUNK–xⁿ=nhx^n–1+h²JUNK=h(nx^n–1+h(JUNK))
where again I factored the h out because I was thinking ahead. Now the limit:
     lim_h→0(f(x+h)–f(x))/h=lim_h→(h(nx^n–1+h(JUNK)))/h=lim_h→0nx^n–1+h(JUNK)=nx^n–1.
Therefore, f(x)=xⁿ is differentiable, and its derivative, f´(x), is nx^n–1.

QotD
If f(x)=1/x, use the definition of derivative to find f´(x).

So f(x+h)–f(x)=1/(x+h)–1/x=(x–(x+h))/((x+h)x)=–h/((x+h)x) and lim_h→0(f(x+h)–f(x))/h=lim_h→0[–h/((x+h)x)]/h=lim_h→0–1/((x+h)x)=–1/(x·x)=–1/x².
Interesting aspects of this computation: if f(x)=1/x, then f(x+h)=1/(x+h). You must understand the grammar (?) of functions to do this. And combining the fractions and converting to a simple fraction: you must know how to do this sort of algebra.

An absurd example
What is

          1        1 
        ----  –  ------
        3x+4      x2+6
 lim  -------------------
x→2          x–2

"Plugging in" x=2 gets 0 on the bottom, and on top, if we substitute correctly, 1/(3x+4) becomes 1/10 and 1/(x²+6) becomes 1/10. So the top is (1/10)–(1/10) which is 0. Surprise (not!): this is a 0/0 situation. We use some algebra. My feeling is great dislike for compound fractions, that is, fractions within fractions. I find them difficult to understand and difficult to manipulate. My advice is to change them into "simple" fractions, where only one division sign will appear. But we need to do this carefully.

The top is 1/(3x+4)– 1/(x²+6) which is
(x²+6)–(3x+4)
-------------------
 (x²+6)(3x+4)
Notice that this fraction is sitting "on top" of (x–2). We have something like this:

   A         A      1            A
 -----     ----- · ---        -------  
   B         B      C           B·C          A
------- = --------------  =  ---------- =  -----    
   C              1                           B·C
             C · ---             1
                  C

Limits and algebra
If lim_x→af(x)=L₁ and If lim_x→ag(x)=L₂ then
    •  lim_x→af(x)+g(x) exists and equals L₁+L₂;
    • lim_x→af(x)·g(x) exists and equals L₁·L₂;
    • (when L₂ is not 0) lim_x→af(x)/g(x) exists and equals L₁/L₂.

QotD (definitely less absurd)
(Please accept my guarantee that this is not absurb, at least for a little while [2 lectures].)What is

       1     1     
      --- – --- 
       x2    w2 
 lim  --------- ?
x→w    x–w

   w2–x2    (w–x)(w+x)       w+x
--------- = ---------- = - ------
x2w2(x–w)    x2w2(x–w)       x2w2

A weird but not absurd limit
f(x)=|x+5|/(2x+10). So what happens as x→7, for example? The x+5→7+5=12. The absolute value for x's near 7 give x+5 near 12. The results are all positive, so |x+5|→12. The bottom, 2x+10, approaches 2·7+10, which is 24. The resulting limit value is 12/24.

There's trouble near –5, though. So what about lim_x→–5|x+5|/(2x+10)?

Here there's a real difficulty with absolute value. Recall that |w| is w if w>0 and |w| is –w if w<0. So |x+5| is x+5 if x+5>0. and |x+5| is –(x+5) if x+5<0. But x+5>0 is the same as x>–5, and x+5<0 is the same as x<–5. So put all this together:
suppose f(x)=|x+5|/(2x+10).
    If x>–5, f(x)=(x+5)/(2x+10)=1/2;
    if x<–5, f(x)=–(x+5)/(2x+10)=–1/2.

To the right is a graph of f. I don't "know" what the value of f(–5) should be (computer programs handle this differently, but many report that trying to evaluate the function at –5 is an error).

Such functions occur when trying to model real situations (hey: "hit" a plate -- that will be a shock! Or drop a chunk of salt into a container of water -- that's a shock to the salt concentration!). So getting some language to describe the behavior is useful.

As x gets close to –5 from the left, the values of f(x) are all –1/2. So people say that lim_x→–5^–f(x), the left-hand limit of f at –5, exists, and they say that the value of this limit is –1/2.

As x gets close to –5 from the left, the values of f(x) are all 1/2. So people say that lim_x→–5⁺f(x), the right-hand limit of f at –5, exists, and they say that the value of this limit is 1/2.

You may be confused at first with the superscript/exponent + and –. The "+" up there always means "from the right" and the "–" up there will always mean from the left. In the expression x→–5⁺ the minus before the 5 means look 5 units to the left of 0. The plus sign in the exponent means see what happens to the right of that number (that is, –5), but look close to it.

A piecewise linear graph and some limits

I drew a graph similar to what's shown to the right. The idea was to use the geometric information on the graph to find out what we know about function values and limits. I also wanted to urge students to understnad the possibilities of our algebraic "language". So to the right is a graph of y=f(x). Some questions:

