Diary for Math 151:04-06, fall 2003

The final exam for this course is on Monday, December 15, from 4 to 7 PM.

DateWhat happened (outline)
Wednesday,
November 12
Please hand in solutions to problems 4.7 #26 and 4.10 #36. If you need hints, let me know.

We had a contest between two teams of students. Each team did an optimization (finding a max or finding a min) problem. Real big prizes were given out.

The (wonderful) Optimization Contest

Problem
What is the rectangle of largest area which can be drawn inside the triangle with vertices (0,0), (3,0), and (0,2)?
Why do you know it is largest?

The team which competed consisted of Ms. Sandoval, Ms. Smallwood, Mr. Szabo, Mr. Morris, and Ms. Meleka. They all earned some glory, I think.

They sketched the picture shown. The rectangle has opposite corners (x,y) and (0,0), with side lengths x and y. Its area is therefore xy. What can we say about the relationship between x and y? The point (x,y) is on the line connecting (0,2) and (3,0), and therefore if we find an equation for this line, x and y must satisfy this equation, and this will be a relationship between x and y. One equation satisfied by both (0,2) and (3,0) is x/3+y/2=1. Therefore y=2(1-x/3). The area becomes A(x)=x·2(1-x/3)=2x-(2/3)x2. What x's will be acceptable in this problem? If we look at the picture, x should be in the interval [0,3]. So we need to maximize A(x)=2x-(2/3)x2 when x is in [0,3]. Some candidates can be found by looking for critical numbers. A'(x)=2-(4/3)x, and the only x for which this is 0 is x=3/2. Then A(3/2)=3/2 seems to be the rectangle of largest area.
Why is it largest?
I will try to give several reasons. The students needed to give only one reason!

• Look at A(x) and the domain, [0,3]. A(0)=0 and A(3)=0 by direct computation, and A(x) must attain its max at either an endpoint or a critical number. Since A(3/2) is positive, A(3/2) must be the maximum value.
• Look at A'(x) to the left and to the right of x=3/2. Notice that 3/2 is the only critical number of A. To the left of 3/2, A'(x)=2-(4/3)x is positive (A'(1)=2/3 for example) and to the right of 3/2, A'(x) is negative (A'(2)=-2/3). So A increases up to x=3/2 and then decreases. Therefore A(x) must have a maximum at x=3/2.
• Let's compute: A''(x)=-4/3 so A(x) is concave down, and we have a local max.
So there are several ways to complete this problem. Problem
What is the triangle of smallest area which can be drawn with vertices at (0,0) and on the positive x- and y- axes so that (3,2) touches the edge of the triangle?
Why do you know it is smallest?

The team which competed consisted of Sunny, Anthony, Alex, Amit, and Sonya ... perhaps they could not afford last names.

We begin with the picture shown. The triangle has height y and base length x. Its area is therefore xy/2. Is there some relationship between x and y? Look at the picture and see that (3,2) is on the same line as (x,0) and (0,y). Well, the slope of the line between (3,2) and (x,0) is -2/(x-3) and the slope of the line between (0,y) and (3,2) is (y-2)/(-3). So these slopes must be equal, and we have (-2)/(x-3)=(y-2)/(-3). Therefore y=[6/(x-3)]+2 and now A(x)=(1/2)x·([6/(x-3)]+2). What is the domain of this function? Look at the picture: x must be greater than 3.

Let's find A'(x). Since A(x)=(3x)/(x-3)+x=(3x+x2-3x)/(x-3)=x2/(x-3) we get A'(x)=[2x(x-3)-x2]/[(x-3)2], and A'(x)=0 when 2x(x-3)-x2=0, which is x2-6=0 so x=0 and x=6. Now we throw out x=0. It need not be considered since it is not in the domain of this problem. x=6 is the only critical number we need to look at, and for x=6, A(6)=12, and we have a 12 by 6 triangle, I think.

How do we know we have found a smallest triangle? Here we can again find several reasons, each of which is enough to justify the conclusion.

• Consider A(x) as x-->infinity. Since A(x)=x2/(x-3), A(x) gets large (A(x)-->infinity). Also, as x-->3+, A(x)-->infinity. Therefore somewhere in between there is a smallest value of A(x), and the only critical point is at x=6, so the critical point must be there.
• Now a reason related to A'(x). We know A'(x)=[x·(x-6)]/[(x-3)2]. In the interval from 3 to infinity, A'(x) has a negative sign for 3<x<6 and it has a positive sign for 6<x. Therefore A(x) is decreasing to the left of x=6 and is increasing afterwards. x=6 must represent a local (and absolute!) minimum for A(x).
• Now consider A''(x). I probably wouldn't, but you can compute it and then simplify the result: you will get A''(x)=18/(x-3)3. Therefore if x>3, A''(x) is positive, so A(x) is concave up in our domain, and its value at x=6 must represent a minimum.

One of the teams got the first prize, and the other got the second prize. The prizes were professionally wrapped, and were incredibly precious: historically, most Kings of England would never have been able to get (or taste) anything like these prizes!

First, in each problem there seemed to be three different ways to check that the numbers found were the desired extreme values. I would classify them as Zeroth Derivative Test (depends on the function itself!), First Derivative Test and Second Derivative Test. Of course, sometimes one "test" may be more or less useful than others, depending on computational difficulties, etc. You need to decide which of them is easiest to use.
Second the alert student will notice some similarities and some differences between the two problems done. The economists call the function to be extremized the objective function. Any relationships between variables are called constraints. The problems done above are sort of dual to each other. The constraint/objective functions almost reverse each other in the problems, as does the max/min nature. This is used in many real world settings.

I then discussed some common difficulties which I saw in yesterday's QotD. Please look at the comments there.

And on to antiderivatives, which we had already discussed in connection with the MVT. Now I discussed this very briefly again. F(x) is an antiderivative of f(x) if F'(x)=f(x). The MVT tells us that if we know F(x) is an antiderivative of f(x) in some interval a<x<b, then all antiderivatives of f(x) in that interval are F(x)+C, where C is a constant.

Therefore all antiderivatives of, say, -7x3 are (first guess, then verify, then write "+C") (-7/4)x4+C. In fact, all derivative "tables" and derivative information is, by reversing the order of looking at them, also antiderivative tables. I mentioned a few specific functions (the antiderivative of xn is (1/(n+1))xn+1+C for n not 0, and 1/x has antiderivative ln(x)+C for x positive, and 1/(1+x2) has antiderivative arctan(x)+C etc.). I actually did one problem.

A silly 6 foot tall calculus instructor throws a ball straight up into the air with initial velocity of 15 ft/sec. When does the ball hit the ground? (We assume the calc instructor has moved away from the ball's path, I guess).
Well, if H(t)=the ball's height in feet at time t, measured in seconds from when the ball started up, we know that H(0)=6 and H'(0)=15. These equations are sometimes called the initial conditions of the problem. We also know that gravity pulls the ball down. Well, then, what do we know? H''(t) is the acceleration of gravity. In the "English system" of measurement, gravity on Earth is about 32 ft/sec2. Of course here is something tricky: gravity is directed down, so in fact we know H''(t)=-32. The minus sign is rather important. If we omit it, the ball will steadily gain speed and never come down. Since H''(t)=-32, H'(t)=-32t+C. But H'(0)=15, so this C is 15. Since H'(t)=-32t+15, we know that H(t)=-16t2+15t+C. Again, the other initial condition implies this C is 6. So H(t)=-16t2+15t+6. When is H(t)=0? We used the quadratic formula: t=[-15+/-sqrt((15)2-4·(-16)(6))]/2(-16). Which root should we use? If we try the + sign, we will get a negative time (the bottom is negative .. tricky). Therefore the ball hits the ground when t=[-15-sqrt((15)2-4·(-16)(6))]/2(-16). This is (if you must!) [480+sqrt(609)]/32, or, approximately 1.239935168, sigh.

The Question of the day was: suppose that y''(x)=x2+1/(x2) and we know that y(1)=3 and y'(1)=-3. Find y(x).
The first antiderivative gives us y'(x)=(1/3)x3-1/x+C. Notice: 1/(x2 is actually x-2: you need to love exponents here! Now y'(1)=3 gives us (1/3)-1+C=-3, so C=-(7/3) I think. Now we know y'(x)=(1/3)x3-1/x-(7/3) so that (antidifferentiating again!) y(x)=(1/(12))x4-ln(x)-(7/3)x+C. But we can now use y(1)=3 and get: (1/(12))14-ln(1)-(7/3)1+C=3. ln(1)=0 at least. The equation for C becomes:(1/12)-(7/3)+C=3. Thus (!!) C=63/12. And y(x)=(1/(12))x4-ln(x)-(7/3)x+(63/12). (Yeah, yeah: that last thing is 21/4. I guess.

Notice that the circle and the parabola both go through (0,0) and both have the x-axis as tangent lines there. Should they maybe bend the same at x=0 to match up well? Can you find an equation y=stuff(x) for the lower half of the circle, and then figure out how to match it up really well?

Maple allows one to graph these things. If you type

```with(plots):
F:=a->implicitplot({y=x^2,x^2+(y-a)^2=a^2},x=-1.5..1.5,y=-.1..2,color=black, thickness=2,scaling=constrained,grid=[75,75]);
```
Then the statement F(.3) causes the first picture below to be shown (too small!), and F(.8) gives the second picture (too large!).
Monday,
November 10
Our next exam will be in one week. I urged students to look at the draft of the formula sheet for the second exam, and give me comments about it. I also asked that student look at the review problems for the second exam and, if they wish, help to assemble an answer collection for these problems.

Next I must report the presence of my evil twin, NEHPETS, who escaped from confinement and taught the rest of the class. As I later determined during discussion with several students, each class costs about \$16.07 in tuition money for each student, more than a good movie, and the efforts of NEHPETS do not deserve such payment. Refunds can be applied for.

Problem #1
So the class discussed some optimization problems. We first talked about how to find the rectangle of largest area enclosed between the curve y=1-x2 and the x-axis. Here is a picture of such a rectangle. It takes some geometrical thought to see that the largest rectangle will be placed symmetric with the x-axis with one edge on the x-axis. Once that is observed, we can take the formula for the area of a rectangle, A=length·width, and observe that, as the rectangle is presented, width (if that is the horizontal measurement) is 2x and length is 1-x2. Therefore the area will depend only on x, and A(x)=x(1-x2). Here we can describe all rectangles by taking x in [0,1]. Therefore we have now changed the problem into a "purely" mathematical one:
Find the maximum of A(x)=x-x3 for x in the unit interval, [0,1].
We developed general theory to cover this. Either the max value is achieved at one of the endpoints (x=0 or x=1) or at an interior critical point. Well, A(0)=0 (plug in x=0 into the formula) and A(1)=0 (again, plug in x=1 into the formula). Since it seems unlikely that all of the rectangles have 0 area, the max rectangle will occur at a critical point in [0,1], with positive area. But A'(x)=1-3x2, and since the derivative exists at all points, the only critical points will be where A'(x)=0. The x's satisfying this are just +\-1/sqrt(3). We can throw out -1/sqrt(3), because this number is not in our domain for this problem. If x=1/sqrt(3), A(1/sqrt(3))=1/sqrt(3)(1-(1/sqrt(3))2)=2/(3sqrt(3)), a positive value, so this is the maximum area.

I then remarked that there were theoretical ways to insure we have a local max (or a local miniumum). Here are some of them:

How to tell if you have a local maximum or minimum
The First Derivative Test
MAX VERSION If a<x0<b, and x0 is the only critical point of f in the interval, and if f'(x) is positive to the left of x0 and negative to the right of x0, then x0 is a local maximum of f.
Reason f is increasing to the left and decreasing to the right of x0.

MIN VERSION If a<x0<b, and x0 is the only critical point of f in the interval, and if f'(x) is negative to the left of x0 and positive to the right of x0, then x0 is a local minimum of f.
Reason f is decreasing to the left and increasing to the right of x0.

The Second Derivative Test
MAX VERSION Suppose that f'(x0)=0. If also f''(x0)<0, then x0 is a local maximum of f.
Reason f is concave down at x0, so x0 must be the top of a little hill.

MIN VERSION Suppose that f'(x0)=0. If also f''(x0)>0, then x0 is a local minimum of f.
Reason f is concave down at x0, so x0 must be the bottom of a little valley.

Why didn't we use any of these results here? We don't need to use either of these in connection with this problem, because we can directly compute and compare f's values. In fact, we may rarely need these results. The Second Derivative Test may seem initially appealing, but I can tell you from personal experience that computing and then evaluating the second derivative of a complicated function is rarely appealing. Also, the Second Derivative Test may unfortunately not give you any information. For example, the function x5 has a critical point at 0 but its second derivative is 0, so the second derivative test does not apply.

Problem #2
We discussed the problem of finding the shortest (or longest!) line segment touching the positive x and y axes and the point (3,2). That was a disaster, truly one of the finest efforts of NEHPETS. This is a serious problem, and it is not clear how well it was presented.

