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PROFESSOR: Hi.
00:00:33.680 --> 00:00:38.210
Our lesson today is called
the 'Inverse Logarithm'.
00:00:38.210 --> 00:00:41.970
And what it will do, among other
things, is give us a
00:00:41.970 --> 00:00:46.450
very nice chance to revisit
inverse functions in general,
00:00:46.450 --> 00:00:50.050
only now applied specifically
to the natural logarithm
00:00:50.050 --> 00:00:53.230
function that we talked
about last time.
00:00:53.230 --> 00:00:55.950
So I call today's lesson, as I
say, 'Inverse Logarithms'.
00:00:55.950 --> 00:00:59.030
And to see what's coming up
here, simply recall that last
00:00:59.030 --> 00:01:02.980
time, we invented, so to speak,
a new function called
00:01:02.980 --> 00:01:04.620
the 'natural log of x'.
00:01:04.620 --> 00:01:06.070
This had nothing to
do with exponents.
00:01:06.070 --> 00:01:06.810
It was what?
00:01:06.810 --> 00:01:11.290
This was the function whose
derivative with respect to 'x'
00:01:11.290 --> 00:01:15.460
was '1/x' and passed through
the point (1 , 0).
00:01:15.460 --> 00:01:18.320
That uniquely defined this
particular function.
00:01:18.320 --> 00:01:19.920
The point being what?
00:01:19.920 --> 00:01:22.650
That this particular function,
since the domain is for
00:01:22.650 --> 00:01:25.640
positive 'x', '1/x'
is positive.
00:01:25.640 --> 00:01:28.880
The curve is always rising,
which means that the function
00:01:28.880 --> 00:01:31.020
itself must be one to one.
00:01:31.020 --> 00:01:34.030
And because the function is one
to one, it means that the
00:01:34.030 --> 00:01:35.640
inverse function exists.
00:01:35.640 --> 00:01:38.970
This is no different from any
other example of forming f
00:01:38.970 --> 00:01:42.650
inverse, given a one to one
function called 'f'.
00:01:42.650 --> 00:01:44.590
So what we do is now,
and remember how
00:01:44.590 --> 00:01:46.000
you invert this function.
00:01:46.000 --> 00:01:49.160
If you want to do this thing
in slow motion, you first
00:01:49.160 --> 00:01:53.400
rotate this through 90 degrees,
then flip it over.
00:01:53.400 --> 00:01:59.410
If you want to do it faster, the
inverse graph of this is
00:01:59.410 --> 00:02:03.940
the reflection of this curve
with respect to the line 'y'
00:02:03.940 --> 00:02:04.880
equals 'x'.
00:02:04.880 --> 00:02:09.000
In any event, if we do this, we
find that the graph of the
00:02:09.000 --> 00:02:12.590
function 'y' equals
'inverse log x' is
00:02:12.590 --> 00:02:14.530
this particular curve.
00:02:14.530 --> 00:02:16.740
And notice the correspondence.
00:02:16.740 --> 00:02:20.560
If this is the point (1 , 0) on
the log curve, the inverse
00:02:20.560 --> 00:02:22.890
point is (0 , 1).
00:02:22.890 --> 00:02:27.880
In other words, the inverse
log of 0 is 1.
00:02:27.880 --> 00:02:28.840
OK.
00:02:28.840 --> 00:02:30.470
Now, the idea is something
like this.
00:02:30.470 --> 00:02:33.880
Just to have a review,
again, of how--
00:02:33.880 --> 00:02:35.930
see, the punchline I want to
make for today's lesson, right
00:02:35.930 --> 00:02:39.490
at the beginning, is that the
nice thing about studying
00:02:39.490 --> 00:02:43.510
inverse functions is that once
you know the function of which
00:02:43.510 --> 00:02:46.670
you're taking the inverse, you
automatically get all the
00:02:46.670 --> 00:02:49.350
mileage you want out of
the inverse function.
00:02:49.350 --> 00:02:53.720
Well, by way of illustration,
if the function natural log
00:02:53.720 --> 00:02:57.835
had the usual logarithmic
property, we would suspect
00:02:57.835 --> 00:02:59.960
that its inverse
should have the
00:02:59.960 --> 00:03:02.590
usual exponential property.
00:03:02.590 --> 00:03:05.980
Now what is the usual
exponential property?
00:03:05.980 --> 00:03:09.650
The exponential function is
characterized by what?
00:03:09.650 --> 00:03:14.840
That if you take 'f of 'x1 plus
x2'', that's 'f of x1'
00:03:14.840 --> 00:03:16.470
times 'f of x2'.