• What is f(1)? What is lim_x→1f(x)? (This is the easy one!)
  I see that the point (1,3) is on the graph of y=f(x). A point is on the graph if it has the form (x,f(x)), so f(1)=3. Also, as x gets close to 1, the heights on the graph, the f(x)'s, get close to 3. So lim_x→1f(x)=3.
  The function y=f(x) is continuous at 1. This is because the limit as x approaches 1 of f(x) exists, and 1 is in the domain of f, and the value of f at 1 is equal to the limiting value of f.
• What is f(4)? What is lim_x→4f(x)? (This is not so easy.)
  Well, f(4)=1 since there a big dot at (4,1). What about the limiting behavior as x→4? A key part of the idea of limit is that what matters is behavior near the point but not the behavior at the point. Now as x→4 from either the right or the left, the height, f(x), gets close to 2. I think that lim_x→3f(x) exists, and that the limit is 2.
  Since the limiting value is not equal to f's value at 4, f is not continuous at 1.
• What is f(2)? What is lim_x→2f(x)? (A bit stranger.)
  If I consider points whose first coordinate is 2, points with coordinates (2,?), there are no such points on the graph of y=f(x). So 2 is not in the domain of f. What about lim_x→2f(x)? Here the heights of the function depend on which side x gets close to 2. If x gets close from the left, then the f(x) values are close to 1. So I know lim_x→2^–f(x) exists, and its value is 1. But if x gets close to 1 from the right, the heights, the f(x) values, seem to get close to 0. So now lim_x→2⁺f(x) exists, and its value is 0. Since the left and right handed limits do not agree, the (two-sided) limit, lim_x→2f(x) does not exist.
  Since the limit does not exist, the function can't be continuous at 2.
• What is f(–1)? What is lim_x→–1f(x)?
  Look at the graph. There's a nice thick dot at (–1,2). Therefore I conclude that f(–1)=2. The two-sided limit at –1 of f does not exist. As x gets close to –1 from the left, the f values (f(x) is the height of the curve) get close to 1. So lim_x→–1^–f(x)=1. Look now at the right side of x=1. The f(x) values drop go to 2, so I conclude that lim_x→–1^–f(x)=2.
  f is not continuous at –1. To be continuous at –1, all these must exist and be the same number: f(–1) and the left-hand limit at –1 and the right-hand limit at –1.

My friend Francine...
The Garden State Parkway runs most of the length of New Jersey. Mile 0 is at Cape May, while the other end, mile 172, seems to be close to Montvale. Suppose that my friend Francine leaves Cape May at 7 AM one morning, and drives north on the Garden State Parkway. Further, suppose she arrives at mile 172, the northern end, at, say, 10 AM. Must Francine at some time be at mile 135 (fairly near Busch campus)? The parkway seal here was "borrowed" from a State of New Jersey webpage.

We discussed various curves which could represent the position of Francine on the parkway in terms of miles from the start of the parkway at time t, in terms of hours elapsed from 7 AM. I tried to show that our everyday intuition lead to the graph being increasing (as you travel from left to right, the points on the graph go up). The graph can have level spots, where Francine pulls over for a rest stop. Legally Francine isn't supposed to drive backwards, though.

If we believe that motion is continuous (so Francine does not have a Star Trek transporter or other device) then the graph of Francine's position goes from (7 AM, 0 miles) to (10 AM, 172 miles) and therefore the graph must have on it at least one point with coordinate description (*,135). All of this, by the way, rests on some complicated assumptions, some of them philosophical (why should motion be continuous?). Today, though, I believe that motion is continuous, and therefore at sometime Francine must be at Mile 135. This is all the idea behind the following important result.

Intermediate Value Theorem
Suppose that the function f is defined and continuous on the interval [a,b]. Then the equation f(x)=y has at least one solution for every y which is between f(a) and f(b).

Problem #5 in section 2.7
Here's the problem:
     Show that cos(x)=x has a solution in the interval [0,1].
Since I am a picture person, I frequently try to draw a graph or two to understand what the problem is about. In this case, certainly the graph of y=x is easy to imagine on [0,1]. What about y=cos(x)? Well, it does help, if you don't have access to a graphing device, to know that cos(0)=1. I know that cosine drops down as x increases from 0. What do we know about cos(1)? As one student declared, cos(1) is both less than 1 and greater than 0. That's because Pi/2 is about 1.57 and cosine decreases between 0 and Pi/2, and does not reach 0 until Pi/2. So the graph of y=cos(x) decreases from (0,1) to (1,cos(1)), and the end point is somewhere above the x-axis. The mental picture I have built is shown to the right. I deliberately did not have a graphing device create an "accurate" display of the situation -- I wanted to show what we should be able to do inside our own heads.

The picture now encourages me to believe the assertion of the problem. The text supplies this hint:
     Show that f(x)=x–cos(x) has a zero in [0,1].
The phrase "has a zero in [0,1]" means that there is some root of f(x)=0 in the interval [0,1]. Well, f(0)=0–cos(0)=–1<0 and f(1)=1–cos(1). Since cos(1) is between 0 and 1, 1–cos(1) will be between 1 and 0. So we know that f(1)>0. Since the values at the endpoints are both positive and negative, I know that 0 is between the endpoint values. The Intermediate Value Theorem then applies to show that f(x)=0 has a solution between the endpoints, which are 0 and 1.

Can we locate this root more precisely?
I will show you how to do this next time.

The idea of limit
Today we begin an official discussion of limit. This is a very useful idea. There are lots of computations and manipulations which are related to limits in this course and in many other situations. But I must be very careful about how class time is used, and I must rely on you to read the textbook and do homework problems. I will choose to emphasize ideas in the lectures. I will try to do a sufficient number of examples to illustrate these ideas, but your own efforts will be key to getting enough experience and sufficient familiarity.

Tree growth
Catalogs sell trees with short descriptions which frequently include such information as "The eventual height of this tree is 40 ft and its growth rate is 2 ft/yr." What do these numbers mean? We discussed this for a while.

First approach
If one took the 2 ft/yr literally, then in a century the tree would be 200 ft high, and a bit later (well, quite a bit later!) the tree would knock down the moon. This is probably not realistic.

Another attempt to understand ...
We made a second attempt to understand. The tree would grow at 2 ft/yr, and then, after 20 years (40/2=20) the tree would stop growing. Well, if we plotted the growth curve, it might look like what is shown to the right. To me, a simple piecewise linear curve is really neat and simple. But I don't think this represents behavior of complex organic objects. For example, I don't think a really tiny tree would actually grow two feet taller in its first year. I think it would begin by growing rather slowly. I also don't think that trees grow at a steady rate, and then tree growth would suddenly and totally stop at some specific height. That's probably not what happens.

Maybe more real ...
Reality is sloppier. I bet that the tree begins its growth fairly slowly. Then the growth gets more rapid and steadier, and persists for quite a while until the tree gets near its "ultimate height". The tree growth slows then, but I would expect more growth, at a slower rate, for a while. I think that a more reasonable possibility for a growth curve is shown to the right.

