Thursday, July 20
I repeated what we learned about concavity last time. Then I remarked
that we'd use this to sketch some graphs, probably in more detail than
many sane human being would need or want.
Example 0:
f(x)=(1/4)x^{4}+(1/3)x^{3}x^{2}+1
FUNC
We've seen this function several times. The function tells me that the
domain is all x. The positive coefficient, 1/4, on the top degree
term, x^{4}, which has an even exponent, tells me that
lim_{x>+infinity}f(x)=+infinity and
lim_{x>infinity}f(x)=+infinity. The precise range is
unclear (until we analyze the first derivative).
FUNC´
Now f´(x)=x^{3}+x^{2}2x. This factors easily to
x(x+2)(x1). There are critical numbers (where f´(x)=0 or doesn't
exist) at 0 and 2 and 1. The critical points (substitute the critical
numbers into the original function!) are at (0,1) and (2,5/3) and
(1,7/12). So now I can tell that the range of the original function is
[5/3,inifnity) and (0,1) is a local max and (1,7/12) is a local min
and (2,5/3) is a global or absolute max. The function increases in
the intervals [2,0] and [1,infinity) and decreases in the intervals
(infinity,1] and [0,1]. I can see this by looking at the "heights"
(the second coordinates) of the critical points and the limits at
+/infinity found before. Or I can look at f´(x). Where is it
positive and where is it negative? The same answers will appear.
FUNC´´
So f´´(x)=3x^{2}+2x2. If we use the quadratic
formula then the roots are 1/3+/sqrt(7)/3. These roots show
where the concavity of the graph of y=f(x) changes. For
(infinity,1/3sqrt(7)/3] the graph is concave up. In the interval
[1/3sqrt(7)/3,1/3+sqrt(7)/3], the graph is concave down. And
finally in the interval [1/3+sqrt(7)/3,infinity) the graph is concave
up.
Example 1: f(x)=e^{x2}
FUNC
I think that the domain is all x. The range? Maybe let's wait until we
analyze the derivative. But I do know that the exponential function is
never 0 and is always positive. I also know that
exp(large positive numbers) is large positive (exponential
growth) and exp(large negative numbers) is close to 0
(exponential decay). From this I can conclude that if
x>+infinity or if x>infinity, then exp(x^{2}) will be
exp(large negative number) and therefore will be small. So
lim_{x>+infinity}e^{x2}=0 and
lim_{x>infinity}e^{x2}=0. The xaxis
(with equation y=0) will be a horizontal asymptote.
FUNC´
We use the Chain Rule to compute f´(x)=e^{x2}(2x).
The exponential function is never 0, therefore the only way this
formula can be 0 is when 2x=0. That's x=0, the only critical
number. Please notice that f(0)=e^{0}=1, and therefore (0,1)
is the only critical point. Since the derivative is
e^{x2}(2x) and the values of the exponential function
are always positive, I see that f´(x)>0 for x<0 and
f´(x)<0 for x>0. The function is increasing in the interval
(infinity,0] and decreasing in the interval [0,+infinity). This also
agrees with the location of the point (0,1) on the graph and the
asymptotic behavior deduced above at +/infinity. The point (0,1) is
a local and absolute maximum. The range of the function is (0,1].
FUNC´´
Since f´(x)=e^{x2}(2x) we know (product rule
and Chain Rule) that
f´´(x)e^{x}(2x)^{2}+e^{x2}. Hey:
rewrite this as e^{x2}((2x)^{2}2). Since the
exponential function is never 0 (and is always positive) the sign of
the second derivative depends on the sign of
(2x)^{2}2=4x^{2}2. Let's see: this is 0 when
x=+/[1/sqrt(2)]. It is positive when x>1/sqrt(2) and is positive
also when x<1/sqrt(2). It is negative when
1/sqrt(2)<x<1/sqrt(2). Therefore, as we say in math classes,
the graph will be concave up in the intervals (infinity,1/sqrt(2)]
and [1/sqrt(2),infinity) and concave down in the interval
[1/sqrt(2),+1/sqrt(2)]. The inflection points occur at
(1/sqrt(2),e^{1/2}) and (1/sqrt(2),e^{1/2}) (plug the
xvalues into the original function!).
Note This isn't a miserable random function. If you ever do any statistics, this is the function whose graph is the wonderful "bellshaped curve". The inflection points indicate dispersal from the mean (in this case the mean is 0).
Example 2: f(x)=(6+7e^{5x})/(3+e^{5x})
FUNC The function is given by a formula which is a
quotient so we should be concerned about which x's given 0 on the
bottom. The bottom is 3+e^{5x}. The values of the exponential
function are always positive so the bottom is always at least
3. Therefore the domain of this function is all numbers. The range
maybe is a bit mysterious, and we'll need some help from the
derivative. What about what happens as x>+infinity? Here the
6+7e^{5x} on the top is "dominated" by the big term,
7e^{5x}. On the bottom as x>+infinity, 3+e^{5x} pays
no attention to the 3, and is essentially e^{5x}. Therefore I
bet that lim_{x>+infinity}(6+7e^{5x})/(3+e^{5x})=lim_{x>+infinity}7e^{5x}/e^{5x}=7. Now
as x>infinity, the exponential term e^{5x}>0
("exponential decay") so that
lim_{x>infinity}(6+7e^{5x})/(3+e^{5x})=lim_{x>infinity}6/3=2.
Therefore the lines y=7 and y=2 are both horizontal asymptotes of
y=f(x), the first on the right and the second on the left.
FUNC´
Since
f(x)=(6+7e^{5x})/(3+e^{5x}) I'll use the quotient rule to get the derivative. Therefore
7·5e^{5x}(3+e^{5x})5e^{5x}(6+7e^{5x}) 105e^{5x}+35e^{10x}(30e^{5}+35e^{10x}) f´(x) =  =  (3+e^{5x})^{2} (3+e^{5x})^{2}and there is neat (?) cancellation so that f´(x)=75e^{5x}/(3+e^{5x})^{2}. When is this equal to 0? Well, it can only be 0 when the top is 0, but the top is a constant multiplied by an exponential. Therefore the derivative is never 0. And since we know that values of exp are all positive, we can deduce that f´(x) is always positive, so that f(x) is always increasing (!).
75(5)e^{5x}(3+e^{5x})^{2}2(3+e^{5x})5e^{5x}75e^{5x} 75(5)e^{5x}(3+e^{5x})2·5e^{5x}75e^{5x}  =  (3+e^{5x})^{4} (3+e^{5x})^{3}Here I've gotten rid of a factor of (3+e^{5x}) on the two pieces of the top and on the bottom. The top is:
Note This curve is what's called a logistic curve. It is a curve which matches growth when there are limited resources. A small chunk of this curve, say for x between .4 and +.4, looks a bit like exponential growth, but the global behavior has constant asymtotic values. This is what happens when the slime mold (?) eats everything around and exactly balances with the energy and nutrition coming in. I hope you can imagine more interesting scenarios.
Example 3: f(x)=(5x^{2}3)/(x^{2}+2)
I was not able to do this in class due to lack of time. So let me
outline what happens.
FUNC If f(x)=(5x^{2}3)/(x^{2}+2), the bottom. x^{2}+2, is
always at least 2. It is never 0. Therefore the domain of f(x) is all
x. As x>+infinity, the dominant term in the top of the formula
defining f(x) is 5x^{2}, and the dominant term in the bottom
is x^{2}. Therefore the limit of f(x) as x>+infinity
is 2. Since the x's all appear only with even powers, this graph is
symmetric with respect to the yaxis, and
lim_{x>infinity}f(x) is also 2. Therefore y=2 is a
horizontal asymptote of y=f(x) (on "both" sides!).
FUNC´
After some algebra, f´(x) turns out to be
26x/(x^{2}+2>^{2}. The bottom is always positive, so
the only way we could have a critical number is when the top is 0, and
this occurs only when x=0. The value of f(0) is 3/2. The derivative
is negative for x<0 and is positive for x>0. So (0,3/2) is a
local (and absolute) minimum. The function is decreasing on
(infinity,0] and is increasing on [0,infinity).
FUNC´´
Much more algebra gives the following for
f´´(x):
26(3x^{2}2)/(x^{2}+2)^{3}. Now
f´´(x)=0 when 3x^{2}2=0 so this is when
x=+/sqrt(2/3). There are two inflection points, and f(x) is
concave up in the interval between them, and concave down
elsewhere. If you look at the graph you'll see that this
information can be confirmed there.
Example 4: f(x)=x^{1/3}(x2)
Again, I was not able to do this in class due to lack of time. One
"feature" of this function is that it isn't differentiable at all
x. So it has something interesting separating it from the other
examples. Also even plotting the darn thing with a machine was more
difficult, since most of the standard plotting devices sort of assume
that curves will be smooth. I had the machine draw the
picture shown with some deviousness.
FUNC
The nasty (or interesting?) thing here is the power of
onethird. Since 3 is odd (!!) the domain of x^{1/3} is all
numbers, and therefore the domain of f(x) is all numbers. Another way
of writing f(x) results from distributing the multiplication:
f(x)=x^{4/3}2x^{1/3}. If x is really huge in
magnitude (either positive or negative) then I bet the highest power
"rules". So f(x) when x is huge is just about x^{4/3}. the
fourth power of the cube root makes things positive. I bet that as
x>+infinity, f(x)>+infinity, and as x>infinity,
f(x)>+infinity.
FUNC´
The derivative, after some algebraic "massaging", turns out to be
[2(2x1)]/[3x^{2/3}]. Hey! What are the critical numbers?
Well, they are either where f&!80;(x)=0 (and that is when 2x1=0 so
x=1/2) or when x=0 (because there is a
power of x in the bottom and we can't divide by 0). Actually, the
domain of f´(x) is all nonzero numbers. The original function
f(x) is not differentiable at x=0. Since f(0)=0 and
f(1/2)=[1/2^{1/3}](1/2), I see that f(x) has a local and
absolute minimum at x=1/2. At x=0, the critical point is neither a
maximum nor a minimum. The sign of f´(x) is negative on both
sides of 0, so f(x) decreases on both sides of 0.
FUNC´´ The second
derivative turns out to be
4(x+1)/[9x^{5/3}]. For x<1, this is a negative number
divided by a negative number, so f´´(x)>0 and f(x) is
concave up. For x between 1 and 0, the top "changes" sign to
positive, but the bottom is still negative: f´´(x)<0 and
f(x) is concave up. For x>0, both the top and bottom of the formula
4(x+1)/[9x^{5/3}] are positive, so the quotient is
positive. Now f´´(x)>0 so f(x) is concave up.
There are two inflection points.
Example 5: f(x)=x^{2}/e^{5x}
FUNC Remember that the exponential function is always
positive, so the bottom (denominator) of the formula defining this
function is never 0. Therefore the domain is all numbers. Notice that
as x>infinity, the bottom, e^{5x}>0 and the top,
x^{2}>+infinity. f(0)=0, and the fraction is always
nonnegative. These facts together show that the range of f(x) is all
nonnegative numbers. The hard thing to decide is what happens as
x>+infinity. The derivative can help with this.
FUNC´ Since
f(x)=x^{2}/e^{5x}, I know that
f´(x)=
2xe^{5x}x^{2}e^{5x}5 (2x5x^{2})  =  (e^{5x})^{2} e^{5x}which we get by factoring out and dropping one e^{5} from all the terms, top and bottom. Where is this 0? We need only consider the top for this: 2x5x^{2}=0 means x(25x)=0 so this means x=0 or x=2/5. These are the critical numbers. Now substitute into f(x) and get the critical points:
But what happens to f(x) as x>+infinity? The graph shown above and the calculus work so far don't entirely convince me that I know. Maybe as x>+infinity, f(x) decreases to .000345 (it doesn't, but maybe it could. Well, here is a trick to show you what happens, or maybe to convince you that f(x) does actually >0 as x>+infinity. Let me pull a little bit of the exponential away on the bottom. I'll write things like this:
x^{2} x^{2} 1  =  ·  e^{5x} e^{4x} e^{x}The factor x^{2}/e^{4x} looks more or less like x^{2}/e^{5x}: in particular, for large enough x's this is positive and decreasing. The other factor, 1/e^{x} must >0 as x>+infinity. So when x gets large, we can write x^{2}/e^{5x} as a product of something that's at worst bounded multiplied by something that >0. So the limit of x^{2}/e^{5x} must be 0 as x>+infinity. And the xaxis (with equation y=0) is a horizontal asymptote of y=f(x).
(210x)e^{5x}5e^{5x}(2x5x^{2}) (210x)5(2x5x^{2})  =  (e^{5x})^{2} e^{5x}The bottom is always positive, so the top will determine the sign of the second derivative and where the second derivative is equal to 0. The top "simplifies" to 25x^{2}20x+2. The roots of the top (using the quadratic formula) are (20)+/sqrt[((20)^{2}4(25)2] all divided by 2(25). This is (2/5)+/(1/5)sqrt(2). So there will be one inflection point between the minimum and the relative maximum, and one inflection point to the right of the relative maximum. This does agree with the graph. y=f(x) is concave up on the intervals (infinity,(2/5)(1/5)sqrt(2)] and [(2/5)+(1/5)sqrt(2),+infinity) and it is concave down on the interval [(2/5)(1/5)sqrt(2),(2/5)+(1/5)sqrt(2)].
I handed out a takehome semiexam.
We discussed some graphs from homework problems. I don't have the record here, but please look at the answers to the takehome semi exam.
Then I asked students to work together and discuss some data. Several groups reported on each data collection, and tried to expression in words and symbols what the data showed. ALso, we tried to draw a convincing and appropriate graph for each data set. The data sets were chosen to describe some variable changing with respect to another variable. The changes themselves had some changes in time, which, in the cases displayed, had some meaning. Therefore I wanted to show that a function and it derivative and its second derivative could be important and useful in describing phenomena.
Data set #1
Here is 1931 inflation data. C(t) is the consumer price
index. Describe what you can about C´(t) and
C´´(t). Describe the behavior of $C(t)$ in words. Draw a
plausible and qualitatively correct graph of C(t).
Inflation rate  