00:03:16.470 --> 00:03:17.710
So you think of it
in terms of the
00:03:17.710 --> 00:03:19.880
usual exponential notation.
00:03:19.880 --> 00:03:24.790
If you have, say, 10 raised to
the 'x1 plus x2' power, that's
00:03:24.790 --> 00:03:27.810
the same as 10 to the
'x1' power times
00:03:27.810 --> 00:03:29.450
10 to the 'x2' power.
00:03:29.450 --> 00:03:33.070
In other words, if our ln
function is genuinely a
00:03:33.070 --> 00:03:36.090
logarithmic function, we would
expect the inverse of that
00:03:36.090 --> 00:03:39.280
function to be a genuine
exponential function.
00:03:39.280 --> 00:03:42.560
And just to work with inverses
again, let's see if this is
00:03:42.560 --> 00:03:44.110
indeed true.
00:03:44.110 --> 00:03:47.310
Let's see if it's really true
that the 'inverse log of 'x1
00:03:47.310 --> 00:03:50.530
plus x2'' is the 'inverse
log of x1' times the
00:03:50.530 --> 00:03:51.890
'inverse log of x2'.
00:03:51.890 --> 00:03:53.910
You see, the whole
idea is what?
00:03:53.910 --> 00:03:55.600
Let's give these things names.
00:03:55.600 --> 00:03:59.770
Let 'y1' be the 'inverse log
of x1' and 'y2' be the
00:03:59.770 --> 00:04:01.930
'inverse log of x2'.
00:04:01.930 --> 00:04:03.830
Now, since we've already
studied the natural log
00:04:03.830 --> 00:04:07.290
function, the most natural thing
to do here is to invert
00:04:07.290 --> 00:04:13.000
these, namely 'y1' equals the
inverse 'natural log of x1' is
00:04:13.000 --> 00:04:16.529
the same as saying 'x1' is
the 'natural log of y1'.
00:04:16.529 --> 00:04:19.640
And similarly, this is the same
as saying that 'x2' is
00:04:19.640 --> 00:04:21.529
the 'natural log of y2'.
00:04:21.529 --> 00:04:24.460
And now we add equals to equals,
and we get that 'x1
00:04:24.460 --> 00:04:29.180
plus x2' is 'natural log y1'
plus 'natural log y2'.
00:04:29.180 --> 00:04:32.330
But allegedly, we understand the
properties of the natural
00:04:32.330 --> 00:04:35.020
log, because if we didn't
understand those, it would be
00:04:35.020 --> 00:04:37.660
kind of futile to be studying
the inverse function.
00:04:37.660 --> 00:04:39.870
What do we know about
the natural log?
00:04:39.870 --> 00:04:44.140
We know that the 'natural
log of 'y1 times y2', by
00:04:44.140 --> 00:04:46.610
definition of a logarithmic
function, is 'log
00:04:46.610 --> 00:04:48.850
y1' plus 'log y2'.
00:04:48.850 --> 00:04:53.020
In other words, 'natural log y1'
plus 'natural log y2', by
00:04:53.020 --> 00:04:55.920
the property of being
logarithmic, is just the
00:04:55.920 --> 00:04:58.880
'natural log of 'y1
times y2''.
00:04:58.880 --> 00:05:02.640
And now, taking this
relationship and inverting it
00:05:02.640 --> 00:05:05.670
to say this is the same
as saying what?
00:05:05.670 --> 00:05:10.690
The 'inverse natural log of 'x1
plus x2'' is equal to 'y1'
00:05:10.690 --> 00:05:12.050
times 'y2'.
00:05:12.050 --> 00:05:16.400
And if we now observe that 'y1'
was the 'inverse natural
00:05:16.400 --> 00:05:20.890
log of x1', and 'y2' is the
'inverse natural log of x2',
00:05:20.890 --> 00:05:24.290
we have the result that we
claimed we would have at the
00:05:24.290 --> 00:05:25.700
beginning here.
00:05:25.700 --> 00:05:29.680
Now again, here's one example
where it's not the result that
00:05:29.680 --> 00:05:31.210
I'm not interested in.
00:05:31.210 --> 00:05:35.230
But I'm interested in showing,
again, what we meant when we
00:05:35.230 --> 00:05:39.100
said that once we learn a new
concept, we can bring to it
00:05:39.100 --> 00:05:41.240
all of our old knowledge.
00:05:41.240 --> 00:05:43.510
You see, all of these things
come from our basic
00:05:43.510 --> 00:05:44.300
definition.
00:05:44.300 --> 00:05:46.230
Well, let's carry on
one step further.
00:05:46.230 --> 00:05:48.670
You see, this was an arithmetic
property.