In fact, I think what's shown to the right is, itself, only an imitation of real tree growth. Growth will depend on climate and weather (moisture, temperature), and nourishment (how much of what kind of minerals, etc.), and competition (in the shadow of something else?), and such factors as where is the lumber harvester. Each tree will likely be somewhat different, and what's presented and described is an average of many observations.

How could tree growth around, say, year 13, be calculated? Well, suppose H(t) is the height of the tree in feet, at time t years after the tree seed germinates. Then we could measure the height of the tree once a year. So we'd have values for H(12) and H(13) and H(14), etc. The growth rate in feet per year for the year [13,14] would be (H(14)–H(13))/(14–13). This number would represent the average growth rate of the tree during that year. Of course, we could want more precise information. I could imagine someone measuring the tree height every 3 months, a quarter of a year. The interval [13,13.25] would be closer to 13. And I would think that the quantity (H(13.25)–H(13))/(13.25–13) would be a better idea of what might be the tree growth rate "at" 13. Of course, this fraction is what's called the average growth rate in the interval [13,13.25]. We could imagine an enthusiast desiring better information. This person could want 10,000 measurements of the tree height during the year, etc. This information could be used to give more precise information about tree growth rate. Now think about real life. I don't believe that 10,000 tree measurements are likely. I think that approximate average rates of growth are all we can expect, and the idea of some totally precise "growth rate" is fictional. But, even though it is fictional, such an ideal growth rate may be useful.

Rock dropping
I used feet and pounds, an antique system of measurements in this discussion. This may be confusing. I am sorry.
Suppose we drop a rock. Then the rock will fall approximately s(t)=16t² ft after t seconds. If we wanted to know the average velocity that the rock fell in the one second after t=3, we could compute:

s(4)–s(3)  16(42)–16(32) 
-------- = ------------= 16(16–9) = 112 
 4–3             1

 s(3.07)–s(3)    16(9.042049)–16(9)      16(.042049) 
------------- = --------------------- = ------------- = 96.112 
  3.007–3              .007                .007

Let's now use algebra. Algebra allows almost everyone to be clever. So we can ask: what is the average velocity of the rock during the time interval [3,3+h] if h is a small positive number? The computation would be something like this:

s(3+h)–s(3)    16(3+h)2–16(32)   16(9+6h+h2)–16·9    16(6h+h2)
------------ = --------------- = ---------------- = ---------- =16(6+h)
    h              h                  h               h

This is actually a rather remarkable computation, and just because you may have seen such things before is no reason to ignore the wonder: you must be like a child .... Let's check the formula. If h=.007, then 16(6+h) is 16(6.007)=96.112 (that's what we got before). To me, the most remarkable part of the computation is that we have a quotient, a tiny distance change on top, and a tiny time change on the bottom. Amazing, somehow the "tiny"'s sort of cancel, and we are left with an average velocity, 16(6+h), which has a sort of stability property as h gets smaller. In fact, if you don't think too hard, as h gets smaller ("h→0") then 16(6+h) gets close to 16(6)=96 ("16(6+h)→96").

So the average velocity over the interval [3,3+h] approaches 96 as the length of the time interval goes to 0. People abbreviate this by declaring that 96 is the instantaneous velocity of the rock at time 3. As far as I know, human beings don't actually measure or observe instantaneous velocities. These instantaneous things are lies, o.k., we shouldn't call them lies, they are useful fictions or mental constructions. Most of this course will be about such useful fictions.

Local linearity
Last time we considered the curve y=3^x and zoomed in on it around the point (0,1). We observed that the curve seemed more and more to appear like a straight line. Hey: I even gave this phenomenon a name: local linearity. The line goes through the point (0,1), and therefore complete information about it only depends on knowing its slope. Well, if we think that the line is about the same as the curve, on a very very small scale, then maybe another point on the line would approximately be (h,3^h) if h is a small positive number. The slope is then the difference in the second coordinates divided by the difference in the first coordinates: (3^h–1)/h (on the bottom, h–0 is just h). In this case, unlike the low-degree polynomial investigated with the rock, the h's don't obviously somehow fade away. We guessed using numerical evidence that as h→0, this quotient also "stabilizes" near the value 1.1. This behavior is not obvious to me algebraically, and, indeed, if I wanted to verify it for you right now, showing the limiting behavior would take a great deal of effort.

A function which isn't locally linear
Look at the graph to the right, and specifically look "near" (0,0). Here I am deliberately omitting the scale marks on the axes, because they don't matter for the purposes of this discussion. If we zoom in around (0,0), we get something which looks much the same. This is a self-similar object and scale doesn't matter. (Self-simlarity is a key part of what are called fractals, one natural (?) appearance of which is here).

There's no magnification which will make the graph to the right look like a straight line through the point (0,0). So this is a graph of a function which is not locally linear. In this course, most of the functions we will consider will be locally linear, so you should see that some rather simple graphs will just be thrown out of consideration.

Please realize that the graph to the right is associated with |x|, and is not an abstract, weird invention.

Observations and ideals
We, humans, can observe average rates of change. We can measure how balls roll down inclined planes. We can, with considerable more ingenuity and equipment (go over to the Chem Department!) make observations of chemical reactions on a near-nanosecond scale. As far as I know, we look at average rates of change. A nice invention is the instantaneous rate of change, which will be a principal object of study in this course. As far as I know, we do not observe instantaneous rates of change. We can only approximate them by real measurements. This course studies the useful fictions related to instantaneous rates of change.

Notation
I introduced some notation to help us talk about average and instantaneous rates of change. We will want to study the behavior of some function f(x) when x gets close to a, and, in particular, we will want to tell people that (in "good" cases) f(x) gets close to b when x gets close to a. Well, arrows are usually used to abbreviate "gets close to". So we want to declare that as x gets close to a, f(x) gets close to b, and we will write for this "As x→a, then f(x)→b." An even more compact (lazier!) way of stating this is to write lim_x→af(x)=b and this is how I'll communicate this complicated "gets close to" idea.