We discussed for a while how to proceed. The sketch is one I drew after the most important decision was made, which is what should the controlling variable be? (More technically, how should the situation be parameterized?) In this case, there are certainly several choices. One choice could be the angle made by the line segment with the positive x-axis. Another choice, which we adapted, is to describe the whole situation in terms of x, the first coordinate of the point on the x-axis and the line segment. I called the total length, L(x). It naturally is a sum of two "subsegments". The subsegment from (x,0) to (3,2) has length sqrt((x-3)2+4). The subsegment from (3,2) to (0,y) has length sqrt(9+(y-2)2) but we need to describe this length in terms of y. There are some similar triangles, so that (x-3)/2=3/(y-2), so that y=3/(x-3)+2, and the length of the second subsegment in terms of x is sqrt(9+9/(x-3)2). Therefore we have the interesting formula L(x)= sqrt((x-3)2+4)+sqrt(9+9/(x-3)2). More interesting is that the domain of L(x) needs some thought. The picture eventually tells us that the domain is x>3. Now our purely mathematical problem has become:
Find the minimum of L(x)= sqrt((x-3)2+4)+sqrt(9+9/(x-3)2) on the domain 3<x<infinity.
This is different from the first problem. That was more straightforward: a function to be maximized on a closed interval. Here we need to worry about various things. For example, if we changed the word "minimum" in this problem to "maximum" then we would have an optimization problem with no solution!!! We'll see that L(x) has no upper bound at all. For example, if x-->infinity, then the length of the first subsegment, sqrt((x-3)2+4), gets very large. The other subsegment can't cancel this growth, since both of them are postive. Therefore L(x) has no maximum in this domain. What happens at x-->3+, that is, x is getting close to the other edge of the domain? Now length of the second subsegment, sqrt(9+9/(x-3)2), gets large (because (x-3)2 is in the bottom of the fraction). So I now see that L(x) has some sort of behavior looking like what is shown here (a not-so-accurate sketch). Of course, we could use Maple or some other computational tool to make a better sketch, but:

• There will be times in your life that computational resources will not be available: for example, our final exam (!) or when you are stranded on a desert island.
• More likely, there will be times in your life when the available computational resources are not adequate or are expensive or ... many things: and this has happened to me.
In any case, it is useful and important to be able to do as much as possible by "pure thought".

What about L'(x)? We computed, and got something like this: L'(x)=(x-3)/sqrt((x-3)2+4)-9(x-3)-3/sqrt(9+9/(x-3)2). Some considerable thought shows that as x-->infinity, L'(x)-->1 and as x-->3+, L'(x)-->-infinity. Therefore by the Intermidiate Value Theorem, L'(x) must have at least one x0 for which L(x0)=0. That x0 (I think there is only one!) will give the shortest length. (Maple reports that there is only one number in L's domain with L'(x)=0, and that number is [approximately] 4.597222619.)
Here is a further embarrassing comment on technology (or at least my use of such!): I used Maple as I described to find the critical number. But I worked fast, or tried to! Maple has a command called fsolve which tried to find approximate roots of functions. So I typed fsolve(DL); (I had already told the program to cal the derivative of L, DL). The program returned a number near 1.4022 -- what! How could there be a minimum length inside the box? Well, I don't think there can be, but there sure can be an x in [0,1] where L'(x)=0. That's a different question. I had not thought about this. In fact, the routine fsolve allows initial guesses (think about Newton's method!) and when I typed fsolve(DL,x=4) I got the approximate root that I reported previously. Use technology carefully!

This problem is similar to numerous problems arising in physics, most immediately in connection with optics ("Snell's Law"). Sometimes you can't just "solve" in any simple way for the minimum or maximum, and numerical methods must be used for an approximate solution.

General ideas (?)
I usually try to do the following:

1. Draw a picture, as well as I can.
2. Label the picture, paying special attention to label any interesting variables.
3. Write any equations connecting the variables.
4. Try to decide what should be the "controlling" (parameterizing) variable, and writing things in terms of that variable.
5. Changing the situation to a "simple" optimization problem in one variable.
6. Look at what you've got: does it make sense?
7. Solve the optimization problem (are you sure that you have an extreme value, and the correct kind of extreme value?).
I admit that I very rarely write all the details of this process, but I honestly report that in almost all cases this is what I do. Maybe the description helps you.

Problem #47
Then I took an 8 inch by 11 inch piece of paper, cut square corners from it, folded up the edges, and asked what size corners should we cut out to get the maximum volume. This is complex, and takes some thought. I'll say that the corners are "x by x". Then the resulting sides of the three dimensional "parallelepiped" (?) are 11-2x and 8-2x and x. The volume is V(x)=(11-2x)(8-2x)x. What is the domain? The edges should all be non-negative, so the domain is [0,4]. V(0)=0 and V(4)=0, so that the max volume will be in between somewhere. The derivative turns out to be 12x2-76x+88, and the roots are (19+/-sqrt(97))/6. These are 4.80814 and 1.52519, approximately. The root of interest is 1.52519: the other one is not in the domain. Since at the endpoints the volume is 0, and this volume is positive, this volume is the max volume.
Note From personal experience, I can assure you that this problem is done correctly. I have actually cut out (paper dolls, anyone?) squares from papers, folded, etc. There does seem to be a unique largest one. I think the one identified here is that one.

The Question of the day was suppose that two non-negative numbers add up to 10. What is the maximum that the product of the square of one number multiplied by the other number could be?
Let's see: suppose x is in [0,10]. Define y by x+y=10. The problem requestions the maximum value of x2y. But y=10-x, and therefore we have a "math question":
Find the maximum of f(x)=x2(10-x) if x is in [0,10]. But the max will occur at endpoints or critical points of f inside the interval. f(0)=0 and f(10)=0 (that's is fairly clear!) so we much search for critical points. Now f'(x)=2x(10-x)-x2=20x-3x2=x(20-3x). This is 0 when x=0 (alreadly considered!) and when x=20/3. Indeed, f(20/3)=[(20/3)]2[10-(20/3)] is a positive number, so that this must be the max value.

1. I didn't use any decimal approximations, and I didn't simplify the final answer. No one asked me to do any of these things, and I am lazy (!~?). At least three students who did this problem in any optherwise correct fashion made erros "simplifying" and approximating.
2. I didn't use the First or Second Derivative Tests to check that my answer was a max (this could have been done -- both of them are applicable here). But this problem asks me to maximize a continuous function on a closed interval, [0,10], and I know if I just check the function's values at the endpoints and the critical points, then I will have "captured" the max value. If the problem involved finding extreme values on, say, the collection of x's for which x>0, then I would have to work more. For example, what is the minimum value of f(x)=3x2+5/x when x>0? And why have you found a minimum?
3. Some students answered the QotD with "x=20/3". As I wrote, this answer is not responsive to what is asked: it gives the "x" but not the "f(x)". Read and answer the question, please.
4. Some (at least three) students tried to answer this question by, in effect, computing f(x) for each of the integers in [0,10] and giving me the largest result. The question as asked did not restrict consideration to integers, and therefore this approach is not correct.

Friday,
November 7
Our next exam will be given on Monday, November 17.

Some links to help preparing for the exam

On this page are a number of links to pages which have practice problems with answers completely worked out. You can glance at material you understand well, and test yourself. You can look at worked-out problems in other areas.
Wednesday,
November 5
Please do and hand in problems 4.3 #42 and 4.5 #48 on Thursday.

We sketched the graphs of a few more functions. We began with problem #24 of section 4.5 which asked for a graph of y=ln(x4+27). First I urged people to use a graphing calculator to get an approximate idea of what the graph looks like. The result was something like what's shown at the right. We can use derivatives to confirm some of the aspects of the curve's shape.

If f(x)=ln(x4+27), then f'(x)=4x3/(x4+27). Since the bottom of this is always positive, the sign of f'(x) is determined by the sign of the top. f'(x) is positive when x is positive and negative when x is negative. Therefore f(x) is increasing when x is positive and decreasing when x is negative. 0 is the only critical point, and it is a local (and, actually, an absolute) minimum. There aren't any maxima.

If f'(x)=4x3/(x4+27), then f''(x)=(12x2(x4+27)-4x3·4x3)/[x4+27]2. Since the bottom is the square of something, it is always positive (it can't be 0!). The sign of f''(x) is determined by the sign of the top. We write 12x2(x4+27)-4x3·4x3=12x6+27x2-16x6=-4x6+12·27x2=-4x2(x4-81). This is a textbook problem, so therefore the roots are easy: 0 and 3 and -3. The function is positive for x between -3 and 0 and between 0 and 3, so that 0 is not a point of inflection. The function must change concavity at a point of inflection. Since the f''(x) is negative for x<-3 and is also negative for x>3, the graph does change concavity at +/-3, and these are points of inflection. We seem to have confirmed the appearance of the graph on the calculator.

We then discussed the rates of growth of log, and found that this function grows very slowly for large x. For example, the function we just graphed: if x>0, the tangent line slopes up. But although the function goes up and up and up (is never bounded!) it goes up really slowly. When x=10100, we approximated the slope of the tangent line, and it was really, really flat (about 4·10-100, a very very small positive number).

We also discussed the logical linage between the "pointwise" information f'(x)=0 and the existence of a local max or min. I urged students to think about certain graphs to convince themselves that knowing some numbers are 0 is not enough information to conclude that the point is a local extremum or an inflection point.
 Here is y=x4. f'(0)=0 and f''(0)=0. Although f''(0)=0, 0 is not an inflection point. You need to check the concavity on both sides of 0 (and the function is concave up on both sides of 0!). Here is y=x5. f'(0)=0 and f''(0)=0. Although f'(0)=0, 0 is not a local extremum (max or min). You need to check the behavior on both sides of 0 (and the function is increasing on both sides of 0!). f(x)=x2/3. The behavior at the origin is interesting. f'(0) does not exist, so that the function has a critical number at 0. It turns out that 0 is a local min of f (and, actually, an absolute min). But 0 is not an inflection point, although 0 is weird: the function is concave down on both sides of 0.

I think we looked at (e100x)/(1+e100x). On many plotting devices, this looks very much like the function which is 0 for x<0 and 1 for x>0, which of course is not too continuous. Such functions occur in considering physical problems whose time scales are very different from "ours", for example, small-scale chemical processes. The graph is increasing everywhere, and it has one point of inflection. The can be seen by looking at the first and second derivatives. The first derivative is [100e100x(1+e100x)-(100e100x)(100e100x)]/[(1+e100x)2] and the top simplifies just to100e100x which is always positive. The bottom is always positive, so this function is always increasing.

Then I tried to do problem #40 from section 4.5, to sketch a graph of y=cos(x)/(2+sin(x)). I urged people to first graph the function in a window of width 2Pi (the function has 2Pi periodicity). It looks like what Maple gave me on the right. And we guessed that there were two critical points, and (at least!) two inflection points. By taking the first and second derivatives we verified these claims. (More algebra and more simplification of the first and second derivatives.) If f(x)= cos(x)/(2+sin(x)), then f'(x)=[-sin(x)·(2+sin(x))-(cos(x)·cos(x))]/[(2+sin(x))2]. Again the bottom is always positive, and the top does simplify a bit: -sin(x)·(2+sin(x))-(cos(x)·cos(x))=-2sin(x)-(sin(x))2-(cos(x))2=-2sin(x)-1. (There are only a few trig identities that I keep in the top of my brain, and (sin(x))2+(cos(x))2=1 is one of them.) Now what can we say about the sign of -2sin(x)-1? I class I actually sketched a graph of this function. To me this is the simplest reasoning, because I tend to think in pictures. I learned from the graph that -2sin(x)-1 is 0 at two places (where sin(x) is -1/2) and is positive in a certain range and negative in a certain range. After some thought, we found that the derivative is positive between (7Pi)/6 and (11Pi)/6, and is 0 at both (7Pi)/6 and (11Pi)/6. Elsewhere in the interval [0,2Pi] the derivative is negative. Therefore our initial "feelings" about {in|de}crease and local extrema are correct. Since f'(x)=[-2sin(x)-1]/[(2+sin(x))2], we can compute the second derivative: f''(x)=[(-2cos(x)·(2+sin(x))2)-2(2+sin(x))(cos(x)))/[(2+sin(x))4]. This can be simplified by canceling just one power of 2+sin(x) everywhere. Then f''(x)=[(-2cos(x)·(2+sin(x)))-2(cos(x)))/[(2+sin(x))3]. The top is -2cos(x)(1+sin(x)), and the top is 0 at Pi/2 and 3Pi/2, and further analysis here will confirm completely our initial "suspicions" about inflection points and concavity. Please think about this:
We can use a graphing device to suggest what the detailed features of a graph are, and we can use the logic of calculus to confirm these features. We can constantly confirm and check for consistency.
This is very useful.

The Question of the day was the following:
Farmer Brown has a long straight stone wall. He also has 100 meters of fence which he will use to make a rectangular pen, one side of which will be the stone wall. What dimensions should he use to get the maximum area?