January  February  March  April  May  June 
7.02%  7.65%  7.69%  8.82%  9.47%  10.12% 
The inflation rate is, essentially, already C´(t). So C´(t) is negative and decreasing (remember to use both magnitude and sign to think about this). Therefore C´´(t)<0. As we move from left to right (time increases) the slope of C(t) is decreasing.
Data set #2
Here is 1998 inflation data. C(t) is the consumer price
index. Describe what you can about C´(t) and
C´´(t). Describe the behavior of C(t) in words. Draw a
plausible and qualitatively correct graph of C(t).
Inflation rate  

May  June  July  August  September  October 
1.69%  1.68%  1.68%  1.62%  1.49%  1.49% 
Again, the inflation rate is, essentially, already C´(t). So C´(t) is positive and decreasing (remember to use both magnitude and sign to think about this). Therefore C´´(t)<0. As we move from left to right (time increases) the slope of C(t) is decreasing although it is certainly always positive.
Data set #3
Here is 1965 inflation data (enough already!). C(t) is the consumer
price index. Describe what you can about C´(t) and
C´´(t). Describe the behavior of C(t) in words. Draw a
plausible and qualitatively correct graph of C(t).
Inflation rate  

January  February  March  April  May  June 
0.97%  0.97%  1.29%  1.62%  1.62%  1.94% 
The inflation rate is, essentially, already C´(t). So C´(t) is positive and increasing: this is a real data set (as far as I know) so it isn't "perfect"  it is {mostlysort of} increasing! Therefore C´´(t)>0. As we move from left to right (time increases) the slope of C(t) is increasing.
Data set #4
Salt (NaCl, sodium chloride) has a saturation amount (the most salt
that can be dissolved) which varies with temperature. Here is the
amount of salt in grams which can be dissolved in 100 mL (milliliters)
of water (H_{2}O) at various temperatures:
Saturation amounts as a function of ^{o}C  

0  10  20  30  40  50 
35.7  35.8  36.0  36.3  36.6  37.0 
Suppose S(c) is the saturation amount as a function of the
temperature. Describe what you can about S´(c) and
S´´(c). Describe the behavior of S(c) in words. Draw a plausible
and qualitatively correct graph of S(c).
So this is maybe simpler data (!) than the inflation rate. This
function is increasing (these numbers are a table of the function, not
the derivative!). The derivative is sort of the rate of change of the
numbers, so from 0 to 10 we get .1 (grams per 100 mL per 10 degrees),
and then .2 and then .3 and then .3 and then .4. So certainly the
saturation amounts are increasing: S´(c)>0. The rate of
increase is, itself, generally increasing, so that
S´´(c)>0. I guess the qualitatively correct graph which
best matches this data is the same one drawn before.
I didn't get the fourth possibility, I think. The fifth data set given in the handout has, more or less, very little change in the derivative. The "curve" is decreasing but the rate of decrease is rather steady.
Discussion of concavity
Here are some pictures I drew. I emphasized that I wasn't
proving that the pictures (logically) identified the same
collections of curves: this was true, but the equivalencies were not
too easy to see.
Secant lines drawn through any pairs of points on the graph (in
the interval!) always lie above the graph between the two points.