00:05:48.670 --> 00:05:51.480
Let's see if we can get some
calculus properties about the
00:05:51.480 --> 00:05:55.390
inverse natural log based on our
knowledge of the natural
00:05:55.390 --> 00:05:56.260
log itself.
00:05:56.260 --> 00:06:00.350
Well, for example, in a course
of this type, the most natural
00:06:00.350 --> 00:06:03.600
thing to do, I guess, is given
any function, we would always
00:06:03.600 --> 00:06:06.270
like to be able to talk about
its derivative, provided, of
00:06:06.270 --> 00:06:08.970
course, the function
is differentiable.
00:06:08.970 --> 00:06:12.580
So for example, a typical
question that one might ask is
00:06:12.580 --> 00:06:17.780
find 'dy dx', if 'y' is 'inverse
natural log x'.
00:06:17.780 --> 00:06:21.130
Again, this works the same way
as it did for the inverse trig
00:06:21.130 --> 00:06:24.230
functions, for the inverse of
everything that we've done so
00:06:24.230 --> 00:06:25.390
far in this course.
00:06:25.390 --> 00:06:28.480
We start with this particular
relationship and right away
00:06:28.480 --> 00:06:31.390
translate it into that which
we're more familiar with,
00:06:31.390 --> 00:06:35.770
namely we translate 'y' equals
the 'inverse natural log of x'
00:06:35.770 --> 00:06:38.890
into 'x' equals 'natural
log y'.
00:06:38.890 --> 00:06:41.960
Now you see again, since I know
how to differentiate 'log
00:06:41.960 --> 00:06:45.180
y' with respect to 'y', the
derivative of 'natural log y'
00:06:45.180 --> 00:06:48.400
with respect to 'y' was
by definition '1/y'.
00:06:48.400 --> 00:06:50.900
From this relationship
here, I can find
00:06:50.900 --> 00:06:55.050
that 'dx dy' is '1/y'.
00:06:55.050 --> 00:06:58.950
And knowing that 'dx dy' is
'1/y', using my inverse
00:06:58.950 --> 00:07:02.650
function general theory that
tells me that to find the
00:07:02.650 --> 00:07:07.120
derivative of 'y' with respect
to 'x', all I have to do is
00:07:07.120 --> 00:07:10.210
invert the derivative of 'x'
with respect to 'y', I wind up
00:07:10.210 --> 00:07:13.890
with the fact that 'dy dx'
is equal to 'y' itself.
00:07:13.890 --> 00:07:17.410
In other words, the 'inverse
log x' function is
00:07:17.410 --> 00:07:21.100
characterized by the fact that
it's its own derivative.
00:07:21.100 --> 00:07:23.610
In other words, it's a
non-destructible type of
00:07:23.610 --> 00:07:26.320
function with respect
to differentiation.
00:07:26.320 --> 00:07:29.000
This is a rather powerful
property.
00:07:29.000 --> 00:07:30.010
You see what this thing says?
00:07:30.010 --> 00:07:31.890
It says that the derivative
of 'y' with respect
00:07:31.890 --> 00:07:34.640
to 'x' is 'y' itself.
00:07:34.640 --> 00:07:38.720
By the way, just as an aside, we
can do the converse of this
00:07:38.720 --> 00:07:39.830
particular problem.
00:07:39.830 --> 00:07:43.120
You see, we started with 'y'
equals 'inverse log x' and
00:07:43.120 --> 00:07:46.030
showed that this particular
equation was satisfied.
00:07:46.030 --> 00:07:48.710
Notice that if we start with
this particular equation,
00:07:48.710 --> 00:07:53.190
starting with 'dy dx' equals
'y', separate the variables,
00:07:53.190 --> 00:07:58.940
and lo and behold, we wind up
with 'dy over y' equals 'dx'.
00:07:58.940 --> 00:08:02.360
And without carrying out the
details, I leave this to you,
00:08:02.360 --> 00:08:04.030
because it is straightforward.
00:08:04.030 --> 00:08:08.420
Notice that as we look at the
left hand side here, we
00:08:08.420 --> 00:08:10.650
hopefully will think
immediately
00:08:10.650 --> 00:08:12.550
of the natural logarithm.
00:08:12.550 --> 00:08:14.040
Namely, what do we want here?
00:08:14.040 --> 00:08:19.240
The function whose derivative
with respect to 'y' is '1/y'.
00:08:19.240 --> 00:08:21.320
Again, you see what I'm saying
is notice that if we had
00:08:21.320 --> 00:08:23.910
started with this particular
differential equation, we
00:08:23.910 --> 00:08:27.130
could have shown that a
logarithm, and hence solving
00:08:27.130 --> 00:08:29.560
for 'y' explicitly, would
have led to the
00:08:29.560 --> 00:08:31.020
inverse logarithm also.