Basic method: plug in
Suppose you want to understand lim_x→af(x). If the function is defined by some formula you understand and believe is well-behaved, plug in. That is, evaluate f(a).

Example What's lim_x→3x²? I think it is 3².

This "trick" is important enough so that there is a label which goes with it. A function f is continuous at x=a if the domain of f includes an interval surrounding a, and if lim_x→af(x) exists, and if the value of that limit is f(a).
In other words, the function is continuous at x=a if the limit of the function as x "approaches" a is gotten just by plugging in a to f: computing f(a). Due to many facts about limits that we will mention next time, most familiar functions are continuous in their domains.

Many of the most interesting limits we need to deal with (those involving rates of change, for example) can't be evaluated by just plugging in. Here are two examples: lim_x→0(3^x–1)/x and lim_x→0(sin(x))/x. In both cases, plugging in gives 0/0. In fact, both limits do exist. The value of the first one is ln(3) and the value of the second one is 1. Neither fact is obvious, and some effort is needed. But sometimes limits can be seen by (relatively!) simple algebraic manipulation. We will try to change them into forms which can be evaluated by "plugging in".

Next method: algebraic transformation, with the goal being ...
Sometimes (fairly frequently in this course) we will need to find some limit but the "plugging in" gives us 0/0, and I don't know how to make sense of this quotient. So one thing we can try is to change the given expression algebraically and maybe get something which is equal but which does permit plugging in.

Example I think we looked at the more complicated example:
lim_x→4(x–4)/(sqrt(x)–2).
Here plugging x=4 in just gets you 0/0, which I don't understand. I can see two ways of analyzing the fraction inside the limit.

Method 1 Multiply by the conjugate. Be sure to multiply by the same thing on the top and bottom, or else the result might not be equal! Here this means:

     (x–4)               (x–4)(sqrt(x)+2)            (x–4)(sqrt(x)+2)  
----------------- = ----------------------------- = ------------------- = sqrt(x)+2
  (sqrt(x)–2)          (sqrt(x)–2)((sqrt(x)+2)           x–4

Method 2 Another way of handle the quotient which some people prefer is this:

    (x–4)         (sqrt(x)–2)(sqrt(x)+2)  
-------------- = ------------------------- =sqrt(x)+2
  (sqrt(x)–2)        (sqrt(x)–2)

Use any valid method that works for you, please. Here's another view of conjugation

QotD
What is lim_x→1(x²–5x+4)(x²–3x+2)?

Plugging in gets
(1²–5+4)(1²–3+2) which is a 0/0 form and doesn't tell us anything. But a number of people called out, "Factor" as a hint.
(x²–5x+4)(x²–3x+2)=[(x–4)(x–1)]/[(x–2)(x–1)]=(x–4)/(x–2) when x≠1.
Therefore lim_x→1(x²–5x+4)(x²–3x+2)=lim_x→1(x–4)/(x–2)=(1–4)/(1–2)=3.

The specific function f considered here is one-to-one (1-1). This means there is exactly one distinct input for each output. How can we verify this? If you believe the graph, the fact means that any horizontal line will intersect the curve at most one time. Algebraically, we can do this: y=x³+7 implies that x=(y-7)^1/3. If y is the output, we have just "constructed" the input, x, corresponding to the output, y.

One-to-one
If we have a 1-1 function, we can think about reversing the process, and the reversed process is the inverse function to the original function. Right now the idea is what I'd like to make sure of. There are certainly implementational issues to worry about, and these aren't easy. But I think the idea should precede computations in this case. The ludicrous diagram to the right is supposed to make clear (?) what's going on. If there is exactly one output value corresponding to each input value, then we could think about reversing this association, and creating an inverse function.

In the case of f(x)=x³+7 I know that f is 1-1 because I can solve f(x)=y and get exactly one value of x for each y. The function g(y)=(y-7)^1/3 is a function defined by the algebraic formula I got while solving. It is a formula for the function inverse to f.

If the point (x,y) which is (f's input, f's output) is on the graph of f, then (y,x) will be on the graph of the inverse to f. The switch (x,y) to (y,x) is geometrically done with a flip across the "main diagonal", y=x.

Part of the graph of the function inverse to this f(x)=x³+7 is shown to the right. It is the cubic curve flipped over the main diagonal.

Squaring
Consider squaring: f(x)=x². We know about this function. It is perhaps the most obvious (once the negative numbers are learned) example of a function which is generally (except for 0) not 1-1. Look: f(2)=2²=4 and f(-2)=(-2)²=4 also. Take the graph, a parabola whose vertex is at (0,0), and flip it over the main diagonal. The result is a curve which is not the graph of a function. "Everyone" knows this. Yet at the same time everyone would like to indicate square roots, and would like to tell that the square root of 4 is ... well, it could be -2 or +2. If a unique output is desired, the unique output which people have chosen is the non-negative answer (a classical word for this was "branch", really, apparantly because the flipped over curve might look like it has several slanting tree things). So sqrt(x) (really, better written with that strange almost-division sign) means the non-negative number whose square is x. So there has been a choice, in terms of the graph shown to the right. "Everyone" has decided to take the top, green, piece of the graph, and if ever the bottom part has to be used, it will be designated -sqrt(x).

Choosing an inverse
In this case the choice of a part of the function's domain and the inverse has been done in our culture years and years ago, and I think it is likely we have all grown up with it so nicely that we haven't thought about it!

