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Professor: So, again
welcome to 18.01.
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We're getting started today
with what we're calling
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Unit One, a highly
imaginative title.
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And it's differentiation.
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So, let me first tell you,
briefly, what's in store in
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the next couple of weeks.
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The main topic today is
what is a derivative.
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And, we're going to look at
this from several different
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points of view, and
the first one is the
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geometric interpretation.
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That's what we'll spend
most of today on.
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And then, we'll also talk about
a physical interpretation
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of what a derivative is.
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And then there's going to be
something else which I guess is
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maybe the reason why Calculus
is so fundamental, and why we
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always start with it in most
science and engineering
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schools, which is the
importance of derivatives, of
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this, to all measurements.
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So that means pretty
much every place.
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That means in science, in
engineering, in economics,
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in political science, etc.
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Polling, lots of commercial
applications, just
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about everything.
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Now, that's what we'll be
getting started with, and then
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there's another thing that
we're gonna do in this unit,
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which is we're going to explain
how to differentiate anything.
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So, how to differentiate
any function you know.
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And that's kind of a tall
order, but let me just
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give you an example.
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If you want to take the
derivative - this we'll see
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today is the notation for the
derivative of something - of
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00:03:11 --> 00:03:15
some messy function
like e ^ x arctanx.
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We'll work this out by
the end of this unit.
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All right?
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Anything you can think of,
anything you can write down,
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we can differentiate it.
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All right, so that's what we're
gonna do, and today as I said,
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we're gonna spend most of our
time on this geometric
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interpretation.
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So let's begin with that.
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So here we go with the
geometric interpretation
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of derivatives.
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And, what we're going to do is
just ask the geometric problem
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of finding the tangent line
to some graph of some
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function at some point.
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Which is to say (x0, y0).
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So that's the problem that
we're addressing here.
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Alright, so here's our
problem, and now let me
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show you the solution.
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00:04:49 --> 00:04:58
So, well, let's
graph the function.
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Here's it's graph.
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Here's some point.
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All right, maybe I should
draw it just a bit lower.
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So here's a point P.
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Maybe it's above the point
x0. x0, by the way, this
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was supposed to be an x0.
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That was some fixed
place on the x-axis.
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And now, in order to perform
this mighty feat, I will use
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another color of chalk.
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How about red?
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OK.
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So here it is.
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There's the tangent line,
Well, not quite straight.
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Close enough.
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All right?
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I did it.
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That's the geometric problem.
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I achieved what I wanted to do,
and it's kind of an interesting
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question, which unfortunately I
can't solve for you in
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this class, which is,
how did I do that?
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That is, how physically did
I manage to know what to do
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to draw this tangent line?
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But that's what geometric
problems are like.
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We visualize it.
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00:06:12 --> 00:06:14
We can figure it out
somewhere in our brains.
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It happens.
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00:06:15 --> 00:06:20
And the task that we have now
is to figure out how to do it
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analytically, to do it in a way
that a machine could just as
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well as I did in drawing
this tangent line.
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So, what did we learn in
high school about what
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a tangent line is?
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Well, a tangent line has an
equation, and any line through
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a point has the equation y -
y0 is equal to m the
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slope, times x - x0.
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So here's the equation for that
line, and now there are two
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pieces of information that
we're going to need to work
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out what the line is.
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00:07:07 --> 00:07:10
The first one is the point.
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00:07:10 --> 00:07:13
That's that point P there.
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00:07:13 --> 00:07:19
And to specify P, given x, we
need to know the level of y,
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which is of course just f(x0).
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That's not a calculus problem,
but anyway that's a very
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important part of the process.
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00:07:28 --> 00:07:31
So that's the first
thing we need to know.
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00:07:31 --> 00:07:39
And the second thing we
need to know is the slope.
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00:07:39 --> 00:07:42
And that's this number m.
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00:07:42 --> 00:07:45
And in calculus we have
another name for it.
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We call it f prime of x0.
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Namely, the derivative of f.
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So that's the calculus part.
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That's the tricky part, and
that's the part that we
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00:07:55 --> 00:07:57
have to discuss now.