Tangent lines to the graph at points in the interval always lie
below the graph on the interval.

Secant lines drawn through any pairs of points on the graph (in
the interval!) always lie below the graph between the two points.

Tangent lines to the graph at points in the interval always lie
below the graph on the interval.

If the second derivative of a function is positive in an interval, then the first derivative of the function is increasing is increasing in that interval. Then, as x "moves" from left to right, the slope of the tangent line to the graph of the function rotates counterclockwise. The result is that the curve bends "up", and is concave up. If the second derivative is negative, the result is the grpah of the function is concave down. It is useful to notice this, and also to notice that there need not be any connection between {inde}creasing behavior and concavity.
If f´´(x)>0 in an interval 
If f´´(x)<0 in an interval 


If f´(x)>0 in an interval 
The function is increasing and the graph is concave up in that
interval. 
The function is increasing and the graph is concave down in
that interval.

If f´(x)<0 in an interval 
The function is decreasing and the graph is concave up in
that interval.

The function is decreasing and the graph is concave down in
that interval.

Points on the graph where the concavity changes are called inflection points. A bunch of examples may be useful now, or they may just increase confusion. Oh wel.
f(x)=x^{3}3x
One good example or two might be useful to
continue the confusion. If f(x)=x^{3}3x, then
f´(x)=3x^{2}3=3(x1)(x+1). There are critical numbers at
+/1, and (using f(x) to get the second coordinates) the critical
points are (1,2) and (1,2). The first derivative's sign changes at
1 and +1, and the function is increasing in the intervals
(infinity,1] and [1,infinity). The function is decreasing in the
interval [1,1]. Now f´´(x)=6x. So when x>0, the graph is
concave up because f´´(x) is positive, and when x<0, the
graph is concave up because f´´(x) is negative. The point
(0,0) is an inflection point. All four of the behaviors in the chart
above can be obersed in different parts of this graph: concave up and
increasing in [1,infinity), concave up and decreasing in [0,1],
concave down and decreasing in [1,0], concave down and decreasing in
(infinity,1].
Notice the tangent line segments, please
There are two pieces of the tangent lines at the local max and the
local min. They are horizontal. The local picture is fairly easy to
understand. But the tangent line through (0,0), which is y=3x, is
less easy to understand. It "looks" tangent to the curve, but on one
side of the curve it is on top and on the other side of the curve it
is on the bottom. At an inflection point, the tangent line cuts
through the curve. This realization may be a bit uncomfortable!
f(x)=x^{4} The graph of f(x)=x^{4} looks like a sort of parabola which is steeper when x is large and flatter when x is near the origin.. Here f´(x)=4x^{3}, and (0,0) is a critical point. Even though f´#180;(x)=12x^{2} and f´#180;(0)=0, the point (0,0) is not an inflection point! The curve is concave up everywhere. We are facing the converse of an implication again, and the converse of a true implication need not necessarily be correct. Here the implication is that if (x_{0},f(x_{0})) is an inflection point, and if f´#180;(x_{0}) exists, then f´#180;(x_{0}) must be 0. That's true. But the other way around is not always correct.  
f(x)=x^{3} Here f´(x)=3x^{2}, and (0,0) is a critical point. We compute f´#180;(x)=6x so the second derivative is positive when x>0 (so the graph is concave up there) and the second derivative is negative when x<0 (so the graph is concave down there). Therefore the point (0,0) is not an inflection point. Notice that the horizontal line at the origin is the tangent line because (0,0) is a critical point, and that line crosses the graph because (0,0) is an inflection point. The critical point is neither a local maximumum nor a local minimum.  
f(x)=x^{1/3} This function increases for all x. The origin is a critical point, but this is because the derivative does not exist at the origin (the geometric tangent line is vertical there, and vertical lines have no slope). The derivative for x not 0 is given by (of course!) f´(x)=(1/3)x^{2/3} and for x not 0, the second derivative has the formula f#180;´(x)=(2/9)x^{5/3}. You can check the signs of the first and second derivative. f(x) is increasing for all x. f(x) is concave up for x<0 and f(x) is increasing for all x. f(x) is concave down for x>0.  
f(x)=x^{2/3} Finally, here is an example of what is called a cusp (the pointy behavior at the origin). Since f´(x)=(2/3)x^{1/3} for x not equal to 0, the function decreases to the left of 0 and increases to the right of 0. And f#180;´(x)=(2/9)x^{4/3} for x not equal to 0. The function is concave down for x<0 and it is concave down for x>0. 
Tuesday, July 18
We did three or four more graphs. Please look at the answers to the
takehome semi exam for work similar
to what was done in class. I tried to emphasize that Step 0 for me under any
realistic circumstances was using a graphing device whenever possible:
a graphing calculator or ... something! Then I would try to get
information about the function in Step
1. And the derivative would be used in Step 2. Later we'll learn
Step 3.
One beauty (!) was f(x)=(1/3)x^{3}(1/5)x^{5}, which had rather complicated behavior near the origin.
Derivatives of derivatives are frequently called higher derivatives. We computed some "higher" derivatives.
The function  x^{8}  5sin(3x)  44e^{9x}  1/x^{3}  

 
The first derivative  8x^{7}  5·3cos(3x)  44·9e^{9x}  (3)x^{4}  This is f´(x) or f^{(1)}(x) or dy/dx or y´.  
 
The second derivative  8·7x^{6}  5·3^{2}sin(3x)  44·9^{2}e^{9x}  (3·4)x^{5}  This is f´´(x) or f^{(2)}(x) or d^{2}y/dx^{2} or y´´.  
 
The third derivative  8·7·6x^{5}  5·3^{3}cos(3x)  44·9^{3}e^{9x}  (3·4·5)x^{6}  This is
f´´´(x) or f^{(3)}(x) or d^{3}y/dx^{3} or y´´´.  
 
The 78^{th} derivative  0  5·3^{78}sin(3x)  44·9^{78}e^{9x}  (3·4·5·6·...·79·80)x^{81}  This is f^{(78)}(x) or d^{78}y/dx^{78}. I've never seen more than two or three primes in a row!  