00:08:31.020 --> 00:08:33.780
But at any rate, notice then
that we can find the
00:08:33.780 --> 00:08:37.750
derivative of the inverse log
just by knowing how to find
00:08:37.750 --> 00:08:40.659
the derivative of the natural
log itself, which is as we
00:08:40.659 --> 00:08:42.549
expect things should be.
00:08:42.549 --> 00:08:47.460
Again, and this is a notational
thing, we do not
00:08:47.460 --> 00:08:50.680
use the language in general
"inverse log x."
00:08:50.680 --> 00:08:53.500
In other words, it's quite
conventional to talk about
00:08:53.500 --> 00:08:54.970
'inverse sine x'.
00:08:54.970 --> 00:08:58.500
You may notice that sometimes
in the text, sometimes in my
00:08:58.500 --> 00:09:01.570
lectures I may have used
the word 'arcsin x'.
00:09:01.570 --> 00:09:07.080
But the general notation
is sine to the minus 1.
00:09:07.080 --> 00:09:09.860
However, when it comes to the
natural logarithm function,
00:09:09.860 --> 00:09:12.480
the inverse is usually
not written as
00:09:12.480 --> 00:09:14.460
'inverse natural log x'.
00:09:14.460 --> 00:09:19.610
It's usually abbreviated, and
let me say this, by the symbol
00:09:19.610 --> 00:09:21.180
'e to the x'.
00:09:21.180 --> 00:09:25.170
Now the reason I say by
the symbol is this.
00:09:25.170 --> 00:09:28.100
If I wanted to, I can say, look,
If I've never heard of
00:09:28.100 --> 00:09:31.910
exponents before, let this
symbol be an abbreviation for
00:09:31.910 --> 00:09:33.550
'inverse log x'.
00:09:33.550 --> 00:09:35.430
Does 'inverse log x' exists?
00:09:35.430 --> 00:09:38.030
Yes, I even drew the graph of
it at the beginning of this
00:09:38.030 --> 00:09:39.360
particular lecture.
00:09:39.360 --> 00:09:43.730
However, again, if you are
tempted to use exponents, the
00:09:43.730 --> 00:09:48.520
notation 'e to the x' is in
keeping in line with the idea
00:09:48.520 --> 00:09:49.730
of our previous lecture.
00:09:49.730 --> 00:09:50.200
I say what?
00:09:50.200 --> 00:09:53.460
This matches the identification
of 'natural log
00:09:53.460 --> 00:09:56.780
x' with the 'log of x'
to the base 'e'.
00:09:56.780 --> 00:09:59.230
Remember, in our previous
lecture, we mentioned that
00:09:59.230 --> 00:10:03.030
this particular function exists
without having to talk
00:10:03.030 --> 00:10:04.160
about a base.
00:10:04.160 --> 00:10:07.780
But if you wanted to identify
it with a traditional
00:10:07.780 --> 00:10:11.180
logarithm, the base you would
have to pick is the base 'e',
00:10:11.180 --> 00:10:14.430
where we showed that 'e' was
some number between 2 and 4.
00:10:14.430 --> 00:10:17.850
In other words, again, it's just
like our 'dy dx' versus
00:10:17.850 --> 00:10:21.160
'dy' divided by 'dx' and
other symbols that we
00:10:21.160 --> 00:10:22.460
invented this way.
00:10:22.460 --> 00:10:27.230
Namely, if I look at 'e to the
x' as being the inverse of the
00:10:27.230 --> 00:10:31.100
natural log, or whether I look
at it as being 'e' raised to
00:10:31.100 --> 00:10:33.870
the 'x' power, where 'e' is that
number that's someplace
00:10:33.870 --> 00:10:37.720
between 2 and 4, I don't get
into any trouble either way
00:10:37.720 --> 00:10:40.550
because of the natural
identification of choosing e
00:10:40.550 --> 00:10:42.440
to be the base of my system.
00:10:42.440 --> 00:10:44.290
But I just mention
that in passing.
00:10:44.290 --> 00:10:46.240
Now you see, the rest
of today's lecture
00:10:46.240 --> 00:10:47.990
will go fairly quickly.
00:10:47.990 --> 00:10:50.630
And the reason that it will go
fairly quickly is that once
00:10:50.630 --> 00:10:53.310
we've established what the
function is that we're talking
00:10:53.310 --> 00:10:56.470
about, every other property of
that function is going to
00:10:56.470 --> 00:10:59.210
follow from the principles that
we've already learned.