Here one dimension is x and the other is y. A relationship between the variables x and y (a constraint) is given by the total length: 2x+y=100. We need to maximize the area, A=xy. Then A=x(100-2x) with 0<=x<=100. At the two ends, A is 0, so the max must occur at an interior critical point, and since A is a differentiable function of x, this happens when A'=0. But A'=(100-2x)-2x. This is 0 when x=25, so that y=100-2(25)=50, and then A=50·25=1250.

We will do more problems of this type next week.

Monday,
November 3
... summer's lease hath all too short a date
For those who came to class: thank you. Today seemed like an extra summer day, a wonderful gift, and Mr. Shakespeare (quoted here from Sonnet #19) was correct. We will have months of chill and gray skies. Now to work.

Today's theme is drawing pictures of functions, especially with the help of the first and second derivatives. We know from the MVT that

 If f'(x)>0 for all x in an interval, then f(x) is increasing in that interval. If f'(x)<0 for all x in an interval, then f(x) is decreasing in that interval.

It is generally much harder to check directly if a function is increasing than to check if its derivative is positive. That's because a serious direct check of the increasing nature of a function would mean checking every pair of numbers and function values on an interval, or verifying the intended comparison in some other way. Looking at the derivative and somehow observing its sign is frequently much easier.

We also will get information from the second derivative.

 If f''(x)>0 for all x in an interval, then f(x) is concave up in that interval. If f''(x)<0 for all x in an interval, then f(x) is concave down in that interval.

 But I should give an illustration of what concavity is. A curve is concave up when it "holds water" -- at least so I was told years ago when I first learned calculus. The purpose of the first picture is to convince you that a function could be concave up and increasing, and could be concave up and decreasing. The second picture illustrates a way of diagnosing where a curve is concave up. If the slopes of the tangent lines (moving from left to right) increase, then the curve will be concave up. But the slopes of the tangent lines are given by the derivative of the function, and that's f'(x). We can insure that f'(x) itself is increasing by requiring that f''(x) be positive. And the second picture attempts to illustrate this logical connection. These pictures illustrate the "complementary" case: concave down. This, I was told years ago, "loses water" (I was never able to understand that!). Again, there is no logical implication between concave {up|down} and {in|de}creasing: one of any of them does not imply any of the others.
Remember that secant lines connect two points on the graph. Sometimes the following is useful to know:

• For concave up graphs, the tangent lines are below the graph and the secant line segments are above the graph.
• For concave down graphs, the tangent lines are above the graph and the secant line segments are below the graph.

 Concavity and {in|de}creasing behavior are not logically related. You must analyze them separately.

We first considered e-x2 and its graph. That is, I asked students to have and use a graphing calculator, and asked that this function be graphed. Mr. Morris remarked that it is a camel, but a dromedary, not a Bactrian camel.

Begin sincere sermon
This is supposed to be what the graph of e-x2 looks like. Can we confirm and make more precise some features of this graph by examining the derivatives? Please note the precise nature of the preceding sentence. I will not ignore the current date and technology (that is, the existence of graphing calculators and such programs as Maple). But you should know that features of the graphs can be confirmed and made more precise using the intellectual tools of calculus, and that sometimes (as a later example will show!) the results of "technology" can be difficult to interpret.
End sincere sermon
If f(x)=e-x2 then f'(x)=(e-x2)(-2x). The exponential function is very nice. It is never 0 and always positive. Therefore the only x for which f(x)=0 is when -2x=0. So x=0 is the only critical number. Now reasoning using the Intermediate Value Theorem says that f (which is certainly continuous!) can have only one sign for x<0 and one sign for x>0 (or else f(x) would have to have to be 0 again). We can check signs at, say. x=1 and x=-1. f is increasing in (-infinity,0) and f is decreasing in (0,infinity). Naturally 0 represents a local (and indeed, absolute!) maximum.

What information can we get from the second derivative? If we use the product and the chain rule correctly, then f''(x)=(e-x2)(4x2-2). Logic similar to the preceding asserts that this is 0 exactly when the non-exponential factor is 0. But 4x2-2=0 when x=+/-sqrt(2). Again, we can check signs in between the 0's of f'', and f will be concave up for x<-1/sqrt(2) and for x>1/sqrt(2). For x between -1/sqrt(2) and +1/sqrt(2), the graph will be concave down. The points where x=+/-1/sqrt(2) are where the concavity of f changes: these are called inflection points. These particular inflection points are related to the standard deviation, which represents dispersal from an average when this function is used in statistics.

Some examples from the textbook
I then did three more textbook problems: 4.3: #26 and #30, and 4.5 #3. I'll try to write up solutions, but I am a bit busy. The first two problems are quite "qualitative" and different from manipulating formulas.

And another example from the textbook
The third, from section 4.5, is more routine (a cubic polynomial whose first derivative is a quadratic which "accidentally" factors!). Here y=2-15x+9x2-x3.

How I prepared the high-tech class example
Then I discussed the following idea: graphing a polynomial. Of course I had chose a rather non-random (!?) polynomial. I really started with the polynomial x20(x-1)31, a polynomial of degree 51, having only two roots, 0 and 1. I used Maple to write an "expanded" form of this polynomial and called it H(x). Then I had Maple antidifferentiate it (again, for a polynomial, this is totally "mechanical") and got Q(x). I presented this polynomial to the class. I asked what its graph looked like. Notice that near, say, 1, there will be things that are almost 0 computationally: very difficult to keep track of using what is called "floating point". In expanded form, Maple, even with 20 digit precision which I requested, can't keep close enough track to actually tell what the graph looks like. We got a mess from the qualitative (and even quantitative!) point of view. The graph is actually quite simple. If the derivative is x20(x-1)31, there is no sign change passing 0 (this turns out to be a point of inflection of f) and there is a sign change (in the derivative) passing x=1, so this is a local extremum for f. In fact, checking signs, for x<1, f'(x)<0 (except for x=0) and for x>0, f'(x)>0. Therefore f is decreasing to the left of 1 (but has a horizontal tangent at x=0) and f is increasing to the right of 1: f has a local minimum at x=1. It actually isn't even that complicated: to the right is a glimpse of how the graph looks qualitatively. Follow this link to see the material I presented in class. You can even find the other inflection point of the polynomial Q(x) by taking the derivative of x20(x-1)31 and finding the root between 0 and 1.
The purpose of this example was to warn you that even very sophisticated technology can be difficult to apply succesfully.
Be careful!

The Question of the day was: suppose f(x)=xex.

1. Compute f'(x). Where is f increasing and decreasing? Does f have any local extreme points? If it does, where are they?
f'(x)=f(x)=ex+f(x)=xex=(x+1)ex. Since the exponential function is never 0, this is only 0 when x=-1, the only critical number of f. Testing the sign of f' is easy, since the exponential factor is always positive. To the left of -1, the function is decreasing and to the right, the function is increasing. Since f(-1)=-1/e, I know that (-1,-1/e) is a local minimum, and, indeed, in this case an absolute minimum.
2. Compute f''(x). Where is f concave up and concave down? Does f have any inflection points? If it does, where are they?
Since f'(x)=(x+1)ex, we see that f''(x)=ex+(x+1)ex=(x+2)ex. The only time this is 0 is when x=-2. To the left of -2 (for x<-2) f'' is negative (check the sign), so f is concave down. To the right of -2 (for x>-2) f'' is positive (check the sign), so f is concave up. When x=-2, concavity changes. Since f(-2)=-2/e2, I know that (-2,-2/e2) is the only inflection point of f.
3. Draw a graph of f. (I suggested using the graphing calculator, and then confirming the picture with what you deduced with f' and f'').

The only part of the picture I haven't completely verified is the claim that the x-axis is a left asymptote for the function. But limx-->-infinityxex=limx-->-infinityx/e-x, and this second form is infinity/infinity (look: x-->-infinity, ao the exponential is growing also). But use L'H on this, to get limx-->-infinity1/-e-x=0, verifying the asymptotic claim.

Please do and hand in problems 4.3 #40 and 4.5 #48 on Thursday.

Thursday,
October 30
(Written on the thirty-first!)
 Do the third problem on the latest workshop for next Thursday.
Wednesday,
October 29
Please hand in problems 4.1: #8 and 4.2 #18 on Thursday.

Warning! Today we will do some very tricky stuff: I repeat that this is the core of the course. Please read the textbook and do the homework problems!

The root of an equation
I asked people if P(x)=1-6x+3x2+4x3 had a root in the interval [0,1]. If we try to apply simple Intermediate Value Theorem logic we will not get enough information, because P(0)=1 and P(1)=2, so the signs of P's values at the endpoints agree, and we can't right now conclude that P must have a root. Now we will use analysis of a more interesting and powerfulkind.

P(x) is the derivative of Q(x)=x-3x2+x3+x4. Why would anyone think of looking at this Q, an "antiderivative"? Well, there is certainly no immediate reason, but maybe the pieces of P like 4x3 and 3x2 might suggest it. Now analyze Q on [0,1]. I compute easily that Q(0)=0, and that Q(1)=0 (this is a very carefully arranged example!). What could Q look like? I distinguished three cases.

1. Q could possibly (I highly doubt it!) be 0 at every point inside the interval [0,1]. In that case there are lots of points where Q'(x)=0 inside [0,1]. (This case is included to make sure that the logical analysis is complete.)
2. Q could possibly have some positive values inside [0,1]. In that case, Q will have a positive maximum in [0,1] and the maximum will be an "interior" maximum, that is, somewhere in between 0 and 1. We discussed this during the last lecture. At such a point, since Q is differentiable, Q'(x) will be 0.
3. Q could possibly have some negative values inside [0,1]. In that case, Q will have a negative minimum in [0,1] and the miniimum will be an "interior" minimum, that is, somewhere in between 0 and 1. We discussed this during the last lecture. At such a point, since Q is differentiable, Q'(x) will be 0.

In all three cases, there is at least one x between 0 and 1 which has Q'(x)=0. Since Q'(x) is the P(x) we started with, that means P(x) must have at least one root inside [0,1], as I had claimed from the beginning. You could graph P with some mysterious device, but now I am sure using "pure thought" that P has such a root (actually it has two roots inside that interval).

The root of some equations
If P(x)=a+bx+cx2+dx3 and if I know that a and b and c and d are real numbers (not further specified!) which satisfy the equation a+(b/2)+(c/3)+(d/4)=0, then P(x)=0 must have at least one root somewhere inside the interval [0,1]. Why is this? (And, of course, the numbers I used in the previous example were selected to obey this constraint.) Well, look at Q(x)=ax+(b/2)x2+(c/3)x3+(d/4)x4. This non-randomly selected function has value 0 when x=0 (because all of the terms have x's in them). When x=1, the condition a+(b/2)+(c/3)+(d/4)=0 exactly shows that Q(1)=0. Well, then, the logic displayed along with the pictures above guarantee that Q'(x), the derivative, must be 0 somewhere inside the interval [0,1]. But since Q'(x) is P(x), I have just shown the existence of the root I suggested at the beginning of this paragraph. Notice that this is more subtle reasoning than the Intermediate Value Theorem: I didn't look for a sign disagreement of P(0) and P(1), but some other features of the derivative of P(x).

Rolle's Theorem
Suppose that f(x) is a differentiable function in the interval [a,b]. Suppose also that f(a)=0 and f(b)=0. Then there must be at least one x inside [a,b] (so a<x<b) with f'(x)=0.
The picture accompanying this statement is how many people remember Rolle's Theorem. The points are at x=a and x=b and the curve is supposed to be the graph of y=f(x) and the dashed lines are two possible candidate lines satisfying the conclusion of Rolle's Theorem: they are horizontal tangents, so the slope, f'(x), is 0 there.

Rolle's Theorem tilted: the Mean Value Theorem
Suppose we rotate the Rolle's Theorem picture a bit. We would get something like what is shown. So some version of a statement of this situation could be this:

 Take two points on the graph of a differentiable function, and join these points by a line. There always will be at least one line tangent to the graph in between these points which will be parallel to this line.

It turns out that this statement is true (a detailed discussion of how to "rotate" to get a verification from Rolle's Theorem is in the text). This statement is a version of the Mean Value Theorem which is one of the two major results of calculus, and the MVT (as I will call it) will contribute to everything we cover in the course from now on. For example, it turns out the MVT can be used to estimate errors in such processes as Newton's method, and, generally, estimation of errors in most of the calculus algorithms I know, when they are really used with numbers begins with the MVT.

A better statement of MVT
I will restate the MVT in a more algebraic fashion. For example, the word "parallel" usually is implement by looking at the slopes of the lines involved. If the two points on the graph of f are, say, (a,f(a)) and (b,f(b)), then the slope of the line connecting these points is [f(b)-f(a)]/[b-a]. The slope of a line tangent to the graph when x=c is f'(c). (My choice of letters is designed to agree with tradition and with your text.) Here is an algebraic way to phrase the MVT:

The Mean Value Theorem
Suppose f is a function differentiable on the interval [a,b]. Then there is at least one point c inside this interval so that f'(c)=[f(b)-f(a)]/[b-a].

