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PROFESSOR: Hi.
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Our lesson today is called
the 'Inverse Logarithm'.
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00:00:38,210 --> 00:00:41,970
And what it will do, among other
things, is give us a
13
00:00:41,970 --> 00:00:46,450
very nice chance to revisit
inverse functions in general,
14
00:00:46,450 --> 00:00:50,050
only now applied specifically
to the natural logarithm
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00:00:50,050 --> 00:00:53,230
function that we talked
about last time.
16
00:00:53,230 --> 00:00:55,950
So I call today's lesson, as I
say, 'Inverse Logarithms'.
17
00:00:55,950 --> 00:00:59,030
And to see what's coming up
here, simply recall that last
18
00:00:59,030 --> 00:01:02,980
time, we invented, so to speak,
a new function called
19
00:01:02,980 --> 00:01:04,620
the 'natural log of x'.
20
00:01:04,620 --> 00:01:06,070
This had nothing to
do with exponents.
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00:01:06,070 --> 00:01:06,810
It was what?
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00:01:06,810 --> 00:01:11,290
This was the function whose
derivative with respect to 'x'
23
00:01:11,290 --> 00:01:15,460
was '1/x' and passed through
the point (1 , 0).
24
00:01:15,460 --> 00:01:18,320
That uniquely defined this
particular function.
25
00:01:18,320 --> 00:01:19,920
The point being what?
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00:01:19,920 --> 00:01:22,650
That this particular function,
since the domain is for
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00:01:22,650 --> 00:01:25,640
positive 'x', '1/x'
is positive.
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00:01:25,640 --> 00:01:28,880
The curve is always rising,
which means that the function
29
00:01:28,880 --> 00:01:31,020
itself must be one to one.
30
00:01:31,020 --> 00:01:34,030
And because the function is one
to one, it means that the
31
00:01:34,030 --> 00:01:35,640
inverse function exists.
32
00:01:35,640 --> 00:01:38,970
This is no different from any
other example of forming f
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00:01:38,970 --> 00:01:42,650
inverse, given a one to one
function called 'f'.
34
00:01:42,650 --> 00:01:44,590
So what we do is now,
and remember how
35
00:01:44,590 --> 00:01:46,000
you invert this function.
36
00:01:46,000 --> 00:01:49,160
If you want to do this thing
in slow motion, you first
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00:01:49,160 --> 00:01:53,400
rotate this through 90 degrees,
then flip it over.
38
00:01:53,400 --> 00:01:59,410
If you want to do it faster, the
inverse graph of this is
39
00:01:59,410 --> 00:02:03,940
the reflection of this curve
with respect to the line 'y'
40
00:02:03,940 --> 00:02:04,880
equals 'x'.
41
00:02:04,880 --> 00:02:09,000
In any event, if we do this, we
find that the graph of the
42
00:02:09,000 --> 00:02:12,590
function 'y' equals
'inverse log x' is
43
00:02:12,590 --> 00:02:14,530
this particular curve.
44
00:02:14,530 --> 00:02:16,740
And notice the correspondence.
45
00:02:16,740 --> 00:02:20,560
If this is the point (1 , 0) on
the log curve, the inverse
46
00:02:20,560 --> 00:02:22,890
point is (0 , 1).
47
00:02:22,890 --> 00:02:27,880
In other words, the inverse
log of 0 is 1.
48
00:02:27,880 --> 00:02:28,840
OK.
49
00:02:28,840 --> 00:02:30,470
Now, the idea is something
like this.
50
00:02:30,470 --> 00:02:33,880
Just to have a review,
again, of how--
51
00:02:33,880 --> 00:02:35,930
see, the punchline I want to
make for today's lesson, right
52
00:02:35,930 --> 00:02:39,490
at the beginning, is that the
nice thing about studying
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00:02:39,490 --> 00:02:43,510
inverse functions is that once
you know the function of which
54
00:02:43,510 --> 00:02:46,670
you're taking the inverse, you
automatically get all the
55
00:02:46,670 --> 00:02:49,350
mileage you want out of
the inverse function.
56
00:02:49,350 --> 00:02:53,720
Well, by way of illustration,
if the function natural log
57
00:02:53,720 --> 00:02:57,835
had the usual logarithmic
property, we would suspect
58
00:02:57,835 --> 00:02:59,960
that its inverse
should have the
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00:02:59,960 --> 00:03:02,590
usual exponential property.
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00:03:02,590 --> 00:03:05,980
Now what is the usual
exponential property?
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00:03:05,980 --> 00:03:09,650
The exponential function is
characterized by what?
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00:03:09,650 --> 00:03:14,840
That if you take 'f of 'x1 plus
x2'', that's 'f of x1'
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00:03:14,840 --> 00:03:16,470
times 'f of x2'.
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00:03:16,470 --> 00:03:17,710
So you think of it
in terms of the
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00:03:17,710 --> 00:03:19,880
usual exponential notation.