Demystification and pattern recognition
I wanted to compute the entries in this table partly to "demystify"
(?) the process, but also to ask people to look for patterns. In the
case of the first function, a polynomial, differentiating pushes down
the highest power by 1 until that highest power gets to be 0 (so the
result then is a constant). All further, higher, derivatives are 0.
What about 5sin(3x)? Here the 5 just gets carried along, but no higher derivative is going to be the zero function. Each derivative will "spit out" (due to the Chain Rule) a 3 to multiply the result. But what about the other parts of the function? Well sine and cosine do a dance (?!) when differentiated. The dance steps are pictured to the right. It is sort of a waltz (forget about the 3x inside for a second, please). Every fourth step brings us back to where we were. So if we wanted the 78^{th} derivative of sine, we would divide 78 by 4. This would be 19: but not exactly  there would a be a remainder of 2. This means that the 78^{th} derivative waltzes around this figure 19 times, and then does two more steps, so that the result would be sine. Wow (maybe). The result would still have the 5 and 78 of the 3's multiplying and the darn 3x inside the function. But the waltz makes the computation easy. (I don't want to make things too darn easy: the derivatives of tan have no simple structure.)
The exponential function is your friend. If sine and cosine's derivatives waltz in a 4step pattern, then, oh my goodness, what exponential does when differentiated is jump up and down, since exp´(x)=exp(x) (the derivative of etothe is etothe!). OF course the 44 is carried along as a multiplicative constant, and each differentiation brings out another multiplication by 9.
1/x^{3} is x^{3}. Differentiation follows the power rule, but because we begin with a negative integer, we never "get" to 0. Actually we just keep going down and down and down. The integers get larger in absolute value and come out "in front" at each differentiation. The sign result alternates between + and , and the oddnumbered derivatives get the minus sign, while the even numbered derivatives get the plus sign. If you desperately needed a compact way to write the derivatives, there is notation available, using the factorial sign (the surprise mark: !) but I am not too interested in efficiency here. I would like you to see the patterns. Higher derivatives for "simple" functions aren't too hard to compute.
I wanted to show a use of higher derivatives. The simplest use might be from a physical problem. So here is one.
Abrupt stop to a car
A car is speeding along at 60 miles per hour (which I remember from
high school is 88 ft/sec). Suppose the car hits a wall and comes to a
stop in a tenth of a second. If the deceleration is assumed to
be constant, what is the deceleration? If you can do such analysis,
maybe you could be an expert witness in car accident cases, and
make big bucks. Ooops, sorry: maybe you could help that the correct
side wins in the court case.
Here a(t)=K, a constant by assumption. I'll use the letter K for a
constant. So I know that a(t)=K, and then by antidifferentiating,
v(t)=Kt+C. But v(0)=88. I'm using distance measured in feet and time
measured in seconds. Thus v(t)=Kt+88. Since v(1/10)=0, we know
K(1/10)+88=0, so that K=880 ft/sec^{2}. The "acceleration" of
gravity is about 32 ft/(sec)^{2}. So let me compare this
deceleration to gravity, an interesting and horrifying comparison. Notice that
880/32 is 27.5: so this deceleration presses the "contents" of the car
(the passengers!) with a force twentyseven and a half times that of
normal gravity. Imagine even briefy your weight multiplying by that
factor. The consequences are most unpleasant. A chapter in a book I
read, Stiff: The Curious Lives of Human Cadavers by
Mary Roach, discusses this situation in detail. Those of you
interested in becoming doctors might want to read this book. But if
you were holding a 10 pound baby in your arms, that baby would, for a
short time, seem to "weigh" 275 pounds. Holding such a load, even for
a very short time, seems perilous. I think the child would be released
at its peril. Indeed, even if you are a big, strong "guy", with
weight, say, 200 pounds, could you hold yourself safely when you might
seem to weigh more than two tons? (27.5·200=5,500)
Monday, July 17
I wrote the Mean Value Theorem again.
Example 1: distance traveled and speed
Suppose I told you that I have driven on a straight road for two hours
with my speed (really, darn it, velocity!) between 40 and 60 miles per
hour. If we assume that my travel is differentiable (I would not want
to travel nondifferentiably, because the times when the derivative
didn't exist would be most unpleasant!) what can be said about the
distance traveled? If f(t) is the position at time t in terms of miles
from the origin (of whatever road I am on) then according to the MVT,
f(time_{2})f(time_{1})/(time_{2}time_{1})=f´(some time in between).
Here
the information I have is that the elapsed time,
time_{2}time_{1}, is 2 hours (time will be measured
in hours, and distance in miles). And I told you that my
velocity is always between 40 and 60, so 40<f´(some time in between)<60. Therefore
the equation above becomes
40<f(time_{2})f(time_{1})/(2)<60
so that (multiplying by 2) we get
40·2=80<f(time_{2})f(time_{1})<60·2=120.
I think that most children aged, maybe 10 or 12 and above will come up
with the same result. Do they have the MVT "wired in" to their brains?
I don't know. I don't know how a child or an adult or any person
(including myself!) thinks. But I wanted to go through this as part of
the effort to convince you that the MVT is a natural part of thought,
at least in certain physical situations.
Example 2
This will be more like something in a math course, so we will be
happy. Suppose f(x)=sqrt(5x^{600}+4). I can compute (the darn
function is arranged so that the computation is easy) f(0)=2 and
f(1)=3. What can one be said about f(.3781334)? This maybe can also be
thought through without needing "higher math" but I want to organize
your head a bit. Well, f(x) is defined by a (relatively!) nice
formula. I bet it is continuous wherever it is defined. And, actually,
since the power inside is even, the "stuff" inside the square root
will be nonnegative, actually positive (at least 4) so the domain of
f(x) is all x. Hey, what do we know using "continuity" about
f(.3781334)? The answer is, not very much. If the only
information is f(0)=2 and f(1)=3 and continuity of f(x), there
is no restriction on f(.3781334). Look at the darn graphs. But:
If f(x)=sqrt(5x^{600}+4) then f´(x)=(1/2)(5x^{600}+4)^{1/2}(5·600x^{599}. Hey, maybe this is a complicated "thing". But the only information I want right now is the sign of this complicated thing when x is in the interval [0,1]. And getting the sign is easy. The only minus is in the exponent (which forces something "downstairs") so I know that f´(x)>0 in [0,1]. This means, according to a result following the MVT, that f(x) is increasing in this interval, so I know that f(0)=2<f(.3781334)<f(1)=3. The desired value of f(.3781334) is between 2 and 3. I will admit that this isn't very much but it is a heck of a lot more than knowing nothing! So the MVT gives us more information.
If  then  The graph looks like ... 