The trig functions and their inverses
The six trig functions are all periodic. Therefore the difficulties of squaring actually become much more obvious. If sin(θ)=1/2, well, then, actually, θ could be Π/6 or 5Π/6 or Π/6-2pi or 5Π/6-2Π or Π/6+2Π or 5Π/6+2Π or ... actually infinitely many other possibilities. Yet if you are interested in numbers, solving equations like sin(θ) =some number for θ might be important (hey, I mentioned in class and meant the following: construct a robot arm, and "watch" its motion -- you'd better be able to understand how the values of the input and output of sine and cosine work!).

arcsine People have agreed to "restrict" the domain of sine to [-Π/2,Π/2]. Then the function is 1-1, changing the inputs [-Π/2,Π/2} to putputs [-1,1} (sine is increasing in that domain, and we will develop techniques to verifty such claims easily). So there is an inverse function. The text calls it sin-1 and I will try not to use that notation since a -1 in the exponent makes me think of reciprocal. The notation I will use is arcsin. Also I should mention that most of the computer languages I know use the designation arcsin to mean what I am describing here (since superscripts are difficult to type!). The graph to the right is arcsin. The endpoints are (-1,-Π/2) and (1,Π/2). Everyone agrees that the domain of arcsin is [-1,1] and the range is [-Π/2,Π/2].

arctan
Tangent, from the inverse function point of view, is in some silly sense twice as bad as sine. ! This is because sine is 2Π periodic and tangent is Π periodic. The tangent domain restriction that people use almost always (99.999% of the time!) is (-Π/2,Π/2). The open interval results from the fact than tangent is not defined at odd multiples of Π/2 (hey, cosine is 0 there). So flip the graph of tangent. The result is arctan (or, in the text, tan^-1).

The domain of arctan is all real numbers. The graph is increasing, and when x gets very large positive, then arctan(x) gets close to Π/2. On the other side, when x gets very large negative, then arctan(x) gets close to -Π/2. There are two (horizontal) asymptotes, y=Π/2 and y=-Π/2. As I mentioned, arctan is used to take big positive and negative numbers and output numbers which are more controlled, between -Π/2 and Π/2. arctan has been more used by me than tan. It is a neat function.

Building powers of 3
I gave an extended discussion of exponents. The aim was to explain my view of how the "exponent laws" were created (invented?). I do not know if, in fact, what I discussed is supported by historical data.

3⁶ means 3·3·3·3·3·3. That is, 3's repeated and multiplied 6 times. Therefore
     RULE 3^{positive integer} means 3's multiplied together that "positive integer" number of times.
Then, surely, people noticed things like 3⁵⁺² is 3⁷, seven 3's multiplied, and this is the same (associativity) as 3⁵·3²: just line things up and multiply them.
     RULE If n and m are positive integers, then 3^n+m=3ⁿ·3^m.
Well, then, this "rule" is neat and easy to remember. But consider the following "equation" (quotes because until you decide what some of the symbols mean, the equation is silly): 3⁶=3⁶⁺⁰=3⁶·3⁰. If this equation is valid, then 3⁰ should mean 1. Making this equation correct would mean that the simple template just above would be true for n and m non-negative integers, and then I (and other people) would have less to remember.
     RULE 3⁰=1.
If we want to extend the n+m equation more, consider the following "equation" (quotes again because we are exploring how the symbols work): 3⁰=3^7-7=3^7+(-7)=3⁷·3^-7. Then since we want 3⁰ to be 1, 1=3⁷·3^-7 so that 3^-7=1/3⁷. Well, then, here's another definition:
     RULE If n is a positive integer, then 3^-n=1/3ⁿ.
So, for example, 3^-2 must be 1/9. Finally, what happens if we repeat exponentiation? Let's consider (3⁴)². What is this? Well, 3⁴ is 3·3·3·3 (four 3's multiplied). Then squaring this gets us (3·3·3·3)·(3·3·3·3). If you count this is eight 3's multiplied: 3⁸. So therefore (3⁴)²=3⁸. Therefore we get another:
     RULE If n and m are positive integers, then (3ⁿ)^m=3^n·m.
But what about, say, 3^7/2? If we believe in the previous result, or, better, we want to use it as a simple template for computation, then consider (3^1/5)⁵. This "should" be 3^(1/5)·5=3¹=3. So therefore 3^1/5 should mean the positive number which is the fifth root of 3, the unique positive number which solves x⁵=3.
     RULE If n is a positive integer, then 3^1/n is the n^th root of 3, that is, the positive solution of xⁿ=3.
People have adapted these rules so that exponentiation is easy to handle.

If b is a positive number, then: bn=b·b·...·b n times. b0=1. bp+q=bp·bq. b-p=1/bp. (bp)q=bp·q. Although maybe these rules are all "obvious" to you, in fact I hope that you see they are designed so people and numbers live together comfortably. To a certain extent, math is a human construction. Certainly mathematical notation is human. The obviousness is partly due to your constant exposure to such equations (since maybe fifth or sixth grade?). I remarked that the repeated exponentiation equation, obvious and clear, is an important part of RSA encryption, a very widely used method for secure electronic communication (for example, commercial web transactions). Google has two and quarter million links to the phrase "rsa encryption".

The definition (?) of 3^sqrt(2)
I mentioned this in class, although I suspect it seemed weird and totally useless to most students. Very briefly, the problem is how to define and compute such things as 3^sqrt(2). The previous discussion showed how to compute, say, 3^77/17 First find the 17^th root of 3 and then take the 77^th power of that root. But sqrt(2) is not rational. It can't be written as a quotient of two integers. Numbers like 3^sqrt(2) will occur in calculus, and we may want to get some approximation of them. So what's done? First, get an approximation of sqrt(2): 1.414213562, for example (!). This is rational (you may not like it, but it is 1414213562/100000000. Then compute 3^1.414213562 which is about 4.728804386. And this is what is done.

The curve y=3^x investigated
I asked people to graph y=3^x on their graphing calculators. Below are some results.

To the right is a graph of y=3x in the window -1<=x<=1, 0<=y<=2. I think the curve looks sort of the way everyone thinks it should. This window has the point (0,30), which is (0,1), in the middle of it.
Please keep track of the dimensions of the window. Here I've "zoomed in". The window is -.25<=x<=.25, .75<=y<=1.25. This is again a square window. The zoom is a factor of 4. The bend in the curve is still visible to me, but the amount of curviness is less.
The zoom is higher. Now the square window has these dimensions: -.01<=x<=.01, .99<=y<=1.01. The curve is, to my weak human eyes, just about flat. The phrase "locally linear" is used for functions like 3x. This phrase means that if the graph of the function is zoomed at enough, the result looks very much like a straight line. In fact, the functions of interest in this course will be exactly the locally linear functions. The straight line, if you look at it carefully (some students did) has slope approximately equal to 1.1. (Hey, (3.05-30)/(.05-0) is about 1.12).
You can go through the same sequence of graphs for y=2x. I'll skip to the third graph. So what's shown to the left is the third result, which is y=2x in the window -.01<=x<=.01, .99<=y<=1.01. Again, this curve is locally linear, and the window is sufficiently small so that the piece of graph visible looks to me like a segment of a straight line. The slope of this line is about .69. (A computed value of (2.05-30)/(.05-0) is about .705.)