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00:03:19,880 --> 00:03:24,790
If you have, say, 10 raised to
the 'x1 plus x2' power, that's
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00:03:24,790 --> 00:03:27,810
the same as 10 to the
'x1' power times
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00:03:27,810 --> 00:03:29,450
10 to the 'x2' power.
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00:03:29,450 --> 00:03:33,070
In other words, if our ln
function is genuinely a
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00:03:33,070 --> 00:03:36,090
logarithmic function, we would
expect the inverse of that
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00:03:36,090 --> 00:03:39,280
function to be a genuine
exponential function.
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00:03:39,280 --> 00:03:42,560
And just to work with inverses
again, let's see if this is
73
00:03:42,560 --> 00:03:44,110
indeed true.
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00:03:44,110 --> 00:03:47,310
Let's see if it's really true
that the 'inverse log of 'x1
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00:03:47,310 --> 00:03:50,530
plus x2'' is the 'inverse
log of x1' times the
76
00:03:50,530 --> 00:03:51,890
'inverse log of x2'.
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00:03:51,890 --> 00:03:53,910
You see, the whole
idea is what?
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00:03:53,910 --> 00:03:55,600
Let's give these things names.
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00:03:55,600 --> 00:03:59,770
Let 'y1' be the 'inverse log
of x1' and 'y2' be the
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00:03:59,770 --> 00:04:01,930
'inverse log of x2'.
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00:04:01,930 --> 00:04:03,830
Now, since we've already
studied the natural log
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00:04:03,830 --> 00:04:07,290
function, the most natural thing
to do here is to invert
83
00:04:07,290 --> 00:04:13,000
these, namely 'y1' equals the
inverse 'natural log of x1' is
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00:04:13,000 --> 00:04:16,529
the same as saying 'x1' is
the 'natural log of y1'.
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00:04:16,529 --> 00:04:19,640
And similarly, this is the same
as saying that 'x2' is
86
00:04:19,640 --> 00:04:21,529
the 'natural log of y2'.
87
00:04:21,529 --> 00:04:24,460
And now we add equals to equals,
and we get that 'x1
88
00:04:24,460 --> 00:04:29,180
plus x2' is 'natural log y1'
plus 'natural log y2'.
89
00:04:29,180 --> 00:04:32,330
But allegedly, we understand the
properties of the natural
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00:04:32,330 --> 00:04:35,020
log, because if we didn't
understand those, it would be
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00:04:35,020 --> 00:04:37,660
kind of futile to be studying
the inverse function.
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00:04:37,660 --> 00:04:39,870
What do we know about
the natural log?
93
00:04:39,870 --> 00:04:44,140
We know that the 'natural
log of 'y1 times y2', by
94
00:04:44,140 --> 00:04:46,610
definition of a logarithmic
function, is 'log
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00:04:46,610 --> 00:04:48,850
y1' plus 'log y2'.
96
00:04:48,850 --> 00:04:53,020
In other words, 'natural log y1'
plus 'natural log y2', by
97
00:04:53,020 --> 00:04:55,920
the property of being
logarithmic, is just the
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00:04:55,920 --> 00:04:58,880
'natural log of 'y1
times y2''.
99
00:04:58,880 --> 00:05:02,640
And now, taking this
relationship and inverting it
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00:05:02,640 --> 00:05:05,670
to say this is the same
as saying what?
101
00:05:05,670 --> 00:05:10,690
The 'inverse natural log of 'x1
plus x2'' is equal to 'y1'
102
00:05:10,690 --> 00:05:12,050
times 'y2'.
103
00:05:12,050 --> 00:05:16,400
And if we now observe that 'y1'
was the 'inverse natural
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00:05:16,400 --> 00:05:20,890
log of x1', and 'y2' is the
'inverse natural log of x2',
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00:05:20,890 --> 00:05:24,290
we have the result that we
claimed we would have at the
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00:05:24,290 --> 00:05:25,700
beginning here.
107
00:05:25,700 --> 00:05:29,680
Now again, here's one example
where it's not the result that
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00:05:29,680 --> 00:05:31,210
I'm not interested in.
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00:05:31,210 --> 00:05:35,230
But I'm interested in showing,
again, what we meant when we
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00:05:35,230 --> 00:05:39,100
said that once we learn a new
concept, we can bring to it
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00:05:39,100 --> 00:05:41,240
all of our old knowledge.
112
00:05:41,240 --> 00:05:43,510
You see, all of these things
come from our basic
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00:05:43,510 --> 00:05:44,300
definition.
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00:05:44,300 --> 00:05:46,230
Well, let's carry on
one step further.
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00:05:46,230 --> 00:05:48,670
You see, this was an arithmetic
property.
116
00:05:48,670 --> 00:05:51,480
Let's see if we can get some
calculus properties about the
117
00:05:51,480 --> 00:05:55,390
inverse natural log based on our
knowledge of the natural
118
00:05:55,390 --> 00:05:56,260
log itself.