Meeting Francine again
We first met my pal Francine on September 17. Francine drives on the Garden State Parkway. Now suppose we know the following. She leaves Cape May (Mile 0 on the GSP) at 7 Am in the morning. At 9:30 AM the same day, she arrives at the other end of the parkway (Mile 172, somewhere in or near Mahwah or Ho-Ho-Kus. There is now enough evidence to convict her of speeding. Why is this? Let s(t)=her position in miles along the GSP with Cape May as the origin, and suppose that t is measured in hours starting at 7 AM. We then know that s(0)=0 and s(2.5)=172. Take a=0 and b=2.5 in the MVT. Then we know that there must be a c between 0 and 2.5 so that s'(c)=(172-0)/(2.5-0)=68.8. The GSP has varying speed limits on different portions of the road, but the current highest speed limit is 65. Since 68.8>65, Francine must have been speeding at some time. Note that this conclusion is slightly unattractive from a "practical" point of view, since it doesn't local or approximate c very much ("Judge, at some time between 7 AM and 98:30 AM she really, really, really was speeding ..."), but we have mathematically verified a violation.

Speed limits
I didn't make these remarks in class, but I should have. If Francine were driving on a highway with a maximum and a minimum speed limit, then the MVT actually gives information about future positions. In this context the statements come close to being trivial, but let me analyze them anyway. Suppose Francine is driving on a restricted access highway, and at 2 in the afternoon, she is at, say, milepost 75. Also suppose the road has posted a minimum speed limit of 40 and a maximum speed limit of 60. If Francine obeys the law, where could she be at 5 in the afternoon? The MVT tells us that [s(5 PM)-s(2 PM)]/[5-2]=s'(t). But s'(t) must be between 40 and 60. So [s(5 PM)-75]/3=something between 40 and 60. Then clearly (well, maybe it is almost clear) that Francine's lowest position at 5 PM is 195 (40·3+75) and Francine's highest possible position is 255 (60·3+75). I guess this seems all sensible, but it is proved, as far as I know, only by using the MVT.

Guessing a function given its derivative: consequences of MVT
Suppose a function f(x) is defined in the interval 0<x<23, and we know that f'(x)=4x6+(5/x4. What can you say about f(x)? Well we thought about this, and after a while a student guessed that f(x) could be (4/7)x7-(5/3)(1/x3). I didn't want to know how the student got this answer, but I did want to know how to check the answer. Of course, it can be checked very easily by differentiating. And it works. I then asked if there were another answer. I was told (4/7)x7-(5/3)(1/x3)+1. How about another? Well, look at (4/7)x7-(5/3)(1/x3)+sqrt(38) or (4/7)x7-(5/3)(1/x3)-.0003, etc. Are these all of the answers? It turns out that all of the answers are (4/7)x7-(5/3)(1/x3) plus some constant. Historically this was not obvious. The simplest physical situation this would refer to is that a velocity is given for a particle moving on a line. The successive positions all differ by a constant, and that constant is always determined by an initial position. The background for believing that we always get one function plus a constant in answer to these sorts of questions is the following:

Fact 1, a consequece of the MVT
Suppose f is a function differentiable on an interval, and that f'(x)=0 for all x in that interval. Then f(x) is constant in the interval.

This is true because if x1 and x2 are different points in the interval, f(x2)-f(x1)=f'(c)(x2-x1) by MVT, and f'(c) has to be 0 so that f(x1)=f(x2), so f always has the same values. No if two functions have the same derivative, their difference will have derivative 0: for example, if F(x)=x2 and G(x)=x2 then (F-G)'(x)=0, so F-G is a constant. Therefore, if we can somehow guess at one F(x) -- in this case, not too hard, since I can (1/3)x3 -- then all F's and G's with this derivative must be (1/3)x3+Constant. These are called antiderivatives of x2.

If derivatives are positive in an interval, we also can learn something. Suppose f'(x) is positive in an interval, and x1<x2 are points in the interval. Then the MVT says f(x2)-f(x1)=f'(c)(x2-x1). Since we are assuming f'(c)>0 and that x2>x1, the right-hand side of the equation is positive, so the left-hand side must be also. Therefore as we move to the right on the number line, the f-values get bigger. Such a function is called increasing.

Fact 2, a consequece of the MVT
Suppose f is a function differentiable on an interval, and that f'(x)>0 for all x in that interval. Then f(x) is increasing in the interval.

There is a "dual" statement. If the derivative is negative, then the function values go down as we travel from left to right.

Fact 3, a consequence of the MVT
Suppose f is a function differentiable on an interval, and that f'(x)<0 for all x in that interval. Then f(x) is decreasing in the interval.

These two statements are useful because determining the signs of functions can be substantially easier than telling that a function is increasing. For example, if f(x)=x15+6x3+17x+cos(5·sin(x)), then f(200) is less than f(300). This is true because f'(x) is 15x14+18x2+17-sin(34·sin(x))·(5(cos(x))), and the powers are even (so always non-negative, and sines and cosines are between -1 and +1 and the 17 overpowers the sin(34·sin(x))·(5(cos(x))) which must be between -5 and +5: the derivative is always positive, so f(200)<f(300).

Sketching graphs and MVT
Here's the graph of a function:

 The first challenge is: what does the graph of the derivative of this function look like? Let's progress from left to right on the graph of the function we were given. The function is decreasing until x=A. The slopes of tangent lines in that region are negative, so the derivative graph will be negative. At x=A, the tangent line has zero slope, and after it, the tangent lines begin to have positive slope. They continue to have positive slope until x=B, and after that they have negative slope. In this graph, the original function is in green. The derivative of this function is in black. Now let's try to sketch a graph of a function whose derivative graph is the graph we were given. This is harder. First, Fact 1 of the MVT implies that there are many such graphs. As soon as you have one (such as (1/3)x3 would be for x2) you would actually have many (such as (1/3)x3+5 and (1/3)x3-2). Physically we are given velocity varying over time, and to get position we need to specify an original position. Here I will ask that the graph of the antiderivative (which is what we are asking) go through the blue dot. Then we use facts 2 and 3 which are consequences of MVT: in an interval where the derivative is positive (such as the interval from the left until x=C) the antiderivative function must increase. Since the original function is 0 at x=C the function we are now sketching must have a horizontal tangent line at x=C. Now progress to the right. The original function is negative, so by Fact 2 from x=C to x=B the new function must decrease. An interesting phenomenon occurs around x=B in the new function. It must have a horizontal tangent because the original function is 0 there, but to both sides the original function is negative, so the new function must decrease on both sides of x=B. This sort of "step" (it looks locally like -x3) is an inflection point. In fact, there are two inflection points. The other one is at x=A when the new function changes from concave down to concave up.

Here is all of the information combined. Things are rather complicated. There are three graphs involved: a function and its first and second derivatives. I have tried coloring parts of the background, and describing features.

On the graph of the function, the letter M indicates a local maximum. The I's indicate inflection points, where the graph changes concavity. The function increases when its first derivative is positive, and it decreases when that first derivative is negative. The information about the second derivative's sign shows up in the concavity of the original function. The original function is concave up when its first derivative is increasing, so that the tangent lines lie below the graph. When the first derivative is decreasing, the tangent lines lie above the graph of the function, and that function is then concave down.

What is likely to be confusing is that in this exercise we began with a graph of the first derivative, then sketched the second derivative, and then sketched a candidate for the original function. I will go into further details on this next time. Please read the text!

The Question of the day was: if you know that f(2)=3 and f'(x)=x2-2x, what is f(5)? Well, we can guess one f(x): (1/3)x3-x2. It is easy to check by differentiation that this is a correct guess. Then by logic similar to what we did earlier, any f with the specified derivative must be (1/3)x3-x2+a constant. What constant should it be? Since f(2) is supposed to be 3, we know that (1/3)23-22+a constant=3, so that the constant must be 3-(1/3)23+22. Then f(x) is exactly (1/3)x3-x2+3-(1/3)23+22 so that f(5) is (1/3)53-52+3-(1/3)23+22. This is the way I would leave the answer, but if you insist on "simplifying", it becomes (125/3)-25+3-(8/3)+4=(117/3)-26=39-18=21. (Wow, to get an integer from such a random problem!)

This is what is called an initial value problem. A simple physical interpretation is that I give the velocity and an initial position at time 2, and then ask for the position at a later time.

Monday,
October 27
The math computer systems were not working for most of the afternoon, so I am late getting this diary entry written. I thank those students, some looking nearly drowned, who came to class today. Neither the weather nor an impending physics test stopped them. I also wish everyone taking the test tonight good luck.

Please do as many of the homework problems in chapter 4 as you can. This material is the heart of the course.
The major themes in this chapter are

Max-min problemsCurve sketching using derivative information

Mathematicians have made a living for almost three centuries using what we're going to discuss!

I began by going over a few more L'Hopital's Rule examples.

1. What is the limx-->infinity[ln(x)]/[x1/300]?
Before using L'H we must check that some appropriate set of hypotheses are satisfied. Here certainly ln(x) gets larger when x gets large "enough" and so does x1/300. That is, the limit of the top as x-->infinity is infinity, and the bottom has the same limiting behavior. It isn't hard, by the way, to see that both the top and the bottom get large. Try a big number (?), like 10100. Then ln(10100)=100ln(10) is approximately 230, and (10100)1/300=101/3 is about 2.15. Well they do get big, but they get big "slowly" and this limit is essentially asking which of the two, the top or the bottom, gets bigger slower.
In any case, this is an indeteriminate form of the type, "infinity/infinity" and we can try to use L'H.
limx-->infinity[ln(x)]/[x1/300]=(using L'H)=limx-->infinity[1/x]/[(1/300)x(1/300)-1]. Let's get rid of the compound fraction in the last expression. We then need to evaluate limx->300x-1+1-1/300. I think I did the exponents correctly, and the result is the limit as x-->infinity of a constant (300) multiplying a negative (-1/300) power of x. The result is certainly 0.
Therefore we conclude that the limit is 0 and x1/300 gets bigger faster than ln(x)! But what about the numbers? We just didn't take numbers that are big enough to see what's happening. Try (10100)100. A little bit of juggling with exponents will show that ln((10100)100)=10,000ln(10) which is about 23,000. And ((10100)100)1/300 is (10100)1/3 which is about 1033. And it is true that 1033 is much much bigger that 23,000. Logs grow more slowly than any positive power of x. (See problem #67 in section 4.4.)
2. I looked again at limx-->infinity[x300]/[ex].
We had done this last time, but I wanted to repeat it, to emphasize the point. You can check that this is again eligible for L'H (again, it is infinity/infinity) and then 300 uses of L'H get us the limx-->infinity[300!]/{ex. The top is a HUGE constant, but it is a constant, and the bottom gets ever larger. The result is 0. Exponential growth is faster than any polynomial growth. (See problem #66 in section 4.4.)
The results comparing log and polynomial and exponential growth are very important in many applications, and should essentially be in your head when you think about growth and decay.
3. I think this is problem #9 of section 4.4: limt-->0(5t-3t)/t.
If I "plug in" t=0, the indeterminate form 0/0 appears, and this makes the problem eligible for L'H. I need to know the derivative of at. I can memorize it (d/dt(at)=atln(a) or if clever, I can deduce it (look: at=(eln(a))t since exponential and logarithm are inverse functions, and then (eln(a))t=e[ln(a)]t, and realize that ln(a) is just a constant, so that d/dt{e[ln(a)]t}=e[ln(a)]t·ln(a)=at·ln(a) by the chain rule).) So we compute: limt-->0(5t-3t)/t=(using L'H)=limt-->0(5tln(5)-3tln(3))/1, and we can evaluate this limit by just plugging in: the answer is ln(5)-ln(3).
4. While the other three examples we fairly realisitic, here is a problem I would only expect to see in a calculus class or a text or an exam: what is limx-->0xsin x?
This is a weird problem. Look at it: x-->0 and sin x-->0 also. It isn't clear to me what happens to the result of exponentiating one with the other. I would like to use L'H, and I need to somehow algebraically change this function into something that looks like a candidate for L'H: that is, something like TOP/BOTTOM which, as x-->0, will look like, say, 0/0 or infinity/infinity. There are really several ways to do this problem. I'll show one strategy here.
If y=xsin x, then ln(y)=ln(xsin x), and then ln(y)=(sin x)ln(x). Now as x-->0, certainly ln(x)-->-infinity ("look" at the graph of ln(x) you should have installed in your head!) and certainly sin x-->0. Can I somehow change (sin x)ln(x) into a quotient? Well, here is a way: (sin x)ln(x)=[ln(x)]/[1/(sin x)]. Those of you who object to this should be advised that I'm doing it because it works which is about the best answer which can be given at this stage (I'm showing it to you because someone showed it to me and that person was shown it by ... all the way back maybe to the Marquis de L'H himself, maybe). Anyway, the fraction [ln(x)]/[1/(sin x)] as x-->0 does look like -infinity/infinity, so I will use L'H, and get (with some care, now, doing the derivative of the bottom!) limx-->[1/x]/[{-1/(sin x)2}·cos x]=limx-->0[(sin x)2]/[x·cos(x)]. This last fraction is again 0/0 as x-->0. Do L'H again, and get limx-->0[2(sin x)(cos x)]/[cos(x)-x·sin x]. All of these darn derivatives need care: for example, the derivative of the bottom uses the product rule. Now what happens as x-->0? The top is again 0 (it is 2·0·0) but the bottom is 1. So the result is finally, 0. But wait! Is it? Only if you are very alert will you notice that we have found limx-->0ln(y), and that it is ln(y) which has limit 0. We can undo ln with an exponential. So y itself has limit exp(0)=1. Wow.
NOTE I don't want to distress you but Maple's response to the command limit(x^(sin(x)),x=0); is the answer 1. Are human beings obsolete?
A preliminary version of what your text calls "Fernat's Theorem" is something like this: At a local maximum, the tangent line to a graph is flat. Well, let's see: I guess "flat" means the line is horizontal, so the derivative is 0 (I should hastily add, if the derivative exists: functions can have corners and other "interesting" (?) features. The other part of this statement of Fermat's Theorem involves the term, "local maximum". I want to spend a little time on that, and try to insure that we have a shared understanding of this term.