00:10:59.210 --> 00:11:03.010
For example, in terms of our
new notation, since the
00:11:03.010 --> 00:11:06.950
derivative of 'e to the x' with
respect to 'x' is 'e to
00:11:06.950 --> 00:11:09.320
the x' itself, remember,
'e to the x' now can
00:11:09.320 --> 00:11:10.300
be viewed as what?
00:11:10.300 --> 00:11:14.500
'e to the x' power, or, to be on
safer grounds, if you want
00:11:14.500 --> 00:11:18.280
to be consistent, it's just an
abbreviation for 'inverse
00:11:18.280 --> 00:11:19.590
natural log x'.
00:11:19.590 --> 00:11:24.030
But at any rate, if 'u' is now
any differentiable function of
00:11:24.030 --> 00:11:26.750
'x', notice that to find the
derivative of 'e to the u'
00:11:26.750 --> 00:11:30.230
with respect to 'x', by the
chain rule, I can say what?
00:11:30.230 --> 00:11:33.780
It's the derivative of 'e to
the u' with respect to 'u'
00:11:33.780 --> 00:11:35.400
times 'du dx'.
00:11:35.400 --> 00:11:38.310
We've just shown that the
derivative of 'e to the u'
00:11:38.310 --> 00:11:41.430
with respect to 'u' is
'e' to the 'u' again.
00:11:41.430 --> 00:11:43.940
And therefore, the derivative
of 'e to the u' with respect
00:11:43.940 --> 00:11:47.820
to 'x' is 'e to the
u' times 'du dx'.
00:11:47.820 --> 00:11:51.070
By the way, this again leads to
another interesting thing
00:11:51.070 --> 00:11:55.260
that makes integrals tough
to handle, in a way.
00:11:55.260 --> 00:11:57.510
We've mentioned this before,
but here's another nice,
00:11:57.510 --> 00:12:01.010
natural environment to bring
this problem up in again.
00:12:01.010 --> 00:12:04.020
When we discussed the second
fundamental theorem of
00:12:04.020 --> 00:12:07.090
integral calculus to show how
one actually would have to
00:12:07.090 --> 00:12:10.500
understand areas to be able
to find a function whose
00:12:10.500 --> 00:12:14.690
derivative was 'e to the
'minus x squared', we
00:12:14.690 --> 00:12:17.920
mentioned, we didn't prove it,
we mentioned that there was no
00:12:17.920 --> 00:12:20.820
familiar function whose
derivative with respect to 'x'
00:12:20.820 --> 00:12:22.920
was 'e to the 'minus
x squared'.
00:12:22.920 --> 00:12:25.560
So this would be a very
difficult problem to handle.
00:12:25.560 --> 00:12:27.160
Now, let's look at
this one instead.
00:12:27.160 --> 00:12:31.430
Let's look at integral '2x 'e
to the minus x squared' dx'.
00:12:31.430 --> 00:12:36.440
To the untrained eye, it would
appear that this is even
00:12:36.440 --> 00:12:37.300
messier than this.
00:12:37.300 --> 00:12:39.610
In other words, this is just 'e
to the 'minus x squared''.
00:12:39.610 --> 00:12:42.900
This one has 'e to the 'minus
x squared'' with a '2x'
00:12:42.900 --> 00:12:45.720
dangling in front of it, and
that looks even tougher.
00:12:45.720 --> 00:12:49.780
But notice that if we look
at our exponent, 'minus x
00:12:49.780 --> 00:12:53.190
squared', the derivative, or the
differential of 'minus x
00:12:53.190 --> 00:12:54.290
squared' is what?
00:12:54.290 --> 00:12:57.180
It's 'minus '2x dx''.
00:12:57.180 --> 00:13:01.340
In other words, the '2x dx' is
precisely what you need to
00:13:01.340 --> 00:13:04.575
reduce this to the form integral
''e to the u' du'.
00:13:04.575 --> 00:13:07.340
In other words, working this
thing out in more specific
00:13:07.340 --> 00:13:11.970
detail, if I let 'u' equal
'minus x squared', 'du'
00:13:11.970 --> 00:13:16.570
becomes 'minus '2x dx'', and
therefore integral '2x 'e to
00:13:16.570 --> 00:13:19.330
the 'minus x squared'' dx'
just becomes what?
00:13:19.330 --> 00:13:22.940
Well, the 'e to the 'minus x
squared'' just becomes 'e to
00:13:22.940 --> 00:13:24.140
the minus u'.
00:13:24.140 --> 00:13:28.400
And '2x dx' is just
'minus du'.