119
00:05:56,260 --> 00:06:00,350
Well, for example, in a course
of this type, the most natural
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00:06:00,350 --> 00:06:03,600
thing to do, I guess, is given
any function, we would always
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00:06:03,600 --> 00:06:06,270
like to be able to talk about
its derivative, provided, of
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00:06:06,270 --> 00:06:08,970
course, the function
is differentiable.
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00:06:08,970 --> 00:06:12,580
So for example, a typical
question that one might ask is
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00:06:12,580 --> 00:06:17,780
find 'dy dx', if 'y' is 'inverse
natural log x'.
125
00:06:17,780 --> 00:06:21,130
Again, this works the same way
as it did for the inverse trig
126
00:06:21,130 --> 00:06:24,230
functions, for the inverse of
everything that we've done so
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00:06:24,230 --> 00:06:25,390
far in this course.
128
00:06:25,390 --> 00:06:28,480
We start with this particular
relationship and right away
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00:06:28,480 --> 00:06:31,390
translate it into that which
we're more familiar with,
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00:06:31,390 --> 00:06:35,770
namely we translate 'y' equals
the 'inverse natural log of x'
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00:06:35,770 --> 00:06:38,890
into 'x' equals 'natural
log y'.
132
00:06:38,890 --> 00:06:41,960
Now you see again, since I know
how to differentiate 'log
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00:06:41,960 --> 00:06:45,180
y' with respect to 'y', the
derivative of 'natural log y'
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00:06:45,180 --> 00:06:48,400
with respect to 'y' was
by definition '1/y'.
135
00:06:48,400 --> 00:06:50,900
From this relationship
here, I can find
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00:06:50,900 --> 00:06:55,050
that 'dx dy' is '1/y'.
137
00:06:55,050 --> 00:06:58,950
And knowing that 'dx dy' is
'1/y', using my inverse
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00:06:58,950 --> 00:07:02,650
function general theory that
tells me that to find the
139
00:07:02,650 --> 00:07:07,120
derivative of 'y' with respect
to 'x', all I have to do is
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00:07:07,120 --> 00:07:10,210
invert the derivative of 'x'
with respect to 'y', I wind up
141
00:07:10,210 --> 00:07:13,890
with the fact that 'dy dx'
is equal to 'y' itself.
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00:07:13,890 --> 00:07:17,410
In other words, the 'inverse
log x' function is
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00:07:17,410 --> 00:07:21,100
characterized by the fact that
it's its own derivative.
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00:07:21,100 --> 00:07:23,610
In other words, it's a
non-destructible type of
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00:07:23,610 --> 00:07:26,320
function with respect
to differentiation.
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00:07:26,320 --> 00:07:29,000
This is a rather powerful
property.
147
00:07:29,000 --> 00:07:30,010
You see what this thing says?
148
00:07:30,010 --> 00:07:31,890
It says that the derivative
of 'y' with respect
149
00:07:31,890 --> 00:07:34,640
to 'x' is 'y' itself.
150
00:07:34,640 --> 00:07:38,720
By the way, just as an aside, we
can do the converse of this
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00:07:38,720 --> 00:07:39,830
particular problem.
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00:07:39,830 --> 00:07:43,120
You see, we started with 'y'
equals 'inverse log x' and
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00:07:43,120 --> 00:07:46,030
showed that this particular
equation was satisfied.
154
00:07:46,030 --> 00:07:48,710
Notice that if we start with
this particular equation,
155
00:07:48,710 --> 00:07:53,190
starting with 'dy dx' equals
'y', separate the variables,
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00:07:53,190 --> 00:07:58,940
and lo and behold, we wind up
with 'dy over y' equals 'dx'.
157
00:07:58,940 --> 00:08:02,360
And without carrying out the
details, I leave this to you,
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00:08:02,360 --> 00:08:04,030
because it is straightforward.
159
00:08:04,030 --> 00:08:08,420
Notice that as we look at the
left hand side here, we
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00:08:08,420 --> 00:08:10,650
hopefully will think
immediately
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00:08:10,650 --> 00:08:12,550
of the natural logarithm.
162
00:08:12,550 --> 00:08:14,040
Namely, what do we want here?
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00:08:14,040 --> 00:08:19,240
The function whose derivative
with respect to 'y' is '1/y'.
164
00:08:19,240 --> 00:08:21,320
Again, you see what I'm saying
is notice that if we had
165
00:08:21,320 --> 00:08:23,910
started with this particular
differential equation, we
166
00:08:23,910 --> 00:08:27,130
could have shown that a
logarithm, and hence solving
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00:08:27,130 --> 00:08:29,560
for 'y' explicitly, would
have led to the
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00:08:29,560 --> 00:08:31,020
inverse logarithm also.