f´(x) is positive on an interval  f(x) is increasing on the interval  As x "moves" from left to right on the horizontal axis, the point (x,f(x)) moves "up" on the graph of y=f(x). 
f´(x) is negative on an interval  f(x) is increasing on the interval  As x "moves" from left to right on the horizontal axis, the point (x,f(x)) moves "up" on the graph of y=f(x). 
I applied these ideas to the graph of f(x)=(1/4)x^{4}+(1/3)x^{3}x^{2}+1. Here f´(x)=x^{3}+x^{2}2x. This is an example from a calculus course so it factors neatly, and f´(x)=x(x+2)(x1). The derivative is 0 at x=0 and x=2 and x=1. It isn't too hard to figure out where the derivative is positive: in the intervals (2,0) and (1,infinity). There the function is increasing. The derivative is negative in the intervals (infinity,2) and (0,1) so the function is decreasing in those intervals. We already this function maybe too well (anything from an exam is such a thing). Let me try to look at something new.
f(x)=(3x+5)/(x^{2}+1)
I look at this function and think about its domain: all x (because the
bottom is never 0 since x^{2}≥0 so it must be at least 1).
I also see that as x>+infinity, the top is mostly 3x and the bottom
is x^{2}, so the function is almost 3/x and this >0 as
x>+infinity. A similar analysis shows that f(x)>0 as x>infinity
also. What about bumps, etc.? A good hint is gotten by looking at a
mechanically produced graph of this function, as shown to the right. I
can get more precise information from the derivative.
We computed f´(x). We used the quotient rule and were
careful. The result, after some algebraic effort, was
(3x^{2}10x+3)/(x^{2}+1)^{2}. I am interested
in the sign of f´(x) and I notice that the bottom is always
a square, so it is always positive (it isn't 0  we saw that
earlier). So let's look at the top:
3x^{2}10x+3
The roots of this quadratic (using the quadratic formula) are
5/3+/sqrt(34)/3. These numbers are about 3.61031 and 0.27698.
We can then substitute them back into the original function, f(x).
The results are the points (3.61031,.38784) and (0.27698,1.37610). I
think one is the top bump and the other is the lower bump.
Here are my conclusions, which I think I'd not be very sure about
without using calculus (because, darn it, the "pictures" produced by
machines can be deceptive and sometimes even incorrect!):
I'll be honest: I doubt many students in Math 135 will need to know even these details about such a function. But if you've go to know, calculus is the best way to be sure you are correct.
Thursday, July 13
We used the MVT to deduce a result about {inde}creasing
functions. First let's state the definitions:
Increasing
A function f(x) is increasing on an interval I if for any pairs
of numbers x_{1} and x_{2} in I, if
x_{1}<x_{2}, then
f(x_{1})<f(x_{2}).
Decreasing
A function f(x) is increasing on an interval I if for any pairs
of numbers x_{1} and x_{2} in I, if
x_{1}<x_{2}, then
f(x_{1})>f(x_{2}).
It could be very difficult to verify either of these definitions directly, since it seems that we'd have to take every pair of points, and check some inequalities for all of them. Well, the MVT allows a much easier method.
f´(x)>0 implies f(x) increasing on an interval
f´(x)<0 implies f(x) decreasing on an interval
Well, suppose f´(x) is positive on I, and x_{1} and
x_{2} are two points on I with
x_{1}<x_{2}. Then by MVT,
f(x_{2})f(x_{1})  =f´(somewhere in between) x_{2}x_{1}If I know that the derivative values are always positive, and if I know that x_{2}x_{1} is positive (which is true if x_{2}>x_{1}) then the equation is
SIGN NOT KNOWN  = POSITIVE # POSITIVE #so that in fact, the SIGN NOT KNOWN, which is the term f(x_{2})f(x_{1}), must also be positive. But then f(x_{2})>f(x_{1}). We have proved that f(x) must be increasing on an interval where the derivative is positive.
A similar verification can be applied to the case of negative derivatives. The usefulness of all this is determining whether functions are positive or negative is usually much easier than checking (directly) whether they are increasing or decreasing.
I applied this to analyze a function like 8e^{3x}+5e^{4x}. Here f(x)=8e^{3x}+5e^{4x} then f´(x)=24e^{3x}+20e^{4x}. Let's see where this derivative is 0: 24e^{3x}+20e^{4x}=0 happens when 20e^{4x}=24e^{3x}. I'll divide by 20 and multiply by e^{3x} using the exponential rules. The result is e^{7x}=24/20=6/5. Taking ln's undresses the exp's, so we know 7x=ln(6/5) and x=(1/7)ln(6/5). What now? I know that lim_{x>+infinity}f(x)=+infinity, because the e^{4x} term gets very large and the other term is positive. Similarly, as x>infinity, the e^{3x} in f(x) gets large. I think that f´(x)>0 for x>(1/7)ln(6/5). I know the derivative can't equal 0, because we found the only number where it equals 0. The derivative can't be negative in that region, because otherwise (Intermediate Value Theorem) it would have to be 0 (not possible!) because it would have both positive and negative values. The logic is a bit extended but I hope you can follow it. Similarly, f(x) is decreasing for x<(1/7)ln(6/5).
The minimum value of the function is f((1/7)ln(6/5)). All other values are larger. I don't really know how to see this easily without calculus.
The first formal 80minute exam was given.
Wednesday, July 12
y=2x+4x3
Various computational devices would produce something like what is
shown to the right as a graph of f(x)=2x+4x3.
Why is this picture correct? I believe it is, but is there some way of getting reassurance about the correctness? I ask this for several reasons:
/ Number if Number≥0 Number=< \ Number if x<0. Second, we learned some qualitative things about the absolute value function which have important implications for the graph of the function: the function is continuous for all numbers, so the graph will be unbroken, and have no jumps or breaks. The function is differentiable everywhere except at the origin. To the left of the origin, the function has derivative 1 (the graph is part of a straight line with slope 1). To the right of the origin, the function has derivative +1 (the graph is part of a straight line with slope +1). At the origin, the function is not differentiable, and the graph shows a corner.
What can I expect about this f(x), which is given by the formula 2x+4x3? It is the sum of continuous functions, so f(x) should also be continuous. The graph should have no breaks or jumps. The function has x in it, and also x3, which "translates" or shifts the absolute value function 3 units to the right. I'd expcet that there should be nice behavior away from 0 and 3, where nice now means differentiable  there should be a tangent line.
Well, the algebra can show some of this. If x<0, then x is x. What about x3? If x is negative, x3 makes it "more negative", so x3 is (x3). therefore if x<0, f(x), which is 2x+4x3, will be 2(x)+4{(x3)}. I fouled this up in class, so let me be careful here: f(x)=2x4(x3)=6x+12.
Now as x "moves" to the right (so x>0) x changes its description. Notice, please, that x3 does not change description "immediately". What do I mean? For example, if x=1,x3 becomes 13=2=2. So if x<3, x3<0, and x3=(x3). Therefore, if 0<x&;t3, f(x)=2x+4x3=2(x)+4{(x3)}=2x4x+12=2x+12.
What happens when x>3? Then x3 is positive, and x3=x3. x itself is of course positive, so x=x. And f(x)=2x+4x3=2(x)+4(x3)=2x+4x12=6x12.
f(x) defined piecewise
Here is another definition of f(x), exactly equivalent to the
original.
/ 6x+12 if x<0 f(x)=< 2x+12 if 0≤x≤3 \ 6x12 if 3<xWhat can I learn with this piecewise definition? Well, I can see that the graph of f(x) is actually made of parts of three lines. To the left of 0, the graph of f(x) is part of a line with slope 6. Between 0 and 3, the graph is part of a line with slope 2, still negative but "shallower" than the other segment. Finally to the right of x=3, the graph is part of a line with slope 6, tilted up. Do the pieces of the lines connect? Well if I consider 6x+12 as x>0^{} the result is 12. And 2x+12 as x>0^{+} also gives 12. Hey: these two pieces connect. And, around x=3, if we look at 2x+12 as x>3^{}, the result is 2(3)+12=6, while if we look at 6x12 as x>3^{+}, the result is 6(3)12=1812=6. These pieces are also connected. Hey: the graph has one piece, and the function is continuous.
What about the derivative? The algebra should reinforce the picture. The infinitesimal rate of change of f(x) for x<0 is the rate of change of the formula 6x+12. I think that rate of change is 6. Between 0 and 3, the rate is 2, from the formula 2x+12. What happens "at" 0? There is no one number describing this rate of change. The derivative does not exist. The function is not differentiable at x=0. A similar thing happens at x=3. So here is a description of the function f´(x). First, this function has the following domain: all real numbers except for 0 and 3. And when x is in the domain of f´(x), we have
/ 6 for x<0 f´(x)=< 2 for 0<x<3 \ 6 for 3<xI hope now that you can see that "algebra" has confirmed the details of the picture of y=f(x).
y=x^{3}2x^{2}+x+5
Now let me look at f(x)=x^{3}2x^{2}+x+5. This is
"easy" (well, maybe). There is one nice algebraic formula defining
f(x). The domain for this polynomial is all real numbers, and
polynomials are continuous and polynomials are differentiable. In
fact, look, f´(x)=3x^{2}4x+1, so the function sure is
differentiable. I should expect that the graph will have no jumps 
it will have one piece. I should expect that the graph will be a nice,
smooth curve, with a neat tangent line at every point. So far, the
picture to the right should be further evidence of these assertions.
But there's some finer structure to the picture which I've not discussed. The curve seems to wiggle in some fashion. How can I be more precise about the wiggling? How can I analyze this wiggling? The wiggling really has to do with the rate of change, so maybe I should look at f´(x)=3x^{2}4x+1 more closely. Well, I was lucky (really!) when I specified f(x), and I did not expect that f´(x) would have a nice algebraic property. I thought that I'd need the quadratic formula for the roots of f´(x): that would mean considering (4)+/sqrt{(4)^{2}4·3·1 (all divided by 2·3. Hey, this is [4+/sqrt(1612)]6=[4+/sqrt(4)]/6=[4+/(2)]/6. The roots are 1/3 and 1. Or I could just have "observed" that f´(x)=3x^{2}4x+1=(3x1)(x1).
Consequences of f´(x)=(3x1)(x1)
I know that the derivative is 0 when x=1. But
f(1)=1^{3}(2)1^{2}+1+5=12+1+5=5. So the graph of the
function should have a horizontal tangent at (1,5). Also I know that
the derivative is 0 when x=1/3. And
f(1/3)=(1/3)^{3}(2)(1/3)^{2}+(1/3)+5=
(1/27)(2/9)+(1/3)+5=(1/27)(6/27)+(9/27)+3=5{4/27}. This is about
5.148. And there's a horizontal tangent at (1/3,5.148)
When I first wrote the polynomial and had a machine graph it, the bumps were not totally "obvious" to me. The initial resolution of the graph wasn't very fine, and I didn't "see" the one bump higher than the other bump. I just saw a sort of wiggle. With the additional evidence of the algebraic analysis, I think I "see" that the graph goes up (as a point moves on the graph from left to right) until x=1/3, and then the graph goes down until x=1, and then finally up once again. This is much more detailed information than the rather casual first graph I saw. For example, I now believe that f(.47) is greater than f(.55): I maybe could not have made such an assertion with confidence before the analysis.
Definition #1
A function f(x) has a relative maximum at c if
The top of the graph is a Relative Max and the bottom is a Relative Min. The endpoints are not eligible to be relative extrema because the function is not defined in an interval containing either of them in its inside. 
There is no Relative Max (again each endpoint is not eligible since the function is not defined in an interval containing that point). There's a Relative Min where indicated, because nearby points (on both sides) have larger function values. 
The bottom of the graph is a Relative Min. There are no points representing a Relative Max. 
There are no Relative Max and no Relative Min in this picture. The highest and lowest points are on the ends which aren't eligible. 
Definition #3
Suppose the function f(x) is defined at c. Then f(x) has a critical number at c if
The tangent line is horizontal at the top and the bottom. At these points, f´(c)=0, so these points are Critical Points. 
The function is not continuous at a point so the derivative doesn't exist there. But points where the derivative doesn't exist are still Critical Points. 
The function is not differentiable at the corner since the left and righthand limits defining the derivative don't agree there. The corner is a Critical Point. 
There are two points where the derivative doesn't exist and these are Critical Points. 
An important inference
These examples and the others I displayed were supposed to help people
to believe that if f´(c) exists and wasn't 0, then nearby values
of f(x) were higher or lower, so that c was not a relative
extremum. The following statement (the "inference" I mentioned) would
be true:
Relative extrema occur only at critical points. 
Here is an "if ... then" statement of the same inference:
If c is a relative max or min, then (c,f(c)) is a critical point. 
Vocabulary
One definition of inference is "the forming of a conclusion
from premisses." Of course, a definition of premiss is "a
previous statement from which another is inferred." Does this help?
A verification of what happens at relative maxes
Suppose f(x) has a relative maximum at c. Also let me suppose that
f(x) is differentiable at c. Then look at the quotient:
[f(c+h)f(c)]/h. I will look at two cases:
h is positive: the secant line is to the right of (c,f(c) Here since f(c) is bigger than f(c+h), I know that the top of the quotient is negative. But the bottom is positive. Therefore the quotient is negative divided by positive, and so it is negative. Secant lines from the right all have negative slopes.  
h is negative: the secant line is to the left of (c,f(c) Still f(c) is bigger than f(c+h), so the top of the quotient is negative. But the bottom is now negative. Therefore the quotient is negative divided by negative, and so it is positive. Secant lines from the left all have positive slopes. 
Now suppose f(x) has a relative maximum at x=c. If f(x) is differentiable at x=c, then the previous remarks show that the derivative is the limit from the left of positive slopes and is also the limit from the right of negative slopes. The twosided limit can exist only when it is 0, so f´(c)=0. Therefore we verified that at a relative max, then either f´(c) does not exist or if f´(c) does exist, then it must be 0.
The result for relative mins is proved in the same way.
Some algebraic examples of critical points
The functions  Sketch of the graphs (A machine helped me with these.) 