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00:07:57 --> 00:08:01
So just to make that explicit
here, I'm going to make a
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definition, which is that f
'(x0) , which is known as the
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00:08:08 --> 00:08:34
derivative, of f, at x0, is the
slope of the tangent line to y
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00:08:34 --> 00:08:47
= f (x) at the point,
let's just call it p.
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All right?
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00:08:50 --> 00:08:56
So, that's what it is, but
still I haven't made any
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progress in figuring out any
better how I drew that line.
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00:09:01 --> 00:09:05
So I have to say something
that's more concrete, because
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I want to be able to cook
up what these numbers are.
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00:09:07 --> 00:09:11
I have to figure out
what this number m is.
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00:09:11 --> 00:09:16
And one way of thinking about
that, let me just try this, so
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I certainly am taking for
granted that in sort of
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non-calculus part that I know
what a line through a point is.
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So I know this equation.
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But another possibility might
be, this line here, how do I
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know - well, unfortunately, I
didn't draw it quite straight,
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00:09:34 --> 00:09:36
but there it is - how do I know
that this orange line is not a
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00:09:36 --> 00:09:45
tangent line, but this other
line is a tangent line?
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00:09:45 --> 00:09:55
Well, it's actually not so
obvious, but I'm gonna
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describe it a little bit.
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It's not really the fact, this
thing crosses at some other
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place, which is this point Q.
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00:10:02 --> 00:10:06
But it's not really the fact
that the thing crosses at two
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place, because the line could
be wiggly, the curve could be
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00:10:07 --> 00:10:11
wiggly, and it could cross back
and forth a number of times.
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00:10:11 --> 00:10:17
That's not what distinguishes
the tangent line.
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00:10:17 --> 00:10:19
So I'm gonna have to somehow
grasp this, and I'll
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00:10:19 --> 00:10:23
first do it in language.
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00:10:23 --> 00:10:29
And it's the following idea:
it's that if you take this
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00:10:29 --> 00:10:37
orange line, which is called a
secant line, and you think of
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00:10:37 --> 00:10:41
the point Q as getting closer
and closer to P, then the slope
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00:10:41 --> 00:10:45
of that line will get closer
and closer to the slope
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00:10:45 --> 00:10:47
of the red line.
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00:10:47 --> 00:10:53
And if we draw it close
enough, then that's gonna
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00:10:53 --> 00:10:54
be the correct line.
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00:10:54 --> 00:10:57
So that's really what I did,
sort of in my brain when
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00:10:57 --> 00:10:58
I drew that first line.
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00:10:58 --> 00:11:01
And so that's the way I'm
going to articulate it first.
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00:11:01 --> 00:11:13
Now, so the tangent line
is equal to the limit of
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00:11:13 --> 00:11:24
so called secant lines
PQ, as Q tends to P.
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00:11:24 --> 00:11:31
And here we're thinking of P as
being fixed and Q as variable.
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00:11:31 --> 00:11:35
All right?
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00:11:35 --> 00:11:39
Again, this is still the
geometric discussion, but now
155
00:11:39 --> 00:11:42
we're gonna be able to put
symbols and formulas
156
00:11:42 --> 00:11:43
to this computation.
157
00:11:43 --> 00:11:56
And we'll be able to work out
formulas in any example.
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00:11:56 --> 00:11:58
So let's do that.
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00:11:58 --> 00:12:03
So first of all, I'm
gonna write out these
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00:12:03 --> 00:12:05
points P and Q again.
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00:12:05 --> 00:12:11
So maybe we'll put
P here and Q here.
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And I'm thinking of this
line through them.
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00:12:12 --> 00:12:16
I guess it was orange, so
we'll leave it as orange.
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00:12:16 --> 00:12:19
All right.
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00:12:19 --> 00:12:24
And now I want to
compute its slope.
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00:12:24 --> 00:12:27
So this, gradually, we'll
do this in two steps.
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00:12:27 --> 00:12:30
And these steps will introduce
us to the basic notations which
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00:12:30 --> 00:12:33
are used throughout calculus,
including multi-variable
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00:12:33 --> 00:12:35
calculus, across the board.