169
00:08:31,020 --> 00:08:33,780
But at any rate, notice then
that we can find the
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00:08:33,780 --> 00:08:37,750
derivative of the inverse log
just by knowing how to find
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00:08:37,750 --> 00:08:40,659
the derivative of the natural
log itself, which is as we
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00:08:40,659 --> 00:08:42,549
expect things should be.
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00:08:42,549 --> 00:08:47,460
Again, and this is a notational
thing, we do not
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00:08:47,460 --> 00:08:50,680
use the language in general
"inverse log x."
175
00:08:50,680 --> 00:08:53,500
In other words, it's quite
conventional to talk about
176
00:08:53,500 --> 00:08:54,970
'inverse sine x'.
177
00:08:54,970 --> 00:08:58,500
You may notice that sometimes
in the text, sometimes in my
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00:08:58,500 --> 00:09:01,570
lectures I may have used
the word 'arcsin x'.
179
00:09:01,570 --> 00:09:07,080
But the general notation
is sine to the minus 1.
180
00:09:07,080 --> 00:09:09,860
However, when it comes to the
natural logarithm function,
181
00:09:09,860 --> 00:09:12,480
the inverse is usually
not written as
182
00:09:12,480 --> 00:09:14,460
'inverse natural log x'.
183
00:09:14,460 --> 00:09:19,610
It's usually abbreviated, and
let me say this, by the symbol
184
00:09:19,610 --> 00:09:21,180
'e to the x'.
185
00:09:21,180 --> 00:09:25,170
Now the reason I say by
the symbol is this.
186
00:09:25,170 --> 00:09:28,100
If I wanted to, I can say, look,
If I've never heard of
187
00:09:28,100 --> 00:09:31,910
exponents before, let this
symbol be an abbreviation for
188
00:09:31,910 --> 00:09:33,550
'inverse log x'.
189
00:09:33,550 --> 00:09:35,430
Does 'inverse log x' exists?
190
00:09:35,430 --> 00:09:38,030
Yes, I even drew the graph of
it at the beginning of this
191
00:09:38,030 --> 00:09:39,360
particular lecture.
192
00:09:39,360 --> 00:09:43,730
However, again, if you are
tempted to use exponents, the
193
00:09:43,730 --> 00:09:48,520
notation 'e to the x' is in
keeping in line with the idea
194
00:09:48,520 --> 00:09:49,730
of our previous lecture.
195
00:09:49,730 --> 00:09:50,200
I say what?
196
00:09:50,200 --> 00:09:53,460
This matches the identification
of 'natural log
197
00:09:53,460 --> 00:09:56,780
x' with the 'log of x'
to the base 'e'.
198
00:09:56,780 --> 00:09:59,230
Remember, in our previous
lecture, we mentioned that
199
00:09:59,230 --> 00:10:03,030
this particular function exists
without having to talk
200
00:10:03,030 --> 00:10:04,160
about a base.
201
00:10:04,160 --> 00:10:07,780
But if you wanted to identify
it with a traditional
202
00:10:07,780 --> 00:10:11,180
logarithm, the base you would
have to pick is the base 'e',
203
00:10:11,180 --> 00:10:14,430
where we showed that 'e' was
some number between 2 and 4.
204
00:10:14,430 --> 00:10:17,850
In other words, again, it's just
like our 'dy dx' versus
205
00:10:17,850 --> 00:10:21,160
'dy' divided by 'dx' and
other symbols that we
206
00:10:21,160 --> 00:10:22,460
invented this way.
207
00:10:22,460 --> 00:10:27,230
Namely, if I look at 'e to the
x' as being the inverse of the
208
00:10:27,230 --> 00:10:31,100
natural log, or whether I look
at it as being 'e' raised to
209
00:10:31,100 --> 00:10:33,870
the 'x' power, where 'e' is that
number that's someplace
210
00:10:33,870 --> 00:10:37,720
between 2 and 4, I don't get
into any trouble either way
211
00:10:37,720 --> 00:10:40,550
because of the natural
identification of choosing e
212
00:10:40,550 --> 00:10:42,440
to be the base of my system.
213
00:10:42,440 --> 00:10:44,290
But I just mention
that in passing.
214
00:10:44,290 --> 00:10:46,240
Now you see, the rest
of today's lecture
215
00:10:46,240 --> 00:10:47,990
will go fairly quickly.
216
00:10:47,990 --> 00:10:50,630
And the reason that it will go
fairly quickly is that once
217
00:10:50,630 --> 00:10:53,310
we've established what the
function is that we're talking
218
00:10:53,310 --> 00:10:56,470
about, every other property of
that function is going to
219
00:10:56,470 --> 00:10:59,210
follow from the principles that
we've already learned.
220
00:10:59,210 --> 00:11:03,010
For example, in terms of our
new notation, since the
221
00:11:03,010 --> 00:11:06,950
derivative of 'e to the x' with
respect to 'x' is 'e to
222
00:11:06,950 --> 00:11:09,320
the x' itself, remember,
'e to the x' now can
223
00:11:09,320 --> 00:11:10,300
be viewed as what?