#1
What are the critical numbers of
f(x)=(1/3)x^{3}+(1/2)x^{2}6x?
So f´(x)=x^{2}+x6=(x+3)(x2), and this is 0 when x=3 or
x=2.  
#2 What are the critical numbers of
f(x)=x^{2}e^{3x}?
Here
f´(x)=2xe^{3x}+x^{2}(3e^{3x})=(2x3x^{2})e^{3x}. Since
the exponential function is never 0, the derivative is 0 only when
2x3x^{2}=0, which is when x(23x)=0.  
#3 What are the critical numbers of
f(x)=x^{2/3}(x2)?
Now f´(x)=(2/3)x^{1/3}(x2)+x^{2/3}(1). Stay alert! Algebra coming. I will rewrite this as
a product, because when I write it as a product,
the result will be equal to zero exactly when either of the factors is
0. 
In general, if I give you a "random" function, finding out where the function is "highest" or "lowest" might be very difficult. But the inference allows one to throw out points which can't be highest or lowest. But here are two more definitions:
Definition #4
A function f(x) has an absolute maximum at c in an interval if
f(c)>=f(x) for all x in that interval.
Definition #5
A function f(x) has an absolute minimum at c in an interval if
f(c)<=f(x) for all x in that interval.
If you look at the examples and at the inference above then you should believe that
Absolute maxes or mins of a function on an interval can occur only at critical numbers or endpoints. 
The geometric idea is that if a function has a nonzero derivative at a number, then the function's graph and its tangent line stick close together, so such a point can't be a relative max or min: the tilted tangent line forces the function to be higher on one side of the number and lower on the other side. So search for relative {maxmin}'s where the derivative doesn't exist (a "corner") or where the derivative is 0 (a "flat point").
Two dots on the graph ...
I drew coordinate axes with two big dots on the xaxis and urged
students to come up and draw a graph going through the two dots. I got
several graphs similar to what is displayed.
I observed that in each case there were points on the graph where the
tangent line was horitizontal. Below are the graphs with those points
and parts of the tangent lines displayed. Only, I was rude and the first graph below, with a bad
point. People really like to draw smooth curves.
The logic of the situation is this: if the function is positive
between the two points, then it has a positive relative max inside the
interval. That must be a critical point (horizontal tangent or not
differentiable). If the function is negative, then ... the same
situation below the xaxis. (Hey, yeah, if the function is never
positive and never negative, then it has lots of horizontal tangents,
because the graph is itself a horizontal line.)
Several versions of Rolle's Theorem
Here's an informal description: if there are two
dots on the xaxis which are part of the graph of a function, then
either the function has a bad place in between or has a flat place
somewhere in between. Since "bad" and "flat" are not usually
considered precise and suitable words in Math 135 (and there are other
implicit assumptions such as continuity), let's rephrase this
statement. Also, the way the result is applied usually assumes that
the function is differentiable everywhere, so the "bad" alternative
vanishes.
Rolle's Theorem (official version) Suppose f(x) is a differentiable function, and f(a)=0 and f(b)=0. Then there is at least one number c between a and b so that f´(c)=0. 
Unless otherwise posted (due to inclement weather, construction, etc), the Turnpike speed limits are as follows: The speed limit is 65 MPH from Mile Marker 1.0 to Mile Marker 97.2. 
Tuesday, July 11
I went backwards and discussed exponential stuff more, because I don't
think I did a good enough job in class last time. The diary notes are
fine, but the classroom performance wasn't: apologies.
We first looked at interest, in the financial sense: simple interest, compound interest, and compound interest with an annual rate and different time periods for compounding. By the way, in reading about this on the web today, I learned the word anatocism, which is a legal word for compound interest. The word was used with this meaning, according to the Oxford English Dictionary, in 1656.
We went over the discussion I wrote about in the previous
diary entry. I hope that the discussion verified the following
formula, which is frequently used:
Suppose we borrow a principal sum of money, P, for t years, at an
annual rate of interest r, which is compounded m times each year. Then
the amount owed is
(1+{r/m})^{mt}
and we can rewrite this as
((1+{r/m})^{1/{r/m}})^{rt}
using the rule (A^{B})^{C}=A^{BC}. The reason for doing this
rewrite is so that we can see what happens as m gets larger and
larger. Then {r/m}>0. And we have the following:
(
(1+LITTLE)^{1/LITTLE}
)^{rt}. We can possibly recognize
(1+LITTLE)^{1/LITTLE} as an approximation to e (see the
discussion during the previous lecture), and this
approximation gets better and better as LITTLE>0. So
the limit is Pe^{rt}. (The variables here, spelling out the word
"pert" actually are used in a great many applications.)
Some differentiations
We differentiated a bunch of examples involving the exponential
function. I remarked that some texts favored exp(x) instead of
e^{x}, because the typography (type setting) is easier without
superscripts! Then we could write exp´(x)=exp(x). We could also
call exp(x) the function, "etothe". Then it is easier to think that
the derivative of etothe is etothe. And a composition such as
e^{cos(x)} can be thought of, in f(g(x)) fashion, as taking x
and changing it into cosine of x, and then changing that into etothe
cosine of x. So the derivative, f´(g(x))g´(x), would be
etothe applied to cosine of x
multiplied by the derivative of the "inside" function, cos(x),
which is sin(x). So the derivative of e^{cos(x)} is
e^{cos(x)}(sin(x)).
And then we turned towards logarithms.
exp
To the right is a graph of the exponential function,
y=e^{x}. I know some things about e^{x}. Of course,
(e^{x})´=e^{x}, because we chose e to work
that way. But I also know that the domain of e^{x} is
all real numbers, and the range of e^{x} is all positive
numbers. Look at the graph. I could take a number on the horizontal
axis, x, and push it up until it "encounters" (hits) the graph. The
coordinates of that point would be (x,e^{x}). Then if I push
to the vertical axis, the result would be e^{x} there.
Why not relabel? Take some point, w, on the vertical axis, and define a new function of w by pushing it to the right until it hits the curve, and then down. The number on the horizontal axis will be called the natural log of w, usually written ln(w).
ln
The standard way to think about this is to pick up the graph of
y=exp(x) and flip it over the "main diagonal line" (y=x). This has the
effect of interchanging the horizontal and vertical
axes. There's also an algebraic interchange: the first and second
coordinates are swapped.
The domain of ln is all positive real numbers, and the range of ln is
all real numbers. There are algebraic log properties which are the
reflection (!) or counterparts of the exponential properties:
Logarithmic rules 

ln(AB)=ln(A)+ln(B) ln(A/B)=ln(A)ln(B) ln(1/B)=ln(B) ln(A^{B})=Bln(A) ln(1)=0 
The "applications" of logarithms in common use which might be encountered outside of a math course include pH (which is minus the log base 10 of the hydrogen ... who let that mole in here?) and allied chemical concepts, the Richter scale (this is a logarithmic scale which describes the strength of earthquakes), and decibels, a logarithmic scale for the loudness of sound. But all of this might be interesting but for us it turns out that there's an essential use for logarithms in calculus. So let's see it.
We will later in the course be interested in reading the Function to Derivative transition in the reverse order. In fact, then the righthand column will be labeled Function and the lefthand column will be labeled Antiderivative. Well, we tried to compute a few antiderivatives, and for powers of x a pattern could be seen.
Function Antiderivative  Derivative Function 