Exponential functions are very commonly used to model such phenomena as radioactive decay and bacterial growth and lots of other things. The local linear approximations are things we will work with. I don't find 1.1 ("slope" of 3^x near (0,1)) and .69 ("slope" of 2^x near (0,1)) particularly nice numbers to work with. I would rather work with 1. We can adjust the local slope of b^x by varying b. There is a unique number b (between 2 and 3) so that the local slope of b^x at (0,1) is 1.
This number is called e. To the right is a graph of 2^x in blue and 3^x in green and e^x in red. The graph does show e^x "sandwiched" between the other two exponentials. It doesn't clearly show that e^x has the desired local slope of 1, but this is actually true. We will later see how to compute e with as much accuracy as desired. e^x is widely known as the exponential function, and frequently referred to as exp(x).

e is approximately 2.71828, and I sort of doubt that more digits will be needed by almost anyone in the class for practical computation. e is not rational, and its decimal expansion does not repeat. If you wish, here are about the first 2 million decimal places of e. People have computed billions of decimal places of both e and Pi. What else is there to do?

Geometry
The real numbers are usually thought of as corresponding to a specific geometric object, the real line. I think of this line as horizontal with 0 sitting "in the middle". 1 is to the right of 0. And this geometric picture brings up the idea of order. In addition to the algebraic structure, there is order: a<b. To me this means (in my picture of the line) that a is to the left of b. Negative numbers are to the left of 0 and positive numbers, including all the positive integers, are to the right of 0. Here are some interesting aspects to note.

Distance including a discussion of | |
The distance between two points is a non-negative real number whose size expresses how far apart the numbers are. This will be important when we study approximation schemes. We'd like to know that the approximation gets "close" to the correct answer, and the closeness will be measure by the distance. Algebraically, if the points correspond to the real numbers a and b, the distance between them is |a-b| and this is the same as |b-a|, so that distance has some symmetry. But I just used absolute value, and here is the piecewise definition of absolute value:
|x|=x if x≥0 and |x|=-x if x<0

Therefore absolute value is always a non-negative number. The absolute value of a number is 0 only when the number itself is 0. And absolute value of a product is the product of the absolute values (this actually is not totally obvious, and needs a bit of thought, I believe). That is, |ab|=|a|·|b| is always true. But addition doesn't work as well. That is, |a+b| and |a|+|b| need not be the same. Try a=4 and b=-1, for example. Then |a+b|=|4-1|=|3|=3 and |a|+|b|=|4|+|-1|=4+1=5. So some care is needed because such expressions will occur in this course and we need to know what are and are not legal manipulations.

Intervals
Suppose we want to discover what numbers x are closer to 9 than a distance of 4. Algebraically this requirement translates to |x-9|<4. We can sort of "unroll" the inequality. The absolute value will be less than 4 if the number itself is both less than 4 and greater than -4. The two inequalities can be compactly written as follows:
-4<x-9<4 which implies 5<x<13.
This is an interval and an interval which does not contain either endpoint is called an open interval. The notation for this interval is (5,13). Intervals which contain both endpoints are called closed. An examples of such an interval is [-4,6], which means the numbers x satisfying -4≤x≤6. There are also half-open intervals, unbounded intervals (with notation using + or - ∞), etc. Please see the textbook.

Warning!
If you wanted to "solve" (better: understand!) the inequality |x-9|>4 you can't just "unroll" it to -4>x-9>4. This inequality has no solutions. There is no number which is simultaneously less than -4 and greater than 4. You can't write this so compactly and using such implications represents an invalid (wrong!) method of solution.

A valid method of solution would involve separately solving the inequalities:
-4<x-9 or x-9>4 which gives 5<x or x>13. This is actually, therefore, two intervals:
(-∞,5) and (13,∞). So the inequality |x-9|>4 has a solution set which is two intervals.

The plane
The conventional way to describe the plane algebraically is to drop down two lines perpendicular to each other: coordinate axes. A point in the plane will then be described by an ordered pair of real numbers. The first coordinate will usually be called the x-coordinate and the second, the y-coordinate. This pair describes ordered distances from the horizontal (the x-axis) and vertical (the y-axis) lines. Please see the text for more about this.

The embarrassment of all this, especially with "new" students, is that (3,8) could describe both a point in R², the plane, and could also describe an open interval (with [missing!] endpoints 3 and 8). The context is supposed to help, but still the notational confusion is possible, and this is lousy.

A (non-vertical) line and its algebraic description
Suppose we wanted an algebraic description of points with coordinates (x,y) which lie on the straight line which goes through (4,3) and (8,13). If (x,y) is such a point, then look at the picture: two right triangles indicated are similar, so the corresponding sides have the same ratios:

13-y   13-3
---- = ----
 8-x    8-4

Distance in the plane, R2 Look, please, at the diagram to the right. In the plane, points correspond to ordered pairs of numbers. So a point p might correspond to an ordered pair, (x1,y1), and q might correspond to (x2,y2). Then the point (x1,y2) is the vertex of a right triangle whose hypotenuse is a line segment connecting p and q. One leg of the right triangle is on a line where all the first coordinates are x1, and the length of that leg is given by the one dimensional formula, |y1-y2|. The other leg is on the line where all the second coordinates are y2, and the length of that leg is |x1-x2|. Then by Pythagoras, the hypotenuse has length sqrt(|x1-x2|2+|y1-y2|2). And usually the absolute values are discarded since we are squaring the quantities. Therefore we officially define: dist(p,q)=sqrt((x1-x2)2+(y1-y2)2) if p has coordinates (x1,y1) and q has coordinates (x2,y2).