Suppose a function f is defined on an interval I, and p is a point inside I. Then p is called a local maximum of f if f(p)>=f(x) for all x in I: that is, f's value at p is at least as big as f's values nearby. The definition of local minimum just changes f(p)>=f(x) to f(p)<=f(x). Together, the local maxes and mins of a function are called the local extreme values, or local extrema of the function. All of the examples we present to learn about local maxima can be changed to examples about local minima if we multiply the functions by -1.

So let's look at some examples. Please note that all of the examples were accompanied by pictures but for various reasons I can't provide pictures here at this time.

• f(x)=-x2. Where does f have local maximums? (In this case and in all that follow, the assumed domain of the function will be "all numbers for which the function makes sense", so for this f the domain is "all real numbers".)
The graph is familiar, an upside down parabola with "top" at (0,0). As one student remarked, the function squares numbers and then puts on a - sign, so the largest value it can have is 0, and other values are negative. The only 0 of this f is at 0, and all other x's have negative values of f. So 0 is the only local maximum of f.
• f(x)=384. Where does f have local maximums?
This is a somewhat silly example, but again agreement on language is very important in what follows. The graph of this f is a horizontal line, and the interesting fact is that every number is a local maximum (and is also a local minimum -- weird but correct!).
• f(x)=x. Where does f have local maximums?
This function, on its natural domain (all real numbers) has no local maximums and no local minimums. Because if p were a local max, then to the right of p there is an x with f(x)=x>p=f(p). So p is not eligible.
• This f(x) is defined piecewise. So f(x)=x if -1<x<1, and f(x)=0 for all other numbers. Where does f have local maximums?
A picture of this function's graph should be drawn. There's a tilted open interval (no endpoints included) going from (-1,-1) to (1,1). The rest of the graph are two closed (including endpoints) horizontal half lines. All the points with x<=-1 are local maxes, and so are the points with x>1 (but not 1 itself, because immediately to its left are higher points on the graph).
Please notice that the tilted interval does not have any local maxes. A student asked after class why the number which is somehow "immediately before" 1 is not a local max, since it would be the highest point on the tilted line. Well, this is an interesting and subtle question. Suppose, say, that "Fred" is the number which is immediately before 1. Then thedistance from 1 to Fred is 1-Fred. What if we halve the distance? We would get the number 1-Fred/2. And f's value at this number is larger than Fred, so Fred can't be a local max. I think we really showed that such a Fred can't exist: there can't be a number immediately before another number! The real numbers themselves are rather weird.
• f(x)=x3-x. Where does f have local maximums?
Again it is best to draw a graph of this f as we have several times before (!). The local extrema will occur where the tangent line is horizontal if we believe the "Fermat" result quoted above. So look where f'(x)=3x2-1 is 0. This is when x=+/-1/sqrt(3). And a local max occurs at -1/sqrt(3).
 Commercial breakSponsored by the Newton-Leibniz Company, purveyors of fine ideas This is really easy. If we didn't have calculus it would be awkward and difficult. Easy is better than hard. Good morning! We send you now back to your regular lecture, with all our sympathy!
We discussed a more precise form of Fermat's result.
Fermat's Theorem Suppose f is defined in an interval I, and p is inside I. If f has a local extremum (a local max or local min) at p, then either f'(p) does not exist or f'(p)=0.

I tried to show why, if f'(p) does exist, it would have to be 0. Here I drew a picture (I always draw a picture whenever possible) If f has a local maximum at p then consider the difference quotient, [f(p+h)-f(p)]/h. If h is a small positive number, this quotient is <=0. You can see this geometrically by looking at the picture, or algebraically by realizing that f(p) is at least as large as f(p+h) and the sign of the quotient can't be negativ e since h is positive. If h is a small negative number, then the graph shows you that the slope of the secant line is <=0, and so does examining the algebra(because h is negative now). Therefore f'(p) is the limit of things that are both <=0 (from the right) and >0 (from the left). The only value f'(p) can have if these limits are to agree is 0. Whew: this argument is worth following since the result is used very often.

Important examples, important definition, and a restatement
I mentioned that the function |x| has a local minimum at 0, but that the derivative of this function doesnot exist at 0 (there's a "corner" at 0). A function f has a critical point at p if either f'(p) does not exist at p or if f'(p) exists and f'(p)=0. So |x| has a critical point at x=0 and x3-x has critical points at x=+/-1/sqrt(3). We restate Fermat's Theorem: local extrema occur only at critical points. This is compact and most people like this restatement.
Warning Look at just f(x)=x3 alone. What are f's critical points? Since f'(x)=3x2 for all x, we just need to look for where 3x2=0, which is only at x=0. But f(0)=0. (I realize it may be difficult to keep the references clear to all of the functions and derivatives floating around: I am sorry.) But to the left of 0 f has negative values, so 0 can't be a local minimum. To the right of 0 f has positive values, so 0 can't be a local maximum. So the critical point 0 of this function is not a local extremuim. The converse of Fermat's Theorem is not valid: a critical points may not be a local extremum!

I then asked how we could determine the maximum and minimum values of a function f defined in an interval a<=x<=b. Ms. Farley with correctly with great precision:

 Given: f defined on [a,b]. Wanted: f's max and min values on this interval. If these values occur where a

I drew some pictures which showed how the max and min values could occur at any of the candidates mentioned by Ms. Farley's algorithm.

The Question of the day was: find the maximum and minimum values of f(x)=x3-2x2+7 on the interval -1<=x<=2.
How did expect people to solve this? I remarked that I did not ask for a graph, and a graph is not necessary. f'(x)=3x2-4x, and the roots of this are 0 and 4/3. These are the critical points of this function and they are inside the interval [-1,2]. Therefore the maximum and minimum values of f must be on the list f(-1), f(0), f(4/3), and f(2). Since f(-1)=4, f(0)=7, f(4/3)=5+(22/27), and f(2)=7, the min value is 4 (achieved at -1) and the max value is 7 (achieved at both 0 and 2). A picture of a graph of this function on the interval [-1,2] is shown to the right.
By the way, the answers 4 and 7 are not the same as the answers -1 and 0 and 2. If someone asked a farmer, how much will each of your watermelons weigh at the end of the growing season, and the answer given was, "August 28" I believe all of you would agree that the answer was not too useful. Please understand the question and try to respond to it.

I had a perfectly splendid QotD involving a pasture along the Raritan River with a bull in it, but people shouted at it.
Please notice I have fallen behind the syllabus, and I will discuss the Mean Value Theorem next time.

Please hand in problems 4.1: #8 and 4.2 #18 on Thursday.

Wednesday,
October 22
More happy volunteers computed some derivatives: Ms. Smallwood and Mr. Gehrmann and Mr. Bruno, who had the assistance of Mr. Rodriguez.

Discussion of exam results
As I wrote in the diary when I returned the first exam:

 Please realize that there will be another exam (100 points) and a final (200 points) and more points from the QotD, textbook problems, and workshop problems. I urge students to study and work for improvement.

Here is what you should know at this stage:

1. Newton's method is a way to (try to) improve a guess at a root of f(x)=0 when f is a differentiable function.
2. The formula for going from an old guess, G, to a new guess, N, is N=G-[f(G)/f'(G)].
3. The geometry of the Newton's method iteration is that the line tangent to y=f(x) (a line of slope f'(G)) is drawn at the point (G,f(G)), and the next guess, N, is obtained by sliding down this tangent line until it hits the x-axis. That intersection is the new guess, N.
A caution
Newton's method can give some surprising answers. It definitely should be used, since under good circumstances it converges very rapidly, but you should be warned that it can also misbehave.

An example
Here is one example of the use of Newton's method with quite a simple function, selected to show some problems. I hope the analysis will not be too hard to understand. If you try a random polynomial you will probably get very messy results, much more difficult to understand and analyze than this example. But the example does show behavior that does happen in general.
The function is f(x)=x/(1+x2). Let's look at the derivative: f'(x)=[1·(1+x2)-2x(x)]/[(1+x2)2]. Here it is worthwhile to simplify, since I would like to work with f'(x). The result is f'(x)=(1-x2)/[(1+x2)2]. The Newton's method iteration for going from an old guess, G, to a new guess, N, is to take G to G-f(G)/f'(G). In this particular case, N=G-f(G)/f'(G) is G-[x/(1+x2)]/[(1-x2)/[(1+x2)2]]. The second term is a compound fraction. The expression can be simplified quite a lot with some algebraic work. In fact, for going from an old guess, G, to a new guess, N, the formula is: N=(2G3)/(G2-1).

Also notice that y=x/(1+x2/) has a root at exactly one value of x, x=0. What I would like to do now is color the points of the curve green if starting from that point the sequence of Newton's method iterations converges to the only root, 0. I would like to color the points red if the iterations do not converge to 0. I would also like to discuss any weird behavior that occurs.

 Here is a picture (qualitatively correct) of x/(1+x2). Probably geometry may be more of a help to us than the algebra alone. In fact, they should be used together, to reinforce each other. The curve drawn is symmetric with respect to the origin. It has pieces only in the first and third quadrants. So the colors I draw should reflect that symmetry. The analysis I present here is more complicated than what I said in class. That was overly simple (!) and I made errors. I apologize! So here we go:

 A green point, then some more green points Certainly, starting with G=0 gets N=0, so that the root itself is green. Are there other green points? Well, drawing a few pictures near O should convince you that if you try starting near 0, the Newton's method iteration rapidly approaches 0. The picture here tries to show that with a few iterations. Some red, then some more red points Suppose we try x=10 as G, our initial guess. Then the formula gives N=(2G3)/(G2-1). How does N compare with G? G seems fairly "large", and the top has twice G cubed, while the bottom only has G squared. I bet that the new guess, N, is to the right of G. In fact, the top of the bump (the maximum, we will officially call it on Monday) is located at (1,1/2) (just see where f'(x)=0 -- remember that f'(x) is (1-x2)/[(1+x2)2], so it is 0 only when the top, 1-x2, is 0, and that is at x=+/-1. So 10 is far to the right. And if G=10, N is even larger, more to the right. For me, looking at the picture is much easier than trying to understand the algebra. And if you look to the right of the top bump, you can see that the tangent lines will all slant "down" (have negative slope) and that therefore for guesses which are bigger than x=1, the new guesses are being driven away from 0, the root. So all of the points on the curve which are bigger than 1 must be red. Of course, a symmetric conclusion is reach for points less than -1, where the iteration drives guesses to the left. Weird things or not so weird at all The algebraic formula N=(2G3)/(G2-1) isn't so good when G=+/-1. We are not supposed to divide by 0. What happens in the picture? The tangent lines at those points are horizontal, and they never intersect the x-axis. So maybe I should color those two points some special color just to show that the red/green distinction doesn't apply to them. I guess I will color them brown, a silly color for a silly point. More red in towards 0 Suppose now we look at the curve a tiny bit to the left of where x=1. Then the tangent line has very small positive slope, and will hit the x-axis on the far left. Above that point, the curve is red, and redness spreads contagiously backwards so the point we started with is red. So some part of the left-hand side of the top bump is red, and similarly, by (anti)symmetry, something on the right-hand side of the bottom bump must be red. Brown: how can there be more brown? Keep traveling left on the top bump, in the red stuff. The next new guess will begin traveling right on a part of the curve where x<0. Eventually, though, the line will intersect the x-axis at x=-1. Do you believe me? well, let us solve N=(2G3)/(G2-1) when N=-1. Then 2G3=1-G2 (by cross-multiplying) and so we need to find numbers for which 2G3+G2-1=0. There is exactly one such number (verify this on your own graphing calculator, please) and its approximate value is .657298, so I think since this number is shoved into -1 it should be brown also. And so should its negative, since that gets pushed into +1. And more red Push to the left from the previous point. Then the Newton's method iteration takes us into the red region directly to the left of the point on the curve where x=-1. And then, if you can imagine it, further to the left we come to a point which gets pushed into the second brown point on the left. So that point should be brown also, and then there should be a part of the curve which is red and then ... This is very complicated. There are actually on each side of the curve infinitely many intervals of red, and each red interval has brown points on its edge, so there are also infinitely many brown points. The red intervals are getting shorter and shorter, though, until: Something new happens If you look at the picture closely, you will see that there is a positive number G so that when we compute N and then compute N again, the two tangent lines are parallel! For this point Newton's method does not go to +/-infinity and does not go to 0: it oscillates repeatedly. We can even compute this number. By symmetry, at this point, N should be -G, so the equation N=(2G3)/(G2-1) becomes -G=(2G3)/(G2-1) and if we cross-multiply we will eventually get 3G3=G. One root is G=0 which gives us nothing new. The other roots are G=+/-sqrt(3). So I have shown these two points in pink: pink for periodic. In general, Newton's method does have points that sort of wander around, and don't converge to anything, and don't "escape" to +/-infinity. Newton's method can be very complicated. I have also indicated the behavior of infinitely alternating red strips and brown points with !? because I (and all other people!) can't draw this very well. This behavior is an example of a fractal set.