00:13:28.400 --> 00:13:29.650
OK?
00:13:33.220 --> 00:13:36.140
I'm sorry, 'u' is 'minus
x squared'.
00:13:36.140 --> 00:13:37.980
'u' is 'minus x squared'.
00:13:37.980 --> 00:13:39.880
So 'minus x squared' is 'u'.
00:13:39.880 --> 00:13:43.120
So this becomes 'minus
e to the 'u du''.
00:13:43.120 --> 00:13:45.970
And that, of course,
is just minus.
00:13:45.970 --> 00:13:47.610
See, the minus will
come outside.
00:13:47.610 --> 00:13:50.200
The integral of 'e to the u'
with respect to 'u' is
00:13:50.200 --> 00:13:51.605
just 'e to the u'.
00:13:51.605 --> 00:13:58.900
And since 'u' is equal to 'minus
x squared', all we're
00:13:58.900 --> 00:14:02.270
saying is that if you
differentiate minus 'e' to the
00:14:02.270 --> 00:14:05.020
'minus x squared', you
wind up with what?
00:14:05.020 --> 00:14:08.790
'2x 'e to the 'minus x
squared'', the reason being
00:14:08.790 --> 00:14:12.390
that you must multiply this by
the derivative of the exponent
00:14:12.390 --> 00:14:13.320
with respect to 'x'.
00:14:13.320 --> 00:14:15.650
The root of the exponent
is 'minus 2x'.
00:14:15.650 --> 00:14:19.180
'Minus 2x' times minus
1 is '2x'.
00:14:19.180 --> 00:14:22.870
If you want to see this in
more concise differential
00:14:22.870 --> 00:14:28.570
notation, you see, what we're
saying is that the integral of
00:14:28.570 --> 00:14:34.240
'e' to some power with respect
to that same power is 'e' to
00:14:34.240 --> 00:14:36.060
that power plus a constant.
00:14:36.060 --> 00:14:38.550
In other words, the function
that one would have to
00:14:38.550 --> 00:14:42.610
integrate 'e to the 'minus x
squared'' with respect to to
00:14:42.610 --> 00:14:45.760
wind up with 'e to the 'minus x
squared'' would be 'minus x
00:14:45.760 --> 00:14:46.860
squared' itself.
00:14:46.860 --> 00:14:50.200
And you see, now, in the next
step, since this is a true
00:14:50.200 --> 00:14:53.800
statement, the differential
of 'minus x squared'
00:14:53.800 --> 00:14:55.590
is minus '2x dx'.
00:15:01.810 --> 00:15:04.810
And now you see, factoring out
the minus sign and multiplying
00:15:04.810 --> 00:15:07.890
through by minus 1, we arrive
at the same result that we
00:15:07.890 --> 00:15:09.180
wound up with before.
00:15:09.180 --> 00:15:12.720
But again, this is the kind of
material that we can drill on
00:15:12.720 --> 00:15:16.590
extensively in the exercises
in this particular unit.
00:15:16.590 --> 00:15:18.020
The technique is what?
00:15:18.020 --> 00:15:21.010
That the basic building block
of the so-called exponential
00:15:21.010 --> 00:15:24.530
function, the inverse natural
log, is that when you
00:15:24.530 --> 00:15:27.560
differentiate it, you
essentially do not destroy it.
00:15:27.560 --> 00:15:30.320
In other words, the derivative
of 'e to the u' with respect
00:15:30.320 --> 00:15:34.640
to 'x' is 'e to the
u' times 'du dx'.
00:15:34.640 --> 00:15:37.550
And I thought that in closing
today's lesson, I might as
00:15:37.550 --> 00:15:42.210
well show you a very powerful
application of this particular
00:15:42.210 --> 00:15:45.100
result in a non-obvious
situation.
00:15:45.100 --> 00:15:48.540
And it's something that
we call second-order
00:15:48.540 --> 00:15:49.570
differential equations.
00:15:49.570 --> 00:15:51.770
I've picked out here a
differential equation.
00:15:51.770 --> 00:15:55.310
Let me just show you what I
have in mind over here.
00:15:55.310 --> 00:15:58.770
Suppose I tell you that 'y'
is a twice-differentiable
00:15:58.770 --> 00:16:02.450
function of 'x' that satisfies
the following identity, that
00:16:02.450 --> 00:16:05.530
the second derivative of 'y'
with respect to 'x' minus 5
00:16:05.530 --> 00:16:09.140
times the first derivative of
'y' with respect to 'x' plus 6
00:16:09.140 --> 00:16:12.150
times 'y' is identically 0.