x^{2}  2x 
{7/2}x^{}  7x 
x^{6}  6x^{5} 
{4/5}x^{6}  4x^{5} 
{1/44}x^{45}  x^{44} 
{1/(8)}x^{8}  x^{9} 
{1/(POWER+1)}x^{POWER+1}  x^{POWER} 
But of course the last entry in the table will only be valid if POWER is not equal to 1. What if we
wanted a function whose derivative was x^{1}, that is, 1/x?
Well, consider the following application of the Chain Rule applied to
the composition of the inverse functions exp and ln (we know that exp
undoes ln  just look at the arrows on the graphs above):
e^{ln(x)}=x. Differentiate this equation. The righthand side
has
derivative 1 certainly, and the lefthand side has derivative
e^{ln(x)}ln´(x)=1. But then x·ln´(x)=1 since
e^{ln(x)} is x. So, solving for ln´ we get the final
entry (for this course!) in the derivative table:
Function  Derivative 

ln(x)  1/x 
A tangent line
What is an equation for the line tangent to y=ln(x) when x=2?
Well, if we have a point and a slope, then we can find such an
equation. The slope is the derivative of ln(x) when x=2. But that's
1/x when x=2. So the slope is (1/2). What about a point? We need
ln(2), and a calculator supplies us with the (approximate) value
.69315, so the point is (2,.69315). And one possible equation is
y.69315=(1/2)(x2).
To the right is a picture of portions of y=ln(x) and
y=(1/2)(x2)+.69315.
Haven't we seen that number before?
We discussed another homework problem, which was finding the
derivative of
f(x)=[3x(x^{4}+x)^{5}]^{3}. I asked for
the 100^{th} derivative of this function. This is written
f^{(100)}(x). This is actually easy because I know that
the top degree in this polynomial is 60 (4·5·3) and
differentiation reduces degrees in polynomials by 1. So after 61
differentiations the zero polynomial will be the result.
If you don't believe me, you can check (not easily!) that f(x)=
60 57 54 51 48 45 42 41 39 38 36 35 x 15x 105x 455x 1365x 3003x 5005x +9x 6435x +90x 6435x +405x 33 32 30 29 27 26 24 23 22 21 5005x +1080x 3003x +1890x 1365x +2268x 455x +1890x 27x 105x 20 19 18 17 16 15 14 13 11 10 7 3 +1080x 135x 15x +405x 270x x +90x 270x +9x 135x 27x +27xGoing on ...
A graph of y=2x+4x3  A graph of y=x^{3}2x^{2}+x+5 

Monday, July 10
We discussed some of the homework
computations.
I distributed copies of the information below. I remarked that the numbers on the chart were invented, but for the purposes of this class we should believe they were real. We discussed the chart for a while (an hour is "a while"!). There's a great deal of information, and I hoped that we could understand the information.
CHIPCO INVESTMENT DOLLARS & PRODUCTION Capital Invested Chips produced Marginal chips produced $ in millions 1,000's of units 1,000's of units per millions $'s 200 3,000 .23 300 3,040 .28 400 3,070 .42 500 3,100 .78 600 3,190 .31 CHIPCO SALES & PROFITS Chips marketed Profit gained Marginal profit 1,000's of units $'s in millions Millions of $'s per 1,000's units 3,000 1.2 .03 3,050 2.8 .02 3,100 3.6 .05 3,150 4.9 .01 3,200 5.1 .02
We decided that the pair of numbers 400 and 3,070 referred to the following phenomenon: if the corporation invests 400 million dollars in capital (building a factory, furnishing it, hiring and training people, etc.) then 3,070,000 chips can be produced.
Then we tried to understand the third column. The word "marginal" is
used in a fairly technical sense, although the use is common in
economics. In the first table above, it refers to the approximate
amount that chip production would increase per each million dollars of
increase in capital investment. Therefore, for example, if $302
million were invested, then (according to this model) chip production
would be 3,040,000 (chip production at the $300 million level)
plus .28(2)(1,000) chips. The 2 comes from the
additional millions of dollars of capital. The 1,000 comes from the
units used for chip production. The .28 is this "marginal"
quantity. In the first table, the marginal quantity is therefore the
approximate amount P/C, relating the change
in chip production to the change in capital investment. It is
sort of a slope, or, more likely, sort of a derivative: indeed, the
use of "marginal" in economics usually means a derivative. In this
model, if $297 million were invested, the approximate expected chip
production would be 3,040,000 (again, chip production at the $300
million level) plus .28(3)(1,000). The novelty
here is the use of the minus sign, since the capital investment is
decreasing rather than increasing.
The marginal production may vary because at different levels, maybe
different new machines have to be purchased, or different types of
employees need to be hired and trained, etc. Things aren't simple. I
hope that students can recognize the third column of the table was
actually a list of values of the derivative of production with respect
to the "independent variable", capital invested. The textbook
economics definition is the amount of production increase if there is
1 more increase of capital (all using the correct units for this
situation).
The second table describes a similar phenomenon, here connecting the
chip amount, C, with the profit derived from these marketing and sale
of these chips. For example, the profit derived from the sale of
3,000,000 chips (the first line of the second table) is $1.2
million. If we now look at the third column, the model predicts a
marginal profit of .03 (in the given units). Using this, if 3,010,000
chips are marketed (that's 10 more 1,000 units of chips) the
additional profit would be .03(10) million dollars, or $300,000. And
if only 2,970,000 chips were marketed, then the profit would be 1.2
million+(.03)(30)(1,000)million. (I think I got all the units
correct.) The third column gives P/C for various
amounts of chip marketing: the change in profits compared to the
change in chips marketed. One of the numbers in the third column is
negative: the .01 at the 3,150 level. Is this realistic? Here is what
could happen with a much more modest "business". I could deliver
newspapers every morning. Say that I have 100 customers, located on
five consecutive blocks of one street. It is easy and efficient for
newspaper delivery: the costs are low. What if I get 1 additional
customer? Well, if the customer lives "across town", with a halfhour
of driving needed, taking on this additional business will actually
probably cost money. There may be reasons that the business is
desired, but on a direct basis the incremental profit is negative
rather than positive.
Of course the validity of such models can certainly be criticized, but
I really wanted to show you these tables to explain what's in the next
paragraph.
The two tables linked together describe a complicated phenomenon. First we "input" capital, M (M is for money), which produces C, a certain number of chips. Then the chips are marketed (and sold, hopefully!) to obtain a certain amount of profit, P. Here we have a composition of functions. For example, suppose we were asked how much profit there is if we put in M=500 million dollars. From the first table we read off C=3,100,000 chips, and from the second table we can then see that P will be 3.6 million dollars.
I hoped that this was all fairly clear. Now I asked what I thought was a difficult question. Suppose we increase M from 500 million dollars to, say, 503 million dollars. What will the model predict the likely profit will be? We can trace this if we are sufficiently alert. The first marginal quantity we need to consider is C/M. For M=500 million dollars, this is .78. So the new chip production is old chip production + increase in chip production, and this will be 3,100,000+(.78)(3)(1,000). Now let us consider the chip/profit table. With C=3,100,000, we see that profit is supposed to be 3.6 million. But we are changing C by adding on the (relatively small) amount of .78(3)(1,000). The relevant marginal quantity here is P/C, on the row where C is 3,100(,000). The marginal amount here is .05, so that the new profit will be the old profit (3.6 million) plus (.05)(.78)(3)(1,000) million dollars. The 3 comes from perturbing the capital investment. The 1,000 comes from my weird units. The really interesting stuff is (.05)(.78): indeed, this represents the marginal profit as capital invested changes, when the capital investment is 500 million dollars. Symbolically, the multiplication might make sense written this way:
P P C  =   M C MSo the C's just seem to cancel out. This multiplication of rates is typical of what happens when a composition changes. This is more complicated than just multiplying fractions, since the fractions (the marginal stuff, the derivatives) need to be "evaluated" on the appropriate rows of the tables. But the key idea is that the rates multiply, but you need to evaluate the rates at the correct inputs.
Further comments
The method used can certainly be criticized. For example, the marginal
cost of chip production at the level M=200 is .23 and at M=300 it is
.28. What do we know about the marginal cost at, say, 350? In my
examples, I was very careful to consider only changes which were quite
small compared to the original quantity. I would not want to use these
ideas far away from the known data. I would need to rethink my
model.
It is easy to get interesting compositions and rates of change in biology. The transmission of influenza through humans and other species (birds, pigs) provides very interesting examples, although I think the biology is more difficult to explain than the economics behind the tables above.
And about the environment ...
Here's a quote from a Stanford web page, and they should know, since
"Silicon Valley" is immediately to the south:
It takes roughly 10 gallons of water to make a single computer chip.That may not sound like much, but multiply it by the millions of chips made each year, and the result is a large and rapidly growing demand for water. A typical semiconductor factory makes about 2 million integrated circuits per month and gulps about 20 million gallons of water, which ultimately must be disposed of as waste. Chips makers also use large amounts of energy and many toxic chemicals, all of which can harm the environment.
What's going on?
I tried to present heuristic evidence that
would allow us to believe the chain rule.
/heuristic/ adj. 1. allowing or assisting to discover. 2. [Computing] proceeding to a solution by trial and error.The Chain Rule Suppose that f and g are differentiable functions. Then F(x)=f(g(x)) is differentiable, and F´(x)=f´(g(x))·g´(x). Or:
Function  Derivative 