 This picture tries to display a completely colored graph. The green points are attracted to the root, 0, by Newton's method. The red points go out to +/- infinity (alternate strips of red go to alternate infinities [here by infinity I just mean eventual travel steadily out to the left or the right]), and each red strip is bordered by brown points which eventually under iteration don't go anywhere (!): they get sent to the wastebasket, outside of the domain of Newton's method. The region labelled !? is too hard to draw, with infinitely many alternations of color. Finally, the two pink points represent points which are alternated with each other periodically by Newton's method. This is already complicated. A "random" function would likely have even more complicated behavior.

I think I began with consideration of a limit like this one: limx-->0[e4x-1]/sin(3x). We had some limits like this in problem 2 of workshop #5 but here I would like to show you a more organized way to compute these limits. Of course I begin my consideration of the limit, which is made up of familiar functions, by just plugging in x=0 in the hope that the functions, which are individually continuous, will be continuous near and at 0. Since this is an invented example, we get 0/0: no hope. We can do the following. Remember that f(a+b)=f(a)+f'(a)b+Err·b. Well, make a into 0 and make b into x. Then f(x)=f(0)+f'(0)x+Err·x where Err-->0 as x-->0. Suppose we try this with f(x)=e4x-1, the "top" function inside the limit above. Since f'(x)=4e4x, we can compute f(0)=e0-1=1-1=0 and f'(0)=4. Then f(0)+f'(0)x+Err·x becomes 0+4x+Err·x. What about the bottom? Here f(x)=sin(3x) so f'(x)=cos(3x)3, and f(0)=sin(3·0)=0 and f'(0)=cos(3·0)3=3. The bottom changes into 0+3x+Err·x. Everything is working out so neatly, of course, because this is an arranged example, but the example was "arranged" so that it illustrates a method which is useful often.

Then limx-->0[e4x-1]/sin(3x) becomes limx-->0[0+4x+Err·x]/[0+3x+Err·x] which (factoring out an x everywhere) is limx-->0[4+Err]/[3+Err]. The error terms are different, but qualitatively they have this common property: as x-->0, both error terms-->0. Now we see "easily" that the limit is 4/3.

The preceding discussion is not a proof, but merely presents heuristic evidence that the result below is true. ("heuristic" is defined in the October 6 diary entry, and it is the appropriate word here.)
L'Hopital's Rule (version 1)
Suppose that f and g have continuous derivatives, and that f(a)=g(a)=0. If limx-->af'(x)/g'(x) exists, then limx-->af(x)/g(x) exists and is the same value.

Warnings
L'Hopital's rule (abbreviated now by L'H is very powerful and also very easy to misuse. One quote I found on the Internet declared, "Giving l'Hopital's Rule to a calculus student is like handing a chainsaw to a three year old."

1. L'H resembles the way we might like to differentiate quotients in our dreams. Please try not to get confused:
• L'H is a method to evaluate limits.
• The quotient rule is a method to compute derivatives.
2. You must check the hypotheses of L'H. A simple limit like limx-->0[x^2-3x-4]/[x3+5x-6] is not equal to limx-->0[2x-3]/[3x2+5], the limit of the quotient of the derivatives. The first limit can be evaluated by direct substitution, due to the continuity and nonvanishing of the denominator: it is 4/6. The second limit can be evaluated similarly and is -3/5. These are not equal. You must check the hypotheses and in work submitted for this class you must show that you have checked them.
I did a few more L'H examples, some fairly ludicrous. Please look at the text for more examples. More serious examples arise frequently in computer science: given various procedures for doing a task, which is likely to be faster. An example of the kind of decisions that need to be made is in problem 5 of workshop #3: which sorting algorithm is faster.

To analyze this kind of limit, we usually look at limx-->infinity. It turns out that L'H works for many different "indeterminate forms": 0/0 is one, and infinity/infinity is another. There are more (please see your text, and see there for more formally stated versions of L'H). Here is one limit which may still look ludicrous, but which I assure you is actually realistic.

What is limx-->infinity[x300]/ex? In order to supply information which may inform or, more likely, distract you (!) I evaluated the top and bottom at x=2. 2300, I said, was about 10100. I was again incorrect, since it is only about 1090. This is 1 followed by 90 zeros. How about ex at x=2? In class I was told "Less than 9", and, indeed, it is about 7.3. So the quotient is approximately 1090/7.3, which looks pretty darn big to me. As x-->infinity, but the top and the bottom -->infinity. I will use L'H. Therefore I have to evaluate the limit limx-->infinity[300x299]/ex (the exponential function is really wonderful). Well, what happens as x-->infinity? Again, this is the indeterminate form infinity/infinity. I try again to use L'H and get limx-->infinity[300·299x298]/ex. Nothing works. Or does it? In fact, if we can imagine using L'H another 298 times, we get limx-->infinity[300!/ex. A few things about the top: if you differentiate a polynomial long enough, it eventually is a constant. And the 300th derivative of x300 is a constant, and it is a constant with some structure: the product of the positive integers from 300 down to 1. Such numbers arise very frequently in calculus and in many many other applications. So frequently that they have their own notation: !. The exclamation point, in context, is read, factorial (it is not read, "excitement mark"). 300! is quite large: about 10614. But it is a constant so that this limit is 0. The actual numbers are all a distraction from the true fact:

The exponential function eventually
grows more rapidly than any polynomial!

The Question of the day was: compute the limit as x-->infinity of [(ln(x))2]/sqrt(x).
As x--> infinity, ln(x)-->infinity and sqrt(x)-->infinity, and so we have an indeterminate form of the infinity/infinity type. Use L'Hopital's rule and try to find the limit as x-->infinity of the quotient: [2(ln(x))·(1/x)]/[1/(2·sqrt(x)]. I generally don't like simplifying if it is only to present a final answer, but this is an intermediate step in a complicated computation and it is a compound fraction, which is ugly to me. So if you change this to a simple fraction you will get [4(ln(x))·sqrt(x)]/x. This itself can be made simpler by getting the powers of x together: there's an x1/2 on top and an x1 on the bottom. So the result is [4(ln(x))]/sqrt(x). As x-->infinity, this still is an indeterminate form of the type infinity/infinity. I will use L'Hopital's rule again to get [4·(1/x)]/[1/(2·sqrt(x)]. This simplifies to 8/sqrt(x) which I see directly -->0 as x-->infinity. You can't use L'Hopital on this last limit since it deson't satisfy the hypotheses.
There are other ways to do this computation correctly, but I emphasize the word "correctly". Unsubstantiated assertions that something -->0 or -->infinity are not useful.
This QotD was supposed to illustrate the following fact:

The logarithm function eventually
grows more slowly than any power of x!

Please hand in tomorrow 3.11:13 (yes, I know this is an odd problem but you need to show how you got the linear approximation and then show some evidence that the interval given is valid) and 4.4:22.

A historical note
L'H is named after Guillaume François Antoine, Marquis de l'Hôpital (1661 - 1704), who published it in his book Analyse des infiniment petits pour l'intelligence des lignes courbes (1692), recognized as the first calculus textbook. It seems, though, that the "rule" and other results were due to Johann Bernoulli, who was paid well by the Marquis. Bernoilli only complained about the lack of attribution to him after the death of the Marquis. See: math for money, isn't it easy?

 I spent some time giving blood after Wednesday's class so I only finished these notes on Friday. Giving blood doesn't really hurt and is not even very inconvenient, and one feels good after doing it. I recommend it if you can do it.

Monday,
October 20
Three happy volunteers (Ms. Brundage and Mr. Guzman and Mr. Stivers) differentiated three of the functions in the review problems for chapter 3. I recommended that people who were not confident about computing derivatives of functions defined by formulas show try a bunch of these problems.

Now I considered again a remarkable equation.

 [REMARKABLE]   f(a+b)=f(a)+f'(a)b+Err·b

This equation was first stated in our class on September 24 which now (to me, at least!) seems like a l-o-n-g time ago. It resulted from unrolling the formal definition of derivative. The most important qualitative aspect of this equation is that Err-->0 as b-->0, so that the term Err·b should go to 0 faster than first order.

A silly example
In this example, f(x)=x2. Let us look at f(a+b). In this case we can compute f(a+b) exactly. It is a2+2ab+b2. We can "match up" the terms here with parts of the REMARKABLE equation above. So f(a)=a2, and since f'(x)=2x, the term f'(a)b is exactly 2ab. The "Err·b" term is b2, and certainly when b is small, this gets smaller faster than b. A numerical example is something like (5.01)2 which we compute exactly as 25.1001, and here a=5 and b=.01. This splits up as 25+.1+.0001=52+2(5)(.1)+(.01>2, and the result of omitting the last term is a very small error.

The official name
The use of f(a)+f'(a)b in place of f(a+b) is called the linear approximation to f at a. There are several reasons it is called the linear approximation. First, algebraic formulas (here thinking of b as the variable) which only involve constant and first degree terms are called linear. Second, consider the tangent line to y=f(x) at x=a. The slope of that line is f'(a). It goes through (a,f(a)). If we increase a by an amount b, the line's height is increased by f'(a)b (because OPP/ADJ is f'(a) and ADJ is b in the right triangle shown). So the linear approximation gives a formula which instead of using the correct value of f(a+b) substitutes the value obtained by "extrapolating" using the value on the tangent line.

Another example, perhaps less silly
What is the square root of a number close to 1? Since square root is a continuous function, such a value of square root should be approximately the square root of 1, which is 1. But how should we "correct" the answer if we want sqrt(1+x) for x small? Here we have f(x)=sqrt(x) and a=1 and b=x. So we need f'(1) which is easy, since f'(x) is (1/2)x-1/2 and f'(1)=1/2. Therefore the linear approximation to f(a+b), that is, to sqrt(1+x), is just 1+(1/2)x. Now I tried to do some arithmetic. The following table is somewhat more extensive than I computed in class with student help. Here my help was Maple and the computations were done to 30 digits accuracy. (!)

Values of x"True" values of sqrt(1+x)Linear approximation values3rd column - 2nd column
1110
1.51.2247448713915890491 1.25-0.0252551286084109509
1.051.02469507659595983831.025-0.0003049234040401617
1.0051.00249688278817106751.025-0.0000031172118289324
1.00051.00024996875781005941.00025-0.0000000312421899406
1.000051.00002499968750781231.000025-0.0000000003124921877
1.0000051.00000249999687500781.0000025-0.0000000000031249922
1.00000051.00000024999996875001.0000005-0.0000000000000312499

Richard Hamming, a very famous applied mathematician of the 20th, declared that

The purpose of computing is insight, not numbers.

So what can we learn from these numbers? Actually, we can observe some characteristics which will always be true, although giving a good argument for them will have to wait a week or two.
• Consider the number of zeros in the first column and in the last column. In the first column, the transition from one number to the next is gotten by "adding" one 0. So the a+b becomes a+(b/10). The last column shows the discrepancy, the difference, between the linear approximation and the true value. Look at the number of zeros. In each step (and this is true data!) the number of zeros increases by two, so while b changes to b/10, the difference gets divided by 100. In other words, the term "Err·b" is actually sort of a constant multiplied by b2: it is a second-order term. This is generally true.
• What about the sign in the last column. It is always negative. Why is that? The curve lies under the tangent, which means that the linear approximation will be an overestimate of the true value.
I did one more numerical example with this function. So what about f(101)? The "true value" of sqrt(101) is about 10.04987, and the linear approximation based at a=1 is 51. The difference is enormous: don't use the linear approximation when the "b" is big. (In fact, if you needed to use a linear approximation for sqrt(101), try a=10. Then this more appropriate linear approximation gives 10.05

I was asked if I could give a quantitative estimate for the error, and again I muttered that I will need to wait a week or two but I would then be able to show one.