00:16:12.150 --> 00:16:15.620
And the question is, if this
equation is to be obeyed, what
00:16:15.620 --> 00:16:17.360
must 'y' be?
00:16:17.360 --> 00:16:20.000
And a very interesting technique
for solving this
00:16:20.000 --> 00:16:22.650
kind of a problem, you see, this
is called a second-order
00:16:22.650 --> 00:16:25.310
differential equation because
the highest derivative that
00:16:25.310 --> 00:16:27.760
appears is the second
derivative.
00:16:27.760 --> 00:16:30.720
OK, a rather powerful
technique using the
00:16:30.720 --> 00:16:34.610
exponential is available to us
for problems of this sort.
00:16:34.610 --> 00:16:36.290
And the idea hinges on this.
00:16:36.290 --> 00:16:40.400
As a trial solution, which I'll
call 'y sub t', let's try
00:16:40.400 --> 00:16:43.920
'e to the rx', where 'r' happens
to be a constant.
00:16:43.920 --> 00:16:47.410
You see, the whole idea is if I
differentiate 'e to the rx'
00:16:47.410 --> 00:16:51.460
with respect to 'x', I get 'e
to the rx' back again, only
00:16:51.460 --> 00:16:55.190
with a factor of 'r', namely the
derivative of my exponent
00:16:55.190 --> 00:16:56.150
in this case.
00:16:56.150 --> 00:16:57.290
See, 'r' is a constant.
00:16:57.290 --> 00:17:00.970
The derivative of 'rx' with
respect to 'x' is just 'r'.
00:17:00.970 --> 00:17:03.720
So notice that the first
derivative of 'y sub t' with
00:17:03.720 --> 00:17:06.550
respect to 'x' is 'r
'e to the rx''.
00:17:06.550 --> 00:17:09.680
The second derivative
of 'y sub t' with
00:17:09.680 --> 00:17:11.339
respect to 'x' is what?
00:17:11.339 --> 00:17:12.579
'r' is a constant.
00:17:12.579 --> 00:17:14.530
I differentiate 'e to the rx'.
00:17:14.530 --> 00:17:17.640
That brings down another factor
of 'r' and leaves me
00:17:17.640 --> 00:17:19.300
with 'e to the rx'.
00:17:19.300 --> 00:17:23.970
You see, the whole idea being,
notice that 'e to the rx' is a
00:17:23.970 --> 00:17:28.580
common factor of 'y sub t', 'y
sub 't prime'', 'y sub 't
00:17:28.580 --> 00:17:29.620
double prime''.
00:17:29.620 --> 00:17:33.030
If I now substitute these
results back into my original
00:17:33.030 --> 00:17:36.950
equation, look what I wind up
with. 'y double prime' becomes
00:17:36.950 --> 00:17:39.080
'r squared 'e to the rx''.
00:17:39.080 --> 00:17:45.230
Minus '5y prime', that's minus
'5r 'e to the rx'', plus '6y',
00:17:45.230 --> 00:17:46.790
'6e to the rx'.
00:17:46.790 --> 00:17:48.650
And that must equal zero.
00:17:48.650 --> 00:17:49.880
And here's the key point.
00:17:49.880 --> 00:17:53.100
'e to the rx' is now
a common factor.
00:17:53.100 --> 00:17:55.700
I factor that out.
00:17:55.700 --> 00:17:59.240
If the product of two numbers
is 0, one of the
00:17:59.240 --> 00:18:01.220
factors must be 0.
00:18:01.220 --> 00:18:04.730
But notice from our graph of the
exponential, the inverse
00:18:04.730 --> 00:18:10.520
logarithm, 'e to the x', 'e to
the rx' can never be negative
00:18:10.520 --> 00:18:12.550
and can never be 0,
in fact. 'e to the
00:18:12.550 --> 00:18:14.190
rx' is always positive.
00:18:14.190 --> 00:18:17.580
Therefore, if 'e to the rx' is
always positive, it must be
00:18:17.580 --> 00:18:21.100
the other factor which is 0.
00:18:21.100 --> 00:18:25.540
But 'r squared' minus '5r' plus
6 is not a second-order
00:18:25.540 --> 00:18:26.920
differential equation.
00:18:26.920 --> 00:18:29.690
It's a second-degree polynomial
equation.
00:18:29.690 --> 00:18:32.940
It's a familiar quadratic
equation, which I can solve
00:18:32.940 --> 00:18:34.010
quite easily.
00:18:34.010 --> 00:18:35.550
In other words, I find what?
00:18:35.550 --> 00:18:39.300
That 'r' must be either 2,
or 'r' must equal 3.