224
00:11:10,300 --> 00:11:14,500
'e to the x' power, or, to be on
safer grounds, if you want
225
00:11:14,500 --> 00:11:18,280
to be consistent, it's just an
abbreviation for 'inverse
226
00:11:18,280 --> 00:11:19,590
natural log x'.
227
00:11:19,590 --> 00:11:24,030
But at any rate, if 'u' is now
any differentiable function of
228
00:11:24,030 --> 00:11:26,750
'x', notice that to find the
derivative of 'e to the u'
229
00:11:26,750 --> 00:11:30,230
with respect to 'x', by the
chain rule, I can say what?
230
00:11:30,230 --> 00:11:33,780
It's the derivative of 'e to
the u' with respect to 'u'
231
00:11:33,780 --> 00:11:35,400
times 'du dx'.
232
00:11:35,400 --> 00:11:38,310
We've just shown that the
derivative of 'e to the u'
233
00:11:38,310 --> 00:11:41,430
with respect to 'u' is
'e' to the 'u' again.
234
00:11:41,430 --> 00:11:43,940
And therefore, the derivative
of 'e to the u' with respect
235
00:11:43,940 --> 00:11:47,820
to 'x' is 'e to the
u' times 'du dx'.
236
00:11:47,820 --> 00:11:51,070
By the way, this again leads to
another interesting thing
237
00:11:51,070 --> 00:11:55,260
that makes integrals tough
to handle, in a way.
238
00:11:55,260 --> 00:11:57,510
We've mentioned this before,
but here's another nice,
239
00:11:57,510 --> 00:12:01,010
natural environment to bring
this problem up in again.
240
00:12:01,010 --> 00:12:04,020
When we discussed the second
fundamental theorem of
241
00:12:04,020 --> 00:12:07,090
integral calculus to show how
one actually would have to
242
00:12:07,090 --> 00:12:10,500
understand areas to be able
to find a function whose
243
00:12:10,500 --> 00:12:14,690
derivative was 'e to the
'minus x squared', we
244
00:12:14,690 --> 00:12:17,920
mentioned, we didn't prove it,
we mentioned that there was no
245
00:12:17,920 --> 00:12:20,820
familiar function whose
derivative with respect to 'x'
246
00:12:20,820 --> 00:12:22,920
was 'e to the 'minus
x squared'.
247
00:12:22,920 --> 00:12:25,560
So this would be a very
difficult problem to handle.
248
00:12:25,560 --> 00:12:27,160
Now, let's look at
this one instead.
249
00:12:27,160 --> 00:12:31,430
Let's look at integral '2x 'e
to the minus x squared' dx'.
250
00:12:31,430 --> 00:12:36,440
To the untrained eye, it would
appear that this is even
251
00:12:36,440 --> 00:12:37,300
messier than this.
252
00:12:37,300 --> 00:12:39,610
In other words, this is just 'e
to the 'minus x squared''.
253
00:12:39,610 --> 00:12:42,900
This one has 'e to the 'minus
x squared'' with a '2x'
254
00:12:42,900 --> 00:12:45,720
dangling in front of it, and
that looks even tougher.
255
00:12:45,720 --> 00:12:49,780
But notice that if we look
at our exponent, 'minus x
256
00:12:49,780 --> 00:12:53,190
squared', the derivative, or the
differential of 'minus x
257
00:12:53,190 --> 00:12:54,290
squared' is what?
258
00:12:54,290 --> 00:12:57,180
It's 'minus '2x dx''.
259
00:12:57,180 --> 00:13:01,340
In other words, the '2x dx' is
precisely what you need to
260
00:13:01,340 --> 00:13:04,575
reduce this to the form integral
''e to the u' du'.
261
00:13:04,575 --> 00:13:07,340
In other words, working this
thing out in more specific
262
00:13:07,340 --> 00:13:11,970
detail, if I let 'u' equal
'minus x squared', 'du'
263
00:13:11,970 --> 00:13:16,570
becomes 'minus '2x dx'', and
therefore integral '2x 'e to
264
00:13:16,570 --> 00:13:19,330
the 'minus x squared'' dx'
just becomes what?
265
00:13:19,330 --> 00:13:22,940
Well, the 'e to the 'minus x
squared'' just becomes 'e to
266
00:13:22,940 --> 00:13:24,140
the minus u'.
267
00:13:24,140 --> 00:13:28,400
And '2x dx' is just
'minus du'.
268
00:13:28,400 --> 00:13:29,650
OK?
269
00:13:29,650 --> 00:13:33,220
270
00:13:33,220 --> 00:13:36,140
I'm sorry, 'u' is 'minus
x squared'.
271
00:13:36,140 --> 00:13:37,980
'u' is 'minus x squared'.