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00:12:35 --> 00:12:40
So the first notation that's
used is you imagine here's
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00:12:40 --> 00:12:45
the x-axis underneath, and
here's the x0, the location
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00:12:45 --> 00:12:47
directly below the point P.
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00:12:47 --> 00:12:51
And we're traveling here a
horizontal distance which
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00:12:51 --> 00:12:53
is denoted by delta x.
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00:12:53 --> 00:12:58
So that's delta x, so called.
176
00:12:58 --> 00:13:06
And we could also call
it the change in x.
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00:13:06 --> 00:13:09
So that's one thing we want
to measure in order to get
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00:13:09 --> 00:13:12
the slope of this line PQ.
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00:13:12 --> 00:13:14
And the other thing
is this height.
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00:13:14 --> 00:13:18
So that's this distance here,
which we denote delta f,
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00:13:18 --> 00:13:21
which is the change in f.
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00:13:21 --> 00:13:29
And then, the slope is just
the ratio, delta f / delta x.
183
00:13:29 --> 00:13:39
So this is the slope
of the secant.
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00:13:39 --> 00:13:42
And the process I just
described over here with this
185
00:13:42 --> 00:13:46
limit applies not just to the
whole line itself, but also in
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00:13:46 --> 00:13:48
particular to its slope.
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00:13:48 --> 00:13:53
And the way we write that is
the limit as delta x goes to 0.
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00:13:53 --> 00:13:56
And that's going
to be our slope.
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00:13:56 --> 00:14:10
So this is slope of
the tangent line.
190
00:14:10 --> 00:14:11
OK.
191
00:14:11 --> 00:14:22
Now, This is still a little
general, and I want to work out
192
00:14:22 --> 00:14:28
a more usable form here, a
better formula for this.
193
00:14:28 --> 00:14:33
And in order to do that, I'm
gonna write delta f, the
194
00:14:33 --> 00:14:36
numerator more explicitly here.
195
00:14:36 --> 00:14:41
The change in f, so remember
that the point P is
196
00:14:41 --> 00:14:43
the point (x0, f(x0)).
197
00:14:43 --> 00:14:51
All right, that's what we got
for the formula for the point.
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00:14:51 --> 00:14:55
And in order to compute these
distances and in particular
199
00:14:55 --> 00:14:58
the vertical distance here,
I'm gonna have to get a
200
00:14:58 --> 00:15:00
formula for Q as well.
201
00:15:00 --> 00:15:05
So if this horizontal distance
is delta x, then this
202
00:15:05 --> 00:15:07
location is (x0
203
00:15:07 --> 00:15:11
delta x).
204
00:15:11 --> 00:15:14
And so the point above
that point has a
205
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formula, which is (x0
206
00:15:20 --> 00:15:25
delta x, f(x0 and
this is a mouthful,
207
00:15:25 --> 00:15:31
delta x)).
208
00:15:31 --> 00:15:33
All right, so there's the
formula for the point Q.
209
00:15:33 --> 00:15:36
Here's the formula
for the point P.
210
00:15:36 --> 00:15:47
And now I can write a different
formula for the derivative,
211
00:15:47 --> 00:15:53
which is the following: so this
f'(x0) , which is the same as
212
00:15:53 --> 00:16:01
m, is going to be the limit as
delta x goes to 0 of the change
213
00:16:01 --> 00:16:07
in f, well the change in f is
the value of f at the upper
214
00:16:07 --> 00:16:11
point here, which is (x0
215
00:16:11 --> 00:16:16
delta x), and minus its value
at the lower point P, which is
216
00:16:16 --> 00:16:23
f(x0), divided by delta x.
217
00:16:23 --> 00:16:24
All right, so this
is the formula.
218
00:16:24 --> 00:16:29
I'm going to put this in a
little box, because this is by
219
00:16:29 --> 00:16:33
far the most important formula
today, which we use to derive
220
00:16:33 --> 00:16:35
pretty much everything else.
221
00:16:35 --> 00:16:38
And this is the way that
we're going to be able to
222
00:16:38 --> 00:16:46
compute these numbers.
223
00:16:46 --> 00:17:06
So let's do an example.
224
00:17:06 --> 00:17:13
This example, we'll
call this example one.