One more example
Finally, I asked what happened to the three-halves powers of numbers near 9. Here f(x)=x3/2, so f'(x)=(3/2)x1/2. If a=9 and b is a small number, we can hope that (9+b)3/2 is close to f(9)+f'(9)b. Since f(9)=27 (hey, I chose 9 because the 3/2's power of 9 is "simple") and f'(9)=(3/2)(3)=4.5, we now know that (9+b)3/2 is approximately 27+(4.5)b. This may or may not be interesting to you, but if you need to really think about many values of x3/2 near 9, this approximation provides some insight. I also can tell you that this approximation is an overestimate of the true value, since x3/2 curves "up" and the tangent line at (9,27) is under the curve. A picture that Maple drew of this situation is to the right. We will soon learn how to tell if the tangent line is underneath or over the curve without looking at the graph.

An implicit function example
Look at the curve in the plane defined by the points whose coordinates satisfy the equation xy4+x2y=6. One point on this curve is (2,1): check me by substituting the numbers into the equation. Now I think there is a point on the curve close to this one with x-coordinate equation to 1.96, so it will be (1.96,?). I would like to find an approximate value for ?. I think the curve is continuous, so an approximate value is about 1, but I can do better. The x-coordinate is "perturbed" by a delta&nbps;x which is -.04, and I would like to know the corresponding linear perturbation in y. The linear approximation idea says that the multiplier will be dy/dx. Let's compute this, using the ideas Mr. Stivers used in his example. So we d/dx the whole equation xy4+x2y=6 and obtain 1·y4+x·4y3y'+2x·y+x2·y'=0. We can solve for y' and we will get: y'=-(y4+2xy)/(4xy3+x2). At the point (2,1) this gives us y'=-5/12, so that delta y=y'·delta x becomes delta y=-(5/12)(-.04). Therefore the new value of y is approximately 1+(5/12)(.04).
Note that the "true" new value for y is about 1.0167, and our approximation is 1.0166, an error less than 1/10,000.
Could you find the points where the tangent line to this curve is horizontal? Where it is vertical?

Magic?

I asked how a calculator computes square roots. Most of the answers I got were, "By pushing the button," but I repeated my question. Here are my suggested answers.
1. Since World War II smaller and smaller genetically modified people have been raised. I think in '47 the two foot high person was the standard of excellence in this direction, and then in '53, the first 1/4 inch person ... and then in the 70's really teensy people were created and put inside calculators and they just used clever guessing to get square roots. They were also put on super-amphetamines to work really fast.
2. If your calculator can manipulate 10 decimal digit numbers, then (since each number position can hold one of 10 choices of digits) the calculator can only be asked to find the square root of 1010 numbers. These square roots were all computed in 1975 in a massive research project similar to sequencing the human genome. The cost was about 47 billion dollars to compute all these square roots, but the task was finally done when Horace and Myrna Whiteapple of Frog's Throat, Georgia, announced ten digits of the square root of 37: 6.0827625302.
3. The calculator uses ancient Babylonian mysteries to compute each square root very efficiently.
Well, no one believed answer #1 although it was very dramatically presented. Sigh. And #2 was (correctly!) criticized because there wasn't enough storage space (and actually it would take rather a lot of time to compute those square roots). The correct answer is actually #3.

Babylonian square roots
People were smart even long ago. Here is a method which was used to approximate square roots about four thousand years ago.

 Finding the square root of A Take an initial guess, B. Replace your initial guess by [B+(A/B)]/2, the average of B and A/B. Continue step 2 until you are satisfied with the accuracy of your approximation.
This looks quite weird. Well, is it? First, if your "guess" is too large (so B2>A) then B>sqrt(A) and (B/A)>(1/sqrt(A)) so that (A/B)<sqrt(A). We are averaging numbers that are on either side of sqrt(A). So maybe we will get closer. The amazing result is that we get very much closer, very fast. Here is an example, again computed by Maple: I wanted to compute the square root of 2. As is well-known (but not by me!) this is 1.4142135623730950488 to 20-digit accuracy. My first guess, A, was 3. Here is a table of next guesses using the Babylonian method, and the difference between the guess and the true value of sqrt(2):

 Next guess 1.83333 0.41911977096023828453164460912 Next guess 1.46212 0.04790764974811707241043248791 Next guess 1.415 0.00078486752170790299608172641 Next guess 1.41421 0.00000021767410253511833468993 Next guess 1.41421 0.00000000000001675206944769266 Next guess 1.41421 0.000000000:27 consecutive 0's!

Notice how rapidly the sequence of guesses converges to sqrt(2): here the number of 0's in the difference between the true value and the guesses seems to double at each step (this is true in general, which is why this method is so nice to use).

I don't know how the Babylonians came up with this method, but I can give you an explanation in the context of this course. Consider the function f(x)=x2-2. Certainly f(x)=0 when x=+/-sqrt(2). So how can we get a sequence of good guesses to the positive root of f? Suppose we have a guess, and then try to improve it. Let's suppose our guess is at x=G. Then the coordinates of the corresponding point on y=x2-2 are just (G,G2-2). The line tangent to y=2-2 at x=G has slope 2G (the derivative of x2-2, which is 2x, at x=G). Therefore OPP/ADJ=2G, so that the adjustment we make to our guess G is to subtract ADJ. But OPP is G2-2, soafter some algebra we see that ADJ is (G2-2)/(2G), and, if we call our new guess N, N=(old guess)-(adjustment)=G-(G2-2)/(2G). Now some more algebra: G-(G2-2)/(2G)=[2G2-(G2-2)]/(2G)=[G2+2]/(2G)=[G+(2/G)]/2. That is, the new guess is the average of G and 2/G, which is exactly the Babylonian method described above with A=2.

As far as I know this is the method which is used for square roots in most computational devices. It is rapid and easy to program, and the error analysis (not presented here) is not hard.

Newton's Method
The Bablyonian square root technique can be generalized to other functions. Try an initial guess, and then improve the guess. A diagram similar to what is above can be drawn for y=f(x), and the improvement method can be written as follows:

Newton's Method  N=G-(f(G)/f'(G))

This "scheme" of forming a new guess from the old is called Newton's Method.

The Question of the day was to explicitly write the "new guess from old" for 21/3, the cube root of 2, and compute the first new guess with 3 as the original guess.
What's the answer? The simplest choice for f(x) seems to be x3-2. Then f'(x)=3x2. So Newton's method becomes: N=G-[(G3-2)/(3G2)]. If G=3, then N turns out to be 56/27, or about 2.074074074. The "true value" of the cube root of 2 is 1.259921050. Here are the first 6 iterations of Newton's method with starting value 3 for this function (the correct digits are boldfaced):

2.074074074, 1.537690539, 1.307076219, 1.261601801, 1.259923288, 1.259921050

The sixth iteration turns out to be correct to 12 decimal places (so the error is less than 10-12).

WARNING! Newton's method certainly can produce weird or wrong results. I'll show some of them on Wednesday. It must be used with some care.

 History and Current Practice The history (from the Wikipedia) "Newton's method was discovered by Isaac Newton and published in Method of Fluxions in 1736. Although the method was described by Joseph Raphson in Analysis Aequationum in 1690, the relevant sections of Method of Fluxions were written earlier, in 1671." Current practice There has been evolution and change in how square roots have been computed. At times in the last few decades, specific hardware has been designed to use Newton's method. The "latest" broadly marketed chip is the Intel 64 bit "Itanium". Here (load carefully: 131 page PDF file!) is something only a true fanatic would want to read. It is the Intel description of how reciprocals and square roots are calculated. The title is Divide, Square Root, and Remainder Algorithms for the IA-64 Architecture. Square root routines in assembly language (software, not hardware!) are described on pages 40 through 70. The introduction says that "iterative methods such as the Newton-Raphson ..." are used, but I found that I needed lots of work to recognize Newton's method in the assembly language routines. The programs are very optimized, so, for example, the repeated nature of the "old guess to new guess" transition is not called as a subroutine, but is written out in detail for a fixed number of repetitions. It is all quite complicated. This is only the assembly language programming. There are references to specific Intel documents proving that the suggested programs give the correct results. There have been many reports of errors in Intel computation, of which this is the most prominent. Perhaps it is better to implement algorithms in "software" because then in the event of errors, the programs can be changed, rather than having to trade in chips because the circuits are wrong.

Thursday,
October 16
Another workshop.
Please hand in a writeup of the third question.
Wednesday,
October 15
These problems are due tomorrow in workshop: 3.8 #50 and 3.10 #24.

I reviewed the derivatives of arcsin x: this is 1/sqrt(1-x2); and arctan x: this is 1/(1+x2); and ln x: this is 1/x.

If y=ax is an exponential function, we could take log (really "ln") of both sides and get ln y= x ln a. Then we could differentiate both sides, remembering the chain rule. We would get (1/y)(dy/dx)=ln a (remember that a and therefore ln a are both constants!). Therefore dy/dx=y·ln a so that the derivative of ax is ax·ln a.
If we were interested in the derivative of 10x we would be looking at 10x(ln(10)). ln(10) is approximately 2.303 (this is enough for most purposes I know, but if you need more digits, here they are: 2.3025850929940456840). In computer science people sometimes consider 2x whose derivative is 2x(ln(2)). ln(2) is approximately .693 (again, enough information for most purposes I know, but ln(2) is .69314718055994530942 [more or less!]). The nicest and simplest ax for calculus purposes is one where the derivative is simplest, where ln a is 1. And that occurs when a=e. So

Of all the exponential functions,
ex is best
because it is its own derivative.

Finally I started on today's topic, related rates. This is essentially applications of implicit differentiation and the chain rule to geometric and physical situations. I just did a few problems.

Example I
Suppose the edge of a cube is growing so that at the time the length of the edge is 2 cm, it is growing at .03cm/sec. How fast is the volume of the cube changing at that time? Here we need to name several variables: e will be the length of an edge of the cube, and V will be the volume of the cube. And we need to "encode" in calculus the further information given. That is something about the rate of change of the volume with respect to time, t, measured in seconds. We are told that de/dt|e=2=.03 (this is read as the rate of change of edge length with respect to time when the edge length is 2). We need to find dV/dt|e=2. Here we have a simple equation relating e and V: V=e3. If we d/dt this equation, being careful to pay respect to the chain rule, we get dV/dt=3e2de/dt. When e=2, we get dV/dt=3(22)(.03). By the way, what is the rate of change of the surface area of the cube at this time? If S represents the surface area, then since a cube has 6 square faces, S=6e2, and dS/dt=12e(de/dt), so that (dS/dt)t|e=2=12·2·.03.

Example II
Here is a slightly more realistic example. A raindrop loses water through evaporation at a rate which is directly proportional to the surface area of the drop. Suppose a raindrop is spherical, and that at the moment the drop has radius .02 cm, it is losing moisture at the rate of -.003 cm3/sec. How fast is it losing moisture when the radius is .005 cm? Here we need either some nice intuition (to be discussed later) or some formulas. The needed formulas are: S, the surface area of a sphere of radius R, is 4PiR2 (the area of 4 "great circles" -- this can be verified using calculus). If V is the volume of a sphere of radius R, then we have been told in the previous green sentence that dV/dt=kS where k is a constant (presumably negative so the raindrop is shrinking). Here we have when R=.02 that dV/dt=-.003, and also kS is k(4Pi(.02)2), so that k=-.003/(4Pi(.02)2). I didn't simplify for reasons which will become apparent soon. We want to know dV/dt when R is .05. But for this we just need kS=-.003/(4Pi(.02)2)·4Pi(.005)2. Golly, the 4 and the Pi cancel, and in fact all we get is -.003(1/4)2. The original rate gets decreased by a factor of 1/16. Probably we could have predicting this, since the rate is directly proportional to the surface area, and the surface area is directly proportional to the square of the radius, so decreasing the radius by a factor of 1/4 causes the rate to change by a factor of 1/16. The whole situation is just juggling various rates, and the derivatives in this case seem almost to obscure the situation instead of making it better.

Example III [Different from what I did in class!]
Revisiting Fred and Jane, bugs. We had considered these bugs earlier. Now here Jane will again travel on the curve y=sqrt(x) so that at time t, her first coordinate will be t. And Fred travels so that his second coordinate at time t will be t, but now Fred travels on yx3 (in class I had Fred traveling on the right half of the parabola y=x2 and this was a bit silly as one student pointed out since it led to a rather simple limit). What happens to the slope of the line connecting Fred and Jane as time gets large? Jane's position is (t,sqrt(t)). Now Fred's position is a bit more interesting. It is (something,t) where something3=t in order for Fred to be on y=x3. Therefore something=t1/3. so that Fred's position is (t1/3,t). What is the slope of the line joining Fred and Jane? It is (t-t1/2)/(t1/3-t). What happens as t-->infinity? Suppose we divide top and bottom by t. We get the equal fraction (1-(1/t1/2))/(1/(t2/3-1). Now as t-->infinity, the terms 1/t1/2 and 1/t2/3 both -->0 since they have positive powers of t "underneath". So again the limiting slope is -1.