00:18:39.300 --> 00:18:43.480
In other words, my claim is that
either 'e to the 2x' or
00:18:43.480 --> 00:18:46.190
'e to the 3x' must
be a solution of
00:18:46.190 --> 00:18:47.790
this particular equation.
00:18:47.790 --> 00:18:52.440
Of course, we can do more with
that, which we will in a later
00:18:52.440 --> 00:18:57.050
part of calculus, not in
this package's work.
00:18:57.050 --> 00:18:59.840
But we're not going to study
differential equations in
00:18:59.840 --> 00:19:01.040
great detail here.
00:19:01.040 --> 00:19:04.340
But for our present purposes,
I think this illustrates how
00:19:04.340 --> 00:19:08.050
one can use the fact that it's
a rather powerful structural
00:19:08.050 --> 00:19:11.630
property when the derivative of
a function with respect to
00:19:11.630 --> 00:19:13.750
'x' is the function itself.
00:19:13.750 --> 00:19:17.240
By the way, as a quick check,
notice that if 'y' equals 'e
00:19:17.240 --> 00:19:20.590
to the 2x', 'y prime' is
twice 'e to the 2x'.
00:19:20.590 --> 00:19:23.850
'y double prime' is
'4 e to the 2x'.
00:19:23.850 --> 00:19:27.200
And therefore 'y double prime'
minus '5y prime'
00:19:27.200 --> 00:19:29.170
plus '6y' is what?
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'.
00:19:35.430 --> 00:19:39.790
And that, in fact, is genuinely,
identically zero.
00:19:39.790 --> 00:19:43.400
In a similar way, checking out
'y' equals 'e to the 3x', we
00:19:43.400 --> 00:19:47.470
get 'y prime' is '3e to the 3x',
'y double prime' is '9e
00:19:47.470 --> 00:19:48.610
to the 3x'.
00:19:48.610 --> 00:19:54.550
Therefore, '9e to the 3x' minus
'15 e to the 3x' plus
00:19:54.550 --> 00:19:58.670
'6e to the 3x' is again,
identically 0.
00:19:58.670 --> 00:20:01.140
And again, you see what this
powerful technique is.
00:20:01.140 --> 00:20:05.500
In general, if 'a' and 'b' are
constants, the substitution 'y
00:20:05.500 --> 00:20:09.360
sub t' equals 'e to the rx'.
00:20:09.360 --> 00:20:13.000
And there'll be a problem in the
exercises on this to give
00:20:13.000 --> 00:20:14.300
you additional drill.
00:20:14.300 --> 00:20:17.770
But that substitution transforms
the second or the
00:20:17.770 --> 00:20:21.700
differential equation, 'y double
prime' plus 'ay prime'
00:20:21.700 --> 00:20:26.750
plus 'by' equals 0, into an
equivalent quadratic equation,
00:20:26.750 --> 00:20:30.010
'r squared' plus 'ar'
plus 'b' equals 0.
00:20:30.010 --> 00:20:32.740
And you see from this equation,
using the quadratic
00:20:32.740 --> 00:20:36.470
formula, we can find the values
of 'r' that satisfy
00:20:36.470 --> 00:20:38.400
this, and that gives
us a couple of
00:20:38.400 --> 00:20:40.040
solutions to the equation.
00:20:40.040 --> 00:20:43.410
Well, at any rate, I
deliberately want this lecture
00:20:43.410 --> 00:20:46.710
to stay short to make the
most important impact.
00:20:46.710 --> 00:20:48.490
And that is again what?
00:20:48.490 --> 00:20:51.660
That once we knew what the
natural log function was, and
00:20:51.660 --> 00:20:55.720
we defined the inverse natural
log function, everything that
00:20:55.720 --> 00:20:59.090
we wanted to know about the
inverse natural log followed
00:20:59.090 --> 00:21:01.380
from the properties
of the log itself.
00:21:01.380 --> 00:21:04.200
And this is basically the
lesson for today.
00:21:04.200 --> 00:21:07.270
We will continue the discussion
of exponential
00:21:07.270 --> 00:21:08.990
functions from a different
point of
00:21:08.990 --> 00:21:10.720
view in our next lecture.
00:21:10.720 --> 00:21:12.850
But until that next
lecture, goodbye.
00:21:15.780 --> 00:21:18.310
ANNOUNCER: Funding for the
publication of this video was
00:21:18.310 --> 00:21:23.020
provided by the Gabriella and
Paul Rosenbaum Foundation.
00:21:23.020 --> 00:21:27.200
Help OCW continue to provide
free and open access to MIT
00:21:27.200 --> 00:21:31.400
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at ocw.mit.edu/donate.