225
00:17:13 --> 00:17:19
We'll take the function
f(x) , which is 1/x .
226
00:17:19 --> 00:17:23
That's sufficiently complicated
to have an interesting answer,
227
00:17:23 --> 00:17:27
and sufficiently
straightforward that we can
228
00:17:27 --> 00:17:32
compute the derivative
fairly quickly.
229
00:17:32 --> 00:17:36
So what is it that
we're gonna do here?
230
00:17:36 --> 00:17:42
All we're going to do is we're
going to plug in this formula
231
00:17:42 --> 00:17:44
here for that function.
232
00:17:44 --> 00:17:47
That's all we're going to do,
and visually what we're
233
00:17:47 --> 00:17:53
accomplishing is somehow to
take the hyperbola, and take a
234
00:17:53 --> 00:18:00
point on the hyperbola, and
figure out some tangent line.
235
00:18:00 --> 00:18:03
That's what we're accomplishing
when we do that.
236
00:18:03 --> 00:18:05
So we're accomplishing this
geometrically but we'll be
237
00:18:05 --> 00:18:07
doing it algebraically.
238
00:18:07 --> 00:18:14
So first, we consider this
difference delta f / delta x
239
00:18:14 --> 00:18:16
and write out its formula.
240
00:18:16 --> 00:18:18
So I have to have a place.
241
00:18:18 --> 00:18:21
So I'm gonna make it again
above this point x0, which
242
00:18:21 --> 00:18:22
is the general point.
243
00:18:22 --> 00:18:25
We'll make the
general calculation.
244
00:18:25 --> 00:18:30
So the value of f at the top,
when we move to the right by
245
00:18:30 --> 00:18:35
f(x), so I just read off from
this, read off from here.
246
00:18:35 --> 00:18:41
The formula, the first thing
I get here is 1 / x0
247
00:18:41 --> 00:18:43
delta x.
248
00:18:43 --> 00:18:46
That's the left hand term.
249
00:18:46 --> 00:18:50
Minus 1 / x0, that's
the right hand term.
250
00:18:50 --> 00:18:54
And then I have to
divide that by delta x.
251
00:18:54 --> 00:18:57
OK, so here's our expression.
252
00:18:57 --> 00:19:00
And by the way this has a name.
253
00:19:00 --> 00:19:10
This thing is called a
difference quotient.
254
00:19:10 --> 00:19:12
It's pretty complicated,
because there's always a
255
00:19:12 --> 00:19:13
difference in the numerator.
256
00:19:13 --> 00:19:16
And in disguise, the
denominator is a difference,
257
00:19:16 --> 00:19:19
because it's the difference
between the value on the
258
00:19:19 --> 00:19:26
right side and the value
on the left side here.
259
00:19:26 --> 00:19:34
OK, so now we're going to
simplify it by some algebra.
260
00:19:34 --> 00:19:35
So let's just take a look.
261
00:19:35 --> 00:19:38
So this is equal to,
let's continue on
262
00:19:38 --> 00:19:41
the next level here.
263
00:19:41 --> 00:19:45
This is equal to 1
/ delta x times...
264
00:19:45 --> 00:19:49
All I'm going to do is put it
over a common denominator.
265
00:19:49 --> 00:19:53
So the common
denominator is (x0
266
00:19:53 --> 00:19:54
delta x)x0.
267
00:19:56 --> 00:19:59
And so in the numerator for
the first expressions I
268
00:19:59 --> 00:20:03
have x0, and for the second
expression I have x0
269
00:20:03 --> 00:20:05
delta x.
270
00:20:05 --> 00:20:08
So this is the same thing as I
had in the numerator before,
271
00:20:08 --> 00:20:11
factoring out this denominator.
272
00:20:11 --> 00:20:17
And here I put that numerator
into this more amenable form.
273
00:20:17 --> 00:20:20
And now there are two
basic cancellations.
274
00:20:20 --> 00:20:33
The first one is that x0 and
x0 cancel, so we have this.
275
00:20:33 --> 00:20:38
And then the second step is
that these two expressions
276
00:20:38 --> 00:20:40
cancel, the numerator
and the denominator.
277
00:20:40 --> 00:20:44
Now we have a cancellation
that we can make use of.