272
00:13:37,980 --> 00:13:39,880
So 'minus x squared' is 'u'.
273
00:13:39,880 --> 00:13:43,120
So this becomes 'minus
e to the 'u du''.
274
00:13:43,120 --> 00:13:45,970
And that, of course,
is just minus.
275
00:13:45,970 --> 00:13:47,610
See, the minus will
come outside.
276
00:13:47,610 --> 00:13:50,200
The integral of 'e to the u'
with respect to 'u' is
277
00:13:50,200 --> 00:13:51,605
just 'e to the u'.
278
00:13:51,605 --> 00:13:58,900
And since 'u' is equal to 'minus
x squared', all we're
279
00:13:58,900 --> 00:14:02,270
saying is that if you
differentiate minus 'e' to the
280
00:14:02,270 --> 00:14:05,020
'minus x squared', you
wind up with what?
281
00:14:05,020 --> 00:14:08,790
'2x 'e to the 'minus x
squared'', the reason being
282
00:14:08,790 --> 00:14:12,390
that you must multiply this by
the derivative of the exponent
283
00:14:12,390 --> 00:14:13,320
with respect to 'x'.
284
00:14:13,320 --> 00:14:15,650
The root of the exponent
is 'minus 2x'.
285
00:14:15,650 --> 00:14:19,180
'Minus 2x' times minus
1 is '2x'.
286
00:14:19,180 --> 00:14:22,870
If you want to see this in
more concise differential
287
00:14:22,870 --> 00:14:28,570
notation, you see, what we're
saying is that the integral of
288
00:14:28,570 --> 00:14:34,240
'e' to some power with respect
to that same power is 'e' to
289
00:14:34,240 --> 00:14:36,060
that power plus a constant.
290
00:14:36,060 --> 00:14:38,550
In other words, the function
that one would have to
291
00:14:38,550 --> 00:14:42,610
integrate 'e to the 'minus x
squared'' with respect to to
292
00:14:42,610 --> 00:14:45,760
wind up with 'e to the 'minus x
squared'' would be 'minus x
293
00:14:45,760 --> 00:14:46,860
squared' itself.
294
00:14:46,860 --> 00:14:50,200
And you see, now, in the next
step, since this is a true
295
00:14:50,200 --> 00:14:53,800
statement, the differential
of 'minus x squared'
296
00:14:53,800 --> 00:14:55,590
is minus '2x dx'.
297
00:14:55,590 --> 00:15:01,810
298
00:15:01,810 --> 00:15:04,810
And now you see, factoring out
the minus sign and multiplying
299
00:15:04,810 --> 00:15:07,890
through by minus 1, we arrive
at the same result that we
300
00:15:07,890 --> 00:15:09,180
wound up with before.
301
00:15:09,180 --> 00:15:12,720
But again, this is the kind of
material that we can drill on
302
00:15:12,720 --> 00:15:16,590
extensively in the exercises
in this particular unit.
303
00:15:16,590 --> 00:15:18,020
The technique is what?
304
00:15:18,020 --> 00:15:21,010
That the basic building block
of the so-called exponential
305
00:15:21,010 --> 00:15:24,530
function, the inverse natural
log, is that when you
306
00:15:24,530 --> 00:15:27,560
differentiate it, you
essentially do not destroy it.
307
00:15:27,560 --> 00:15:30,320
In other words, the derivative
of 'e to the u' with respect
308
00:15:30,320 --> 00:15:34,640
to 'x' is 'e to the
u' times 'du dx'.
309
00:15:34,640 --> 00:15:37,550
And I thought that in closing
today's lesson, I might as
310
00:15:37,550 --> 00:15:42,210
well show you a very powerful
application of this particular
311
00:15:42,210 --> 00:15:45,100
result in a non-obvious
situation.
312
00:15:45,100 --> 00:15:48,540
And it's something that
we call second-order
313
00:15:48,540 --> 00:15:49,570
differential equations.
314
00:15:49,570 --> 00:15:51,770
I've picked out here a
differential equation.
315
00:15:51,770 --> 00:15:55,310
Let me just show you what I
have in mind over here.
316
00:15:55,310 --> 00:15:58,770
Suppose I tell you that 'y'
is a twice-differentiable
317
00:15:58,770 --> 00:16:02,450
function of 'x' that satisfies
the following identity, that
318
00:16:02,450 --> 00:16:05,530
the second derivative of 'y'
with respect to 'x' minus 5
319
00:16:05,530 --> 00:16:09,140
times the first derivative of
'y' with respect to 'x' plus 6
320
00:16:09,140 --> 00:16:12,150
times 'y' is identically 0.
321
00:16:12,150 --> 00:16:15,620
And the question is, if this
equation is to be obeyed, what
322
00:16:15,620 --> 00:16:17,360
must 'y' be?