Example IV
Here we consider a moving bug named Egbert. Egbert travels on the curve x2y-xy4=2 and when he (?) is at the point (2,1) his horizontal velocity, dx/dt, is 1. What is dy/dt at that time? This became the Question of the day. Indeed if we d/dt the blue equation carefully (using the chain rule and the product rule!) we get 2x(dx/dt)y+x2(dy/dt)-(1·y4+x·4y3(dy/dt))=0. Now "insert" x=2 and y=1 and dx/dt=1. It is possible to solve for dy/dt first, and then insert values, but I generally find it simpler if the numbers are put in now. So we have: 2·2·1·1+22·dy/dt-(1·14+2·4·13·dy/dt)=0. This leads to dy/dt=3/4. (I think!)

The graph that is displayed was generated by Maple and shows x2y-xy4=2. We can almost (?) check the answer, because on this curve as Egbert travels, dy/dx=(dy/dt)/(dx/dt) (the chain rule again!), and so dy/dx=(3/4)/1=3/4. Egbert at the point (2,1) should be traveling slightly tilted up, and indeed he is (look at the graph). The line shown is y=(3/4)(x-2)+1, the tangent line to the curve at the point (2,1).

The exams come back ...

I returned the exams. Here is a discussion of the results. Please realize that there will be another exam (100 points) and a final (200 points) and more points from the QotD, textbook problems, and workshop problems. I urge students to study.
Sunday (!),
October 12
About 40 to 50 students showed up. I tried to outline the course so far:
• "Review" of precalculus: functions, graphs, analytic geometry
• Limits: the definition, graphical interpretation, "intuition", algebraic manipulation to detect limits.
• Continuity: definition, graphical interpretation, Intermediate Value Theorem, continuity of familiar functions
• The derivative: a rate of change, slope of a tangent line. Formal definition. The derivative as a function.
• Differentiation: the algorithms and the deriviatives of the common functions.
I spent some time going over review problems and inventing "new" problems. I hope the review session was helpful.
Wednesday,
October 8
I asked Ms. Brundage to do problem 21 in section 3.4: find an equation for a line tangent to y=tan(x) at the point (Pi/4,1). We need to know the slope of the tangent line, and that slope is the derivative of tan(x) at x=Pi/4. But tan's derivative is (sec(x))2, so the slope needed is (sec(Pi/4))2 which is equal to 2. You can check (and you should know!) the wonderful special triangles. Therefore the line desired is y-1=2(x-Pi/4).

Mr. Fredo and Mr. Lopes volunteered to do problem 13 of section 3.5: find dy/dx if y=cos(a3+x3). Here the trick and the difficulty is the inclusion of the symbol, a. The conventional understanding, unless specifically advised otherwise, is that unidentified letters represent constants relative to such variables as x. Therefore the chain rule implies that dy/dx=-sin(a3+x3)·3x2.

Mr. Cedeno happily did problem 19 of section 3.5, which is a double use of the chain rule after a use of the product rule. Here y=(2x-5)4(8x2-5)-3. The outermost (?) part of this formula is a product, so we use the product rule:
dy/dx=d/dx((2x-5)4)·(8x2-5)-3+(2x-5)4·d/dx((8x2-5)-3). Now each "d/dx" uses the chain rule on a power function. The answer to the whole problem will be:
dy/dx=4(2x-5)3(2)·(8x2-5)-3+(2x-5)4·-3(8x2-5)-2(16x).
It may be easy to get lost inside the maze of these derivatives.

Finally, Mr. Pankaj Bajaj did the following problem: suppose y is implicitly defined as a function of x by the equation y·cos(Pi·xy)=1/3. Find dy/dx at the point where x=1/2 and y=1/3. In fact, cos(Pi·(1/2)(1/3))=cos(Pi/6)=sqrt(3)/2. So I really should make the right-hand side of that equation sqrt(3)/6. (Another error of the lecturer!) The intent was to get the equation differentiated implicitly, and that Mr. Bajaj did, using the product rule and then the chain rule and then the product rule:
(dy/dx)·cos(Pi·xy)+y·(-sin(Pi·xy)[Pi·y+Pi·x(dy/dx)]=0.
We then insert the values of cosine and sine given by the special triangles which all students should know. We get: (dy/dx)·sqrt(3)/2+(1/3)·(-1/2[Pi·(1/3)+Pi·(1/2)(dy/dx)]=0. We can solve for dy/dx now.

I remarked:

1. There would be a review session in our regular classroom on Sunday, October 12, at 4:30 PM.
2. There is a review sheet of problems which will be gone over in workshop tomorrow.
3. I would like to urge students to read the questions carefully, and write their answers with some care. I would like to read answers to the questions I actually ask, and not imaginary or invented questions. I cannot read students' minds about what they would or should have put on papers.
4. I have now graded workshops for all three sections. Some students are clearly getting the idea, and can write good narratives to accompany their computations. Some do not write such statements, and therefore might be at a disadvantage when I ask for sentences describing an answer on the exam.

Some links to help preparing for the exam

Here are some pages which have practice problems with answers completely worked out. You can glance at material you understand well, and test yourself. You can look at worked-out problems in other areas. And now on to more, newer and possibly interesting material: well, it is still fairly technical, enlarging the collection of functions we can differentiate. Implicit differentiation lets us find derivatives of lots of inverse functions.

arcsin
The text calls this function sin-1 and some other sources call it asin. I find the "-1" superscript awkward, and it sometimes confuses me when I compute since I get it confused with "one over ...". So I will call the function inverse to sine, arcsine, and abbreviate its value at x as arcsin(x).

What's happening in the pictures (left to right):
The first picture is supposed to be a portion of the graph of sine. It is 2Pi periodic, and its range is [-1,1]. The green line is the "main diagonal", y=x, which also happens to be tangent to y=sin(x) at (0,0). This is because the slope of the tangent line is the derivative of sine, which is cosine, and cos(0)=1. To get the inverse function, we interchange inputs and outputs. Geometrically we flip the graph over the main diagonal, and get the second picture. The tangent line is still tangent, but now, look at the red line. This demonstrates that the flipped graph is not the graph of a function. It fails the vertical line test to be a graph of a function. Thus we need to cut away (!) part of the graph. The "clouds" in blue-green (?) demonstrate what will be cut away. And what's left is shown in the third picture. This is the official graph of y=arcsin(x): domain [-1,1] and range [-Pi/2,Pi/2]. It has arcsin(0)=0, and the tangent lines seem always to slope up, so the derivative should be positive. And if we are ver careful to note it, the lines tangent to sine at +/-Pi/2 are horizontal, so the lines tangent to the flipped curve will be vertical and have no slope so there will be no derivative at +/-1. Consider this process:

1. y=arcsin(x)   The inverse function
2. sin(y)=sin(arcsin(x))   Using the inverseness
3. sin(y)=x   Recognition of inverseness
4. cos(y)(dy/dx)=1   Implicit differentiation
5. dy/dx=1/cos(y)   Solving for dy/dx
6. Since sin(y)=x, x2+(cos(y))2=1, and so cos(y)=+/-sqrt(1-x2). Which sign to take? In the range we are considering between -Pi/2 and Pi/2, cosine is positive. Also the slope of the tangent line is positive for arcsine. Thus we will take the + sign. And cos(y)=sqrt(1-x2).   Solving for dy/dx stuff in x
7. Thus arcsin'(x)=1/sqrt(1-x2)   Statement of formula
We will go through this a few more times today. But in any case, the derivative is 1/sqrt(1-x2). Notice that for -1<x<1, this is positive, and that the derivative formula is not valid at +/-1 as we predicted from the geometric evidence.

Example: The derivative of arcsin(5x3-ex) is [1/sqrt(5x3-ex)](5·3x2-ex) using the chain rule.

arctan
The text calls this function tan-1 and some other sources call it atan. I will call the function inverse to tangent, arcsine, and abbreviate its value at x as arctan(x).

What's the picture supposed to show? The initial picture is y=tan(x). This function is periodic with period Pi, and its domain does not include odd multiples of Pi/2. The function is rather simple looking (!), always tilted up, and has vertical asymptotes at odd multiples of Pi/2. Flipping to get an attempted inverse function reveals lots of problems (I omitted the red line here). The standard restriction is to throw out the "branches" that don't intersect the horizontal axis, and that's what I've attempted to suggest with the blue-green "clouds". Again, y=x is a tangent line to both arctan and tan at (0,0). Arctan is very useful. It "compresses" all of the real numbers into the interval from -Pi/2 to Pi/2, so if you have lots of data and you don't know ahead of time how big (+ or -) the data will be, composing it with arctan will at least control it a bit. Now for the derivative.

1. y=arctan(x)   The inverse function
2. tan(y)=tan(arctan(x))   Using the inverseness
3. tan(y)=x   Recognition of inverseness
4. (sec(y))2(dy/dx)=1   Implicit differentiation
5. dy/dx=1/(sec(y))2   Solving for dy/dx
6. Since tan(y)=x, x2+1=(sec(y))2 (much easier than arcsine!).   Solving for dy/dx stuff in x
7. Thus arctan'(x)=1/(1+x2)   Statement of formula
Example: The derivative of arctan(e(5x4)) is 1/(1+(e(5x4))2· e(5x4)·5·4x3: somewhat of a mess, along with several uses of the chain rule.

The inverse function to exp
Should this be called arcexp? Well, it isn't. The exponential function, ex, models exponential growth. It "takes" x and multiplies e's x number of times (can you understand that statement?). The inverse function to an exponential function is a logarithmic function. This logarithm function is very important. The log functions you may deal with include log10 (used in the definition of pH, and for hand calculation) and log2 (used in some computer science applications). For reasons that will appear very soon, the log function which identifies how many powers of e appear in a number is called the natural log. Your text abbreviates it as ln, and that's what I will use, but note that many sources call it just log. What's ln? It is a function whose domain is all positive numbers and whose range is all real numbers.
This picture shows exp, the exponential function, ex. Since this function is one-to-one, its inverse will be a function: no more blue-green clouds! The green line is the tangent at (0,1) which has slope=1. Then we flip it and get the graph of ln. What about the derivative?

1. y=ln(x)   The inverse function
2. exp(y)=exp(ln(x))   Using the inverseness
3. exp(y)=x   Recognition of inverseness
4. exp(y)(dy/dx)=1   Implicit differentiation
5. dy/dx=1/(exp(y))2   Solving for dy/dx
6. Since exp(y)=x, we're done! This is even easier still   Solving for dy/dx stuff in x
7. Thus ln'(x)=1/x   Statement of formula
I then differentiated y=xsin(x) by doing this: ln(y)=ln(xsin(x))=sin(x)ln(x), so that we can d/dx the equation and get: (1/y)(dy/dx)=(product rule!)cos(x)ln(x)+sin(x)/x, so that dy/dx=y[cos(x)ln(x)+sin(x)/x]=xsin(x)[cos(x)ln(x)+sin(x)/x].

Next time: ax, and why e is the best of all possible a's.

The Question of the day was extremely and painfully stupid. For the first time, I think the majority of the students did not do the question correctly.

I asked students to find dy/dx if y=arcsin([ln(x+3)/arctan(ex)]). This was a terrible and overelaborate concatenation of new functions, and I regret that I asked something so complex.

Let me try to "disassemble" this function. For example, what is the derivative of arctan(ex)? We must use the chain rule. The "outer" function is arctan, and the inner function is ex. Therefore the derivative of arctan(ex) is 1/[1+(ex)2]·ex.

Now what is the derivative of ln(x+3)? This is perhaps a simpler use of the chain rule, and this derivative is 1/(x+3)·1.

What is the derivative of ln(x+3)/arctan(ex)? Here the new ingredient is using the quotient rule. So the derivative of this quotient is
[(1/(x+3))·arctan(ex)-1/[1+(ex)2]·ex·ln(x+3)]

(arctan(ex)2

Now continue to go backwards. We are asked for the derivative of arcsin(ln(x+3)/arctan(ex)? We use the chain rule, and place ln(x+3)/arctan(ex) into the argument of the derivative of arcsin, and multiply by the expression we found above for the derivative of the quotient. I think that the answer is:
(1/sqrt(1-[ln(x+3)/arctan(ex)]2))·([(1/(x+3))·arctan(ex)-1/[1+(ex)2]·ex·ln(x+3)])/(arctan(ex)2.
From the derivative of arcsine               From the quotient rule

I hope this is comprehensible and again I apologize for throwing such an absurd thing at people. See you Sunday, if you would like to be there. Or else Monday for the first exam.