278
00:20:44 --> 00:20:48
So we'll write that under here.
279
00:20:48 --> 00:20:53
And this is equals -1 / (x0
280
00:20:54 --> 00:20:56
delta x)x0.
281
00:20:57 --> 00:21:03
And then the very last step is
to take the limit as delta
282
00:21:03 --> 00:21:09
x tends to 0, and
now we can do it.
283
00:21:09 --> 00:21:10
Before we couldn't do it.
284
00:21:10 --> 00:21:11
Why?
285
00:21:11 --> 00:21:15
Because the numerator and the
denominator gave us 0 / 0.
286
00:21:15 --> 00:21:17
But now that I've made
this cancellation, I
287
00:21:17 --> 00:21:19
can pass to the limit.
288
00:21:19 --> 00:21:21
And all that happens is
I set this delta x to
289
00:21:21 --> 00:21:22
0, and I get -1/x0^2.
290
00:21:25 --> 00:21:32
So that's the answer.
291
00:21:32 --> 00:21:34
All right, so in other words
what I've shown - let me put
292
00:21:34 --> 00:21:36
it up here- is that
f'(x0) = -1/x0^2.
293
00:21:52 --> 00:21:57
Now, let's look at the graph
just a little bit to check this
294
00:21:57 --> 00:22:01
for plausibility, all right?
295
00:22:01 --> 00:22:04
What's happening here, is
first of all it's negative.
296
00:22:04 --> 00:22:08
It's less than 0, which
is a good thing.
297
00:22:08 --> 00:22:16
You see that slope
there is negative.
298
00:22:16 --> 00:22:20
That's the simplest check
that you could make.
299
00:22:20 --> 00:22:24
And the second thing that I
would just like to point out is
300
00:22:24 --> 00:22:29
that as x goes to infinity,
that as we go farther to the
301
00:22:29 --> 00:22:32
right, it gets less
and less steep.
302
00:22:32 --> 00:22:46
So as x0 goes to infinity,
less and less steep.
303
00:22:46 --> 00:22:49
So that's also consistent here,
when x0 is very large, this is
304
00:22:49 --> 00:22:52
a smaller and smaller number
in magnitude, although
305
00:22:52 --> 00:22:54
it's always negative.
306
00:22:54 --> 00:23:00
It's always sloping down.
307
00:23:00 --> 00:23:03
All right, so I've managed
to fill the boards.
308
00:23:03 --> 00:23:06
So maybe I should stop
for a question or two.
309
00:23:06 --> 00:23:06
Yes?
310
00:23:06 --> 00:23:11
Student: [INAUDIBLE]
311
00:23:11 --> 00:23:18
Professor: So the question
is to explain again
312
00:23:18 --> 00:23:22
this limiting process.
313
00:23:22 --> 00:23:26
So the formula here is we
have basically two numbers.
314
00:23:26 --> 00:23:29
So in other words, why is it
that this expression, when
315
00:23:29 --> 00:23:33
delta x tends to 0, is
equal to -1/x0^2 ?
316
00:23:33 --> 00:23:37
Let me illustrate it by
sticking in a number for x0
317
00:23:37 --> 00:23:39
to make it more explicit.
318
00:23:39 --> 00:23:43
All right, so for instance,
let me stick in here
319
00:23:43 --> 00:23:46
for x0 the number 3.
320
00:23:46 --> 00:23:48
Then it's -1 / (3
321
00:23:49 --> 00:23:50
delta x)3.
322
00:23:52 --> 00:23:54
That's the situation
that we've got.
323
00:23:54 --> 00:23:56
And now the question is what
happens as this number gets
324
00:23:56 --> 00:23:59
smaller and smaller and
smaller, and gets to
325
00:23:59 --> 00:24:01
be practically 0?
326
00:24:01 --> 00:24:04
Well, literally what we
can do is just plug in 0
327
00:24:04 --> 00:24:05
there, and you get (3
328
00:24:05 --> 00:24:07
0)3 in the denominator.
329
00:24:07 --> 00:24:08
Minus one in the numerator.
330
00:24:08 --> 00:24:13
So this tends to
-1/9 (over 3^2).