Example
The distance between between (3,-2) and (6,4) is sqrt([3-6]²+[4-(-2)]²)=sqrt(3²+2²)=sqrt(13).

A point (x,y) is on a circle of radius sqrt(13) centered at (6,4) (shown to the right) if sqrt((x-6)²+(y-4)²)=sqrt(13). People usually square this so it looks nicer, and the result is (x-6)²+(y-4)²=13

Functions
The word function is used in a technical sense in calculus, and is one of the most important vocabulary words. It is the logical setting for how things are transformed to other things. In the case of Math 153, the "things" are numbers. So functions will change numbers to numbers. A function is a rule changing numbers to numbers, with the important restriction that each "input" number is assigned a unique output number. You could think of a function as a machine with an input and a unique output associated to each input. The collection of all valid inputs to the machine, those inputs which don't cause the machine to break, is called the domain. The collection of all of the outputs for these valid inputs is called the range.

One example, from the toy model, already tricky!
In the toy model of paper folding I mentioned last time, we got the equation V=x(11-2x)(8-2x). Here x represented the side length of the "excluded" corner squares, and V was the volume of the resulting container. Well, the formula on the right side of the equation defines a function. Usually we write V(x) to show the dependence of V on the input value, x. So V(x)=x(11-2x)(8-2x). What is the domain of this function, V? If we just consider the formula, we'll get one answer: the polynomial has a value for all x's. The "natural domain" of a function defined by the formula x(11-2x)(8-2x) is all x. But if the formula is being used to model the physical situation, then there are definite restrictions on x. Certainly x should not be negative (cut out a square with negative sides?). And x can't be bigger than half of the smaller side length (remember that the paper's original size was 8 by 11 inches). So the domain is (0,4). Some people (many people, actually) argue that it makes more sense for the domain in this model to be [0,4]. I won't continue this discussion now, but we will come back to it. But certainly the physically reasonable domain for V in this problem is not the same as the "natural domain".
The range of V is interesting. If we allow x=0, the range "begins" at 0. How big does it get? I don't know now. So the range of V is [0, I don't know now]. The value of I don't know now is actually the desired largest volume. We'll be able to find this easily with calculus. I don't know any other way to find this largest volume easily.

A more complicated function
If a function is given by a formula, and there are no other restrictions involved, we will usually assume that the function has its natural domain, the largest collection of input numbers for which the formula makes sense. In this course, the most frequently encountered simple restrictions will be: don't divide by 0 and don't take square roots of negative numbers (this is a real calculus course!).

For example, suppose that the function f is defined by the formula f(x)=sqrt(x)+sqrt(3-2x). What is the domain of this function? Since sqrt(x) occurs, I know that x≥0. But sqrt(3-2x) is also present, so 3-2x≥=0, which is equivalent to 3≥2x which is equivalent to 3/2≥x. Both of the restrictions must be satisfied, so that the domain is the closed interval [0,3/2] (both endpoints are included).

The graph of a function
The collection of all points in the plane which correspond to the ordered pairs (x,f(x)), when x is in the domain of the function, is called the graph of the function. I like pictures, so I like graphs. Substantial evidence exists (allocation of neural resources -- brain power!) suggesting that humans can process lots of visual clues efficiently, much more, say, than many numbers or algebraic formulas.

I asked for a graph of the function we just discussed, with two square roots. Here I expected and wanted people to use a graphing calculator! A computer-drawn graph is shown to the right. Please bring your calcululators to class.
I don't know the exact range of this function. We'll come back to this question later.

Another way to specify functions
Very often in real life functions are not given by formulas.

I gave the information in the table to the right. The table declares values for certain functions (A, B, and C) at certain inputs (0, 1, and 2). For example, C(2)=4. We computed the following: C(B(1))=C(0)=-2 and B(C(1))=B(2)=5. These results are not equal. C(1)B(1)=2·0=0 and B(1)C(1)=0·2=0. These results are equal. If f(x)=B(x2+x), then f(0)=B(02+0)=B(0)=-1 and f(1)=B(12+1)=B(2)=5.

x A B C 0 1 -1 -2 1 -3 0 2 2 3 5 4

Here I wanted to stress that tabular information (the collection of "data points") was enough information to do these problems. Notice, please, that while multiplication is commutative, composition of functions is not. Composition is also written with a little circle. So B(C(1)) can be written BoC(1).

Please note that many functions arise from data points and not from formulas. Theory suggests formulas but the data points, measured by engineers and other experimenters, are reality. So don't disregard this method of specifying functions.

It is also possible to specify a function with its graph. So that's what I did.

An example
I drew a curve much like what is displayed to the right. What was drawn? It was a nice ("continuous", technical word to be defined later) curve which "interpolated" (was drawn to connect) the five "data points" (-2,4), (-1,0), (0,-2), (1,-3), and (2,-1). I told students that this was a graph of the function A (or A(x) if you want to give an explicit independent variable).

This is the graph of a function, since it passes the vertical line test: every vertical line intersects the graph at most one time. This is logically the same as requiring that every input yields exactly one output.

What is the domain of this function? That's the set of x's for which there is a point (x,something) on the graph. Look at the collection of vertical lines which touch this graph. The x's which touch or cross this graph are all x's in the closed interval [-2,2].

What is the range of this function? That's the set of y's for which the point (something,y) is on the graph. You can find these by considering all of the horizontal lines which touch or cross the graph. Notice, please, that there are horizontal lines which cross the graph more than once. This is definitely permitted by the definition of function. There can be distinct inputs which have the same outputs. The range is [-3,4].