323
00:16:17,360 --> 00:16:20,000
And a very interesting technique
for solving this
324
00:16:20,000 --> 00:16:22,650
kind of a problem, you see, this
is called a second-order
325
00:16:22,650 --> 00:16:25,310
differential equation because
the highest derivative that
326
00:16:25,310 --> 00:16:27,760
appears is the second
derivative.
327
00:16:27,760 --> 00:16:30,720
OK, a rather powerful
technique using the
328
00:16:30,720 --> 00:16:34,610
exponential is available to us
for problems of this sort.
329
00:16:34,610 --> 00:16:36,290
And the idea hinges on this.
330
00:16:36,290 --> 00:16:40,400
As a trial solution, which I'll
call 'y sub t', let's try
331
00:16:40,400 --> 00:16:43,920
'e to the rx', where 'r' happens
to be a constant.
332
00:16:43,920 --> 00:16:47,410
You see, the whole idea is if I
differentiate 'e to the rx'
333
00:16:47,410 --> 00:16:51,460
with respect to 'x', I get 'e
to the rx' back again, only
334
00:16:51,460 --> 00:16:55,190
with a factor of 'r', namely the
derivative of my exponent
335
00:16:55,190 --> 00:16:56,150
in this case.
336
00:16:56,150 --> 00:16:57,290
See, 'r' is a constant.
337
00:16:57,290 --> 00:17:00,970
The derivative of 'rx' with
respect to 'x' is just 'r'.
338
00:17:00,970 --> 00:17:03,720
So notice that the first
derivative of 'y sub t' with
339
00:17:03,720 --> 00:17:06,550
respect to 'x' is 'r
'e to the rx''.
340
00:17:06,550 --> 00:17:09,680
The second derivative
of 'y sub t' with
341
00:17:09,680 --> 00:17:11,339
respect to 'x' is what?
342
00:17:11,339 --> 00:17:12,579
'r' is a constant.
343
00:17:12,579 --> 00:17:14,530
I differentiate 'e to the rx'.
344
00:17:14,530 --> 00:17:17,640
That brings down another factor
of 'r' and leaves me
345
00:17:17,640 --> 00:17:19,300
with 'e to the rx'.
346
00:17:19,300 --> 00:17:23,970
You see, the whole idea being,
notice that 'e to the rx' is a
347
00:17:23,970 --> 00:17:28,580
common factor of 'y sub t', 'y
sub 't prime'', 'y sub 't
348
00:17:28,580 --> 00:17:29,620
double prime''.
349
00:17:29,620 --> 00:17:33,030
If I now substitute these
results back into my original
350
00:17:33,030 --> 00:17:36,950
equation, look what I wind up
with. 'y double prime' becomes
351
00:17:36,950 --> 00:17:39,080
'r squared 'e to the rx''.
352
00:17:39,080 --> 00:17:45,230
Minus '5y prime', that's minus
'5r 'e to the rx'', plus '6y',
353
00:17:45,230 --> 00:17:46,790
'6e to the rx'.
354
00:17:46,790 --> 00:17:48,650
And that must equal zero.
355
00:17:48,650 --> 00:17:49,880
And here's the key point.
356
00:17:49,880 --> 00:17:53,100
'e to the rx' is now
a common factor.
357
00:17:53,100 --> 00:17:55,700
I factor that out.
358
00:17:55,700 --> 00:17:59,240
If the product of two numbers
is 0, one of the
359
00:17:59,240 --> 00:18:01,220
factors must be 0.
360
00:18:01,220 --> 00:18:04,730
But notice from our graph of the
exponential, the inverse
361
00:18:04,730 --> 00:18:10,520
logarithm, 'e to the x', 'e to
the rx' can never be negative
362
00:18:10,520 --> 00:18:12,550
and can never be 0,
in fact. 'e to the
363
00:18:12,550 --> 00:18:14,190
rx' is always positive.
364
00:18:14,190 --> 00:18:17,580
Therefore, if 'e to the rx' is
always positive, it must be
365
00:18:17,580 --> 00:18:21,100
the other factor which is 0.
366
00:18:21,100 --> 00:18:25,540
But 'r squared' minus '5r' plus
6 is not a second-order
367
00:18:25,540 --> 00:18:26,920
differential equation.
368
00:18:26,920 --> 00:18:29,690
It's a second-degree polynomial
equation.
369
00:18:29,690 --> 00:18:32,940
It's a familiar quadratic
equation, which I can solve
370
00:18:32,940 --> 00:18:34,010
quite easily.
371
00:18:34,010 --> 00:18:35,550
In other words, I find what?
372
00:18:35,550 --> 00:18:39,300
That 'r' must be either 2,
or 'r' must equal 3.
373
00:18:39,300 --> 00:18:43,480
In other words, my claim is that
either 'e to the 2x' or
374
00:18:43,480 --> 00:18:46,190
'e to the 3x' must
be a solution of
375
00:18:46,190 --> 00:18:47,790
this particular equation.