331
00:24:16 --> 00:24:18
And that's what I'm saying
in general with this
332
00:24:18 --> 00:24:21
extra number here.
333
00:24:21 --> 00:24:25
Other questions?
334
00:24:25 --> 00:24:25
Yes.
335
00:24:25 --> 00:24:34
Student: [INAUDIBLE]
336
00:24:34 --> 00:24:40
Professor: So the question is
what happened between this
337
00:24:40 --> 00:24:43
step and this step, right?
338
00:24:43 --> 00:24:45
Explain this step here.
339
00:24:45 --> 00:24:48
Alright, so there were
two parts to that.
340
00:24:48 --> 00:24:53
The first is this delta x which
is sitting in the denominator,
341
00:24:53 --> 00:24:56
I factored all the
way out front.
342
00:24:56 --> 00:24:58
And so what's in the
parentheses is supposed to
343
00:24:58 --> 00:25:00
be the same as what's in
344
00:25:00 --> 00:25:03
the numerator of this
other expression.
345
00:25:03 --> 00:25:07
And then, at the same time as
doing that, I put that
346
00:25:07 --> 00:25:10
expression, which is the
difference of two fractions, I
347
00:25:10 --> 00:25:12
expressed it with a
common denominator.
348
00:25:12 --> 00:25:14
So in the denominator here,
you see the product of
349
00:25:14 --> 00:25:17
the denominators of
the two fractions.
350
00:25:17 --> 00:25:19
And then I just figured out
what the numerator had
351
00:25:19 --> 00:25:22
to be without really...
352
00:25:22 --> 00:25:27
Other questions?
353
00:25:27 --> 00:25:32
OK.
354
00:25:32 --> 00:25:41
So I claim that on the whole,
calculus gets a bad rap, that
355
00:25:41 --> 00:25:47
it's actually easier
than most things.
356
00:25:47 --> 00:25:52
But there's a perception
that it's harder.
357
00:25:52 --> 00:25:56
And so I really have a duty
to give you the calculus
358
00:25:56 --> 00:25:59
made harder story here.
359
00:25:59 --> 00:26:03
So we have to make things
harder, because that's our job.
360
00:26:03 --> 00:26:06
And this is actually what most
people do in calculus, and it's
361
00:26:06 --> 00:26:09
the reason why calculus
has a bad reputation.
362
00:26:09 --> 00:26:15
So the secret is that when
people ask problems in
363
00:26:15 --> 00:26:19
calculus, they generally
ask them in context.
364
00:26:19 --> 00:26:22
And there are many, many
other things going on.
365
00:26:22 --> 00:26:25
And so the little piece of the
problem which is calculus is
366
00:26:25 --> 00:26:28
actually fairly routine and has
to be isolated and
367
00:26:28 --> 00:26:28
gotten through.
368
00:26:28 --> 00:26:32
But all the rest of it, relies
on everything else you learned
369
00:26:32 --> 00:26:36
in mathematics up to this
stage, from grade school
370
00:26:36 --> 00:26:36
through high school.
371
00:26:36 --> 00:26:39
So that's the complication.
372
00:26:39 --> 00:26:41
So now we're going to
do a little bit of
373
00:26:41 --> 00:26:49
calculus made hard.
374
00:26:49 --> 00:26:53
By talking about
a word problem.
375
00:26:53 --> 00:26:57
We only have one sort of word
problem that we can pose,
376
00:26:57 --> 00:27:03
because all we've talked about
is this geometry point of view.
377
00:27:03 --> 00:27:06
So far those are the only kinds
of word problems we can pose.
378
00:27:06 --> 00:27:08
So what we're gonna do is
just pose such a problem.
379
00:27:08 --> 00:27:30
So find the areas of triangles,
enclosed by the axes and
380
00:27:30 --> 00:27:39
the tangent to y = 1/x.
381
00:27:39 --> 00:27:43
OK, so that's a
geometry problem.
382
00:27:43 --> 00:27:46
And let me draw a
picture of it.
383
00:27:46 --> 00:27:52
It's practically the same as
the picture for example one.
384
00:27:52 --> 00:27:54
We only consider the
first quadrant.