Creating some new functions

Define the function B by the equation B(x)=A(x)+2. What does the graph of B look like? What is the domain of B? What is the range of B? Here the "+2" pushes the graph up two units. The resulting graph is shown to the right. The second coordinates of the data points are increased by 2. The domain, the input values, are unchanged. So the domain is [-2,2]. The output values, the range, are all increased by 2, and therefore the range is [-1,6].
Define the function C by the equation C(x)=A(x+2). What does the graph of C look like? What is the domain of C? What is the range of C? This is a bit trickier. The "+2" is on the inside of the function. The values that will emerge certainly will be values of A. This "+2" affects what are the legal inputs to C. The inputs to A are numbers between -2 and 2. Numbers which will be inputs to C are numbers which when 2 is added will be numbers between -2 and 2. If you can unwind this, you will see that such numbers are between -4 and 0. This graph will be the graph of C translated two units to the left. The domain is [-4,0] and the range is [-3,4].
I defined the function D(x) by D(x)=A(2x) and wanted its graph, its domain, and its range. I told students this was the QotD. What is the QotD? I'll try to give a simple computation, usually towards the end of each lecture, which I'll call the Question of the Day. Students get full credit for handing in any answer. They may work with other students, use a calculator, look at their notes or textbooks, but they must hand in some answer. Yes, this a way to take attendance. It is also a way for me to see if people understand what I am doing. This is also a way for students, if they wish, to see if they can do (correctly!) what the lecturer considers to be a "simple computation". I still need to look at the answers (more to come!).
I'll use this background color for material I didn't have time to do in class. Define the function E by the equation E(x)=|A(x)|. What does the graph of E look like? What is the domain of E? What is the range of E? This is definitely more fun than the others. The absolute value is on the outside of A. So all of the numbers which are valid for as inputs to A are going to be good inputs for the function E. If an output is negative, the absolute value flips the sign. Geometrically the numbers make the point on the graph appear symmetrically flipped over the x-axis. The result is what is shown. The domain of this function is [-2,2]. The range changes, and is now [0,4]. To me, the most interesting geometric feature of this graph is the sharp corner (?) at (-1,0). We will analyze such behavior and misbehavior in detail as part of this course.

Mr. Nakamura will discuss trig functions on Tuesday=Monday.

What is 153?
It is 151+. So what is 151 and what is + (in this context)?
151 is the first semester of a three-semester calculus sequence generally taken by students whose intended major(s) urgently need and use calculus. This includes math/stat/cs, as well as physical sciences (physics, chemistry, geology, meteorology), some of the more intricate bio majors (biochemistry, biophysics), as well as (of course!) all of the engineering disciplines. Almost all upper-level engineering courses will use calculus as a basic language. Indeed, I have been told that a beginning engineering student's performance in this first calculus course is well-correlated with eventual success in undergraduate engineering.
+ means more support is offered in this course (hence the "intensive" adjective). Our placement process advises us about the math courses entering students (who want to take math) should take. It is generally accurate in telling us about student preparation. The School of Engineering also uses other information (such as SAT and AP scores, maybe high school math courses) to help assign classes to entering students. Discussions with various people in the Engineering School led us to create 153 and offer it to students who may benefit from the additional support of another class meeting and a somewhat smaller lecture. It is my earnest hope the every student in this course will benefit from it, and will do well in their future studies.

A rough comparison
In New Jersey, a high school calculus course has 50 minute periods, nominally 180 student days: let's say 160. So the time in-class is about 8,000 minutes.
Math 151 covers most of the material of a typical year-long high school calculus course, and then grades people probably in a somewhat more demanding way. Math 151 has two lectures each week. There are 14 weeks, and the lectures are 80 minutes long. Most of the course exposition is in the lecture periods. Two lectures are lost to exams. So 13 times 80 is about 1,040, then doubled is 2,080. Therefore ...
Therefore YOU will be your most important teacher! Working in groups is good. Asking questions is good. Part of the + is extra support is the recitation, which will be used to review material we hope is known and strengthen your ability to work with that material. We will take attendance at every class session, and low attendence will definitely be reflected in the course grade. A discussion of course grading is here. There will also be a weekly workshop, as all sections of Math 151 have. Students will discuss somewhat non-routine problems in groups. Students will be required to write up a detailed solution of a selected problem and hand this in the next week. Why?
• Essentially all engineering students will participate in a group project in one or more upper-level courses, and extended written solutions will be required. Practice now, when the stakes are lower.
• Some of the workshop problems will be intentionally (!) ill-defined, and the methods needed may not be obvious. Real-world problems frequently are not given with the methods of solution, and sometimes what a solution is may not be clear!
• Many job environments have people working in teams. So get used to it now, when the stakes are lower. And maybe use these workshops to begin study sessions outside of class.

Introduction
What's calculus about? The simple answer, which we will use as initial motivation, is to learn how to find and compute lines tangent to curves and areas of regions with curved boundaries. These problems are easy to state and most people like pictures. But truthfully very few of you will have any reason to compute tangent lines and areas after calculus courses. So here's the goal of the course:

We want to teach you how to model, analyze, and, to the extent possible, solve problems involving all sorts of rates of change and accumulation.

• For example, stretch a rubber band. Hooke's Law, which describes some of what happens, declares that the amount of stretching is directly proportional to the force. Well, this means if one ounce stretches the rubber band, say, 1/3^rds of an inch, then two ounces should stretch the band 2/3^rds of an inch. That's easy. But I don't think that putting 10 tons on the rubber band would make it many miles long. The simple model has restricted validity.
• Later in the course I will show you a picture of the solubility of sodium sulfate in water as the temperature varies. The result is quite (unexpectedly to me!) complicated. What model will describe this?
• Disease spread (for example, influenza) has been extensively studied and is quite important (how much vaccine should be made, and what types, etc.). The models are fairly accurate, but ... sometimes are complicated.
• Enzyme-catalyzed chemical reactions can be fairly accurately modeled using techniques of this course. We may see some simple examples later.

A simple toy model

Equal-sized squares are cut from the corners of an 8 inch by 11 inch rectangular piece of paper. The flaps in the resulting piece of paper are folded up. What is the resulting volume which the paper object encloses?

Although this is a rather simple problem, already some interesting features appear. Notice that x=-50 or x=200 makes no sense in this problem. Also notice that V (in units of inches³) could not be -11 or 206,358 (I think!).

And now students took the diagnostic exam.

Maintained by greenfie@math.rutgers.edu and last modified 9/1/2009.