376
00:18:47,790 --> 00:18:52,440
Of course, we can do more with
that, which we will in a later
377
00:18:52,440 --> 00:18:57,050
part of calculus, not in
this package's work.
378
00:18:57,050 --> 00:18:59,840
But we're not going to study
differential equations in
379
00:18:59,840 --> 00:19:01,040
great detail here.
380
00:19:01,040 --> 00:19:04,340
But for our present purposes,
I think this illustrates how
381
00:19:04,340 --> 00:19:08,050
one can use the fact that it's
a rather powerful structural
382
00:19:08,050 --> 00:19:11,630
property when the derivative of
a function with respect to
383
00:19:11,630 --> 00:19:13,750
'x' is the function itself.
384
00:19:13,750 --> 00:19:17,240
By the way, as a quick check,
notice that if 'y' equals 'e
385
00:19:17,240 --> 00:19:20,590
to the 2x', 'y prime' is
twice 'e to the 2x'.
386
00:19:20,590 --> 00:19:23,850
'y double prime' is
'4 e to the 2x'.
387
00:19:23,850 --> 00:19:27,200
And therefore 'y double prime'
minus '5y prime'
388
00:19:27,200 --> 00:19:29,170
plus '6y' is what?
389
00:19:29,170 --> 00:19:35,430
It's '4e to the 2x' minus '10 e
to the 2x', '6 e to the 2x'.
390
00:19:35,430 --> 00:19:39,790
And that, in fact, is genuinely,
identically zero.
391
00:19:39,790 --> 00:19:43,400
In a similar way, checking out
'y' equals 'e to the 3x', we
392
00:19:43,400 --> 00:19:47,470
get 'y prime' is '3e to the 3x',
'y double prime' is '9e
393
00:19:47,470 --> 00:19:48,610
to the 3x'.
394
00:19:48,610 --> 00:19:54,550
Therefore, '9e to the 3x' minus
'15 e to the 3x' plus
395
00:19:54,550 --> 00:19:58,670
'6e to the 3x' is again,
identically 0.
396
00:19:58,670 --> 00:20:01,140
And again, you see what this
powerful technique is.
397
00:20:01,140 --> 00:20:05,500
In general, if 'a' and 'b' are
constants, the substitution 'y
398
00:20:05,500 --> 00:20:09,360
sub t' equals 'e to the rx'.
399
00:20:09,360 --> 00:20:13,000
And there'll be a problem in the
exercises on this to give
400
00:20:13,000 --> 00:20:14,300
you additional drill.
401
00:20:14,300 --> 00:20:17,770
But that substitution transforms
the second or the
402
00:20:17,770 --> 00:20:21,700
differential equation, 'y double
prime' plus 'ay prime'
403
00:20:21,700 --> 00:20:26,750
plus 'by' equals 0, into an
equivalent quadratic equation,
404
00:20:26,750 --> 00:20:30,010
'r squared' plus 'ar'
plus 'b' equals 0.
405
00:20:30,010 --> 00:20:32,740
And you see from this equation,
using the quadratic
406
00:20:32,740 --> 00:20:36,470
formula, we can find the values
of 'r' that satisfy
407
00:20:36,470 --> 00:20:38,400
this, and that gives
us a couple of
408
00:20:38,400 --> 00:20:40,040
solutions to the equation.
409
00:20:40,040 --> 00:20:43,410
Well, at any rate, I
deliberately want this lecture
410
00:20:43,410 --> 00:20:46,710
to stay short to make the
most important impact.
411
00:20:46,710 --> 00:20:48,490
And that is again what?
412
00:20:48,490 --> 00:20:51,660
That once we knew what the
natural log function was, and
413
00:20:51,660 --> 00:20:55,720
we defined the inverse natural
log function, everything that
414
00:20:55,720 --> 00:20:59,090
we wanted to know about the
inverse natural log followed
415
00:20:59,090 --> 00:21:01,380
from the properties
of the log itself.
416
00:21:01,380 --> 00:21:04,200
And this is basically the
lesson for today.
417
00:21:04,200 --> 00:21:07,270
We will continue the discussion
of exponential
418
00:21:07,270 --> 00:21:08,990
functions from a different
point of
419
00:21:08,990 --> 00:21:10,720
view in our next lecture.
420
00:21:10,720 --> 00:21:12,850
But until that next
lecture, goodbye.
421
00:21:12,850 --> 00:21:15,780
422
00:21:15,780 --> 00:21:18,310
ANNOUNCER: Funding for the
publication of this video was
423
00:21:18,310 --> 00:21:23,020
provided by the Gabriella and
Paul Rosenbaum Foundation.
424
00:21:23,020 --> 00:21:27,200
Help OCW continue to provide
free and open access to MIT
425
00:21:27,200 --> 00:21:31,400
courses by making a donation
at ocw.mit.edu/donate.
426
00:21:31,400 --> 00:21:36,148