f(g(x))  f´(g(x))·g´(x) 
Examples
My first
example was something like this (about as simple as I could
imagine):
If F(x)=(x^{2}+7)^{300}, what will F´(x) be? Success
here probably will result from recognizing that the chain rule
applies.
If F(x)=f(g(x)), then g(x) is x^{2}+7 so g´(x)=2x, and f(x) is
x^{300} so f´(x)=300x^{299}. Thus
F´(x)=f´(g(x))g´(x)=f´(x^{2}+7)(2x)=300(x^{2}+7)^{299}(2x).
Whew!
An alternate strategy
You could take the remainder of the summer off and "expand"
(x^{2}+7)^{300}. It is only a polynomial of degree
600. And then it will be easy to differentiate this as a sum of
constants multiplying monomials. That's it: report back next fall on
this method. Be sure to get everything correct.
Now for some realistic comments: almost no one ever bothers to
write all of the intermediate steps. That is, in practice
very few f's and g's are actually identified. What happens is that
people see and differentiate the outside most function (f above), put
in the inner function (g) in that derivative, and then multiply by
g'. For example, consider sin(7x^{3}+x^{2}). What is
its derivative? The outside function is sine, whose derivative is
cosine. So I begin by writing and thinking the following:
cos(what's inside)·the
derivative of what's inside.
The result is
cos(7x^{3}+x^{2})·(7[3x^{2}]+2x).
This expression is
a formula for the derivative of
sin(7x^{3}+x^{2}). Again, I urge you to consider the
significance and necessity (!) of appropriate parentheses in
these expressions. The "argument" of cosine is 7x^{3}+x^{2}
and the cosine expression is then multiplied by the expression
(7[3x^{2}}+2x).
Example
What is the derivative of [sin(x^{2})]^{100}? I emphasize that this is not
a situation where the 100 and the 2 could be combined.
This is a
triple composition, but don't worry about it: just handle each layer as
you see it.
The derivative is 100[sin(x^{2})]^{99}·[sin(x^{2})]·2x.
Maybe using the Chain Rule is like peeling an onion, layer after layer, and each layer when "peeled" (differentiated?) contains a copy of the contents of its unpeeled self. Sigh. Sometimes metaphors are really silly. (Why are you crying? I don't always cry when I peel an onion.)
We did a large number of examples. I tried to mix in the quotient and product rules, and the derivative of tangent. The formulas were horrible, ludicrous, silly, etc.
Exponential growth; exponential decay
I "reviewed" this at jet speed. The reason for the quotes is that I
don't think there was much real review  I just wrote some formulas
on the board.
There are exponential and logarithmic rules which I hope you know. Although many of them are "wired into" calculators, the need for people to have some feeling about exponentials is important in many applications (compound interest is basically exponentiation, and of course pH calculations are logarithms).
Exponential rules 

A^{B+C}=A^{B}·A^{C} (A^{B})^{C}=A^{BC} A^{0}=1 A^{B}=1/A^{B} 
Notice that things aren't the same:
For example, the function f(x)=2^{x} is not the same as
g(x)=x^{2}: f(5)=2^{5}=32 and
g(5)=5^{2}=25.
Payment Principal Interest Principal Cumulative Principal number payment payment repaid interest balance 1 488.18 833.33 488.18 833.33 99511.82 2 492.24 829.27 980.42 1662.60 99019.58 3 496.35 825.16 1476.77 2487.76 98523.23 4 500.48 821.03 1977.25 3308.79 98022.75 5 504.65 816.86 2481.90 4125.65 97518.10 10 years go by ... 116 1267.80 53.71 94822.90 58472.26 5177.10 117 1278.37 43.14 96101.27 58515.40 3898.73 118 1289.02 32.49 97390.29 58547.89 2609.71 119 1299.76 21.75 98690.05 58569.64 1309.95 120 *1309.95 10.92 100000.00 58580.56 0.00 *The final payment has been adjusted to account for payments having been rounded to the nearest cent.I got these numbers from a web page of Bret Whissel. Links on that page (especially the link labeled, "More about this calculator") describe the math used. The key is essentially a decreasing exponential function but adjustments need to be made to be sure that the payments actually all work out neatly.
Rates of change of exponential functions
Well, if f(x)=a^{x}, I'd like to get the derivative of
f(x). So I need to look at lim_{h>0}{f(x+h)f(x)}/h. So:
a^{x+h}a^{x} a^{x}a^{h}a^{x} a^{x}(a^{h}1) a^{h}1 lim  = lim  = lim  = a^{x} lim  h>0 h h>0 h h>0 h h>0 hI can't just plug in h=0 because otherwise I'll be dividing by 0. Then I try some algebraic manipulations to perhaps reveal what is really going on. The only thing I get is a common factor of a^{x} which, since there's no h in it, I can bring "out" of the limit. I need to look at the limit as h>0 of {a^{h}1}/h.
The slope of a secant line
Look at y=a^{x}. When x=0, then y=a^{0}=1, so (0,1) is
on the curve. The situation when h is a small positive number is shown
in the picture. Then (h,a^{h}) is on the curve
y=a^{x}, and the slope of the secant line connecting these
two points is [a^{h}1]/[h0]={a^{h}1}/h.
The second page of the Chipco
information contained pictures like those below. The limit as h>0 of
{a^{h}1}/h is rather subtle, and I first looked at it with
a=2.
Graphs of y=2^{x} and y=.69315x+1  

On the interval 2≤x≤2  On the interval .2≤x≤.2  On the interval .02≤x≤.02 
Graphs of y=3^{x} and y=1.09861x+1  

On the interval 2≤x≤2  On the interval .2≤x≤.2  On the interval .02≤x≤.02 
Now I don't want to deal with weird numbers like 1.09861 and .69315
because I am lazy. 2^{x} is too shallow and 3^{x} is
too steep. Maybe there is a MYSTERY number,
M, so that M^{x} has exactly
slope 1 at x=0. That would be nice (and choosing M
would be very analogous to using radians instead of degrees: the
derivative would be so much easier. Well look at this:
Graphs of y=M^{x} and y=1x+1
On the interval 2≤x≤2
On the interval .2≤x≤.2
On the interval .02≤x≤.02
This is, by the way, 2.704813829. Well, you can see it in the
accompanying table which was handed out, a great souvenir, in class. Give one
to your friends.
This whole chain of ideas was first discovered several hundred years ago by
Euler. In honor of Euler, and also to remember the word "exponential",
the mystery number is called "e", and the attached exponential
function is usually called the exponential function.
The computational strategy shown here for e can be proved to
converge but it does converge rather slowly. For example,
when n=1,000,000, the approximation is 2.718280469 and e is actually
2.718281828459. So even the millionth power is only correct to 4
decimal places. There are faster strategies.
Here is
just about the last derivative formula (except we'll also get a
formula for logs later):
This function is its own derivative because that's how we chose e! The
limit should be 1 for lim_{h>0}{e^{h}1}/h. I hope
you also see a coincidence coming from the discussion of compound
interest. If you compound a 5% loan n times each year for t years, a
formula for the amount is
(1+{.05/n})^{nt}(10,000). But this turns out, as
n>infinity, to be the same as (10,000)e^{.05t}. Hey, isn't
that cool!
Discussion of the hypothetical number M
What can I tell you about M? Well, if h is very small,
(M^{h}1)/h is approximately 1. What's a small
h? We could try maybe h=1/100.
Function Derivative e^{x}
e^{x}
n (1+{1/n})^{n} 1 2.000000000 2 2.250000000 3 2.370370369 4 2.441406250 5 2.488320000 6 2.521626376 7 2.546499699 8 2.565784514 9 2.581174789 10 2.593742460 20 2.653297705 30 2.674318750 40 2.685063838 50 2.691588029 60 2.695970192 70 2.699116318 80 2.701484941 90 2.703332434 100 2.704813829 200 2.711517123 300 2.713764888 400 2.714891744 500 2.715568521 600 2.716020591 700 2.716341924 800 2.716584847 900 2.716772937 1,000 2.716923932 10,000 2.718145926 100,000 2.718268237 1,000,000 2.718280469
Maintained by greenfie@math.rutgers.edu and last modified 7/10/2006.