385
00:27:54 --> 00:27:55
Here's our shape.
386
00:27:55 --> 00:28:00
All right, it's the hyperbola.
387
00:28:00 --> 00:28:03
And here's maybe one of our
tangent lines, which is
388
00:28:03 --> 00:28:05
coming in like this.
389
00:28:05 --> 00:28:12
And then we're trying to
find this area here.
390
00:28:12 --> 00:28:14
Right, so there's our problem.
391
00:28:14 --> 00:28:16
So why does it have
to do with calculus?
392
00:28:16 --> 00:28:18
It has to do with calculus
because there's a tangent line
393
00:28:18 --> 00:28:22
in it, so we're gonna need
to do some calculus to
394
00:28:22 --> 00:28:24
answer this question.
395
00:28:24 --> 00:28:30
But as you'll see, the
calculus is the easy part.
396
00:28:30 --> 00:28:34
So let's get started
with this problem.
397
00:28:34 --> 00:28:37
First of all, I'm gonna
label a few things.
398
00:28:37 --> 00:28:40
And one important thing to
remember of course, is that
399
00:28:40 --> 00:28:41
the curve is y = 1/x.
400
00:28:42 --> 00:28:44
That's perfectly
reasonable to do.
401
00:28:44 --> 00:28:49
And also, we're gonna calculate
the areas of the triangles, and
402
00:28:49 --> 00:28:51
you could ask yourself,
in terms of what?
403
00:28:51 --> 00:28:54
Well, we're gonna have to pick
a point and give it a name.
404
00:28:54 --> 00:28:56
And since we need a number,
we're gonna have to do
405
00:28:56 --> 00:28:56
more than geometry.
406
00:28:56 --> 00:28:59
We're gonna have to do some
of this analysis just
407
00:28:59 --> 00:29:01
as we've done before.
408
00:29:01 --> 00:29:04
So I'm gonna pick a point and,
consistent with the labeling
409
00:29:04 --> 00:29:08
we've done before, I'm
gonna to call it (x0, y0).
410
00:29:08 --> 00:29:13
So that's almost half the
battle, having notations, x and
411
00:29:13 --> 00:29:18
y for the variables, and x0 and
y0, for the specific point.
412
00:29:18 --> 00:29:25
Now, once you see that you have
these labellings, I hope it's
413
00:29:25 --> 00:29:28
reasonable to do the following.
414
00:29:28 --> 00:29:31
So first of all, this is
the point x0, and over
415
00:29:31 --> 00:29:33
here is the point y0.
416
00:29:33 --> 00:29:37
That's something that
we're used to in graphs.
417
00:29:37 --> 00:29:40
And in order to figure out the
area of this triangle, it's
418
00:29:40 --> 00:29:43
pretty clear that we should
find the base, which is that we
419
00:29:43 --> 00:29:45
should find this location here.
420
00:29:45 --> 00:29:48
And we should find the
height, so we need to
421
00:29:48 --> 00:29:55
find that value there.
422
00:29:55 --> 00:29:58
Let's go ahead and do it.
423
00:29:58 --> 00:30:02
So how are we going to do this?
424
00:30:02 --> 00:30:14
Well, so let's
just take a look.
425
00:30:14 --> 00:30:17
So what is it that
we need to do?
426
00:30:17 --> 00:30:21
I claim that there's only one
calculus step, and I'm gonna
427
00:30:21 --> 00:30:25
put a star here for
this tangent line.
428
00:30:25 --> 00:30:28
I have to understand what
the tangent line is.
429
00:30:28 --> 00:30:30
Once I've figured out what the
tangent line is, the rest of
430
00:30:30 --> 00:30:33
the problem is no
longer calculus.
431
00:30:33 --> 00:30:35
It's just that slope
that we need.
432
00:30:35 --> 00:30:38
So what's the formula
for the tangent line?
433
00:30:38 --> 00:30:45
Put that over here. it's going
to be y - y0 is equal to,
434
00:30:45 --> 00:30:48
and here's the magic number,
we already calculated it.
435
00:30:48 --> 00:30:50
It's in the box over there.
436
00:30:50 --> 00:30:58
It's -1/x0^2 ( x - x0).
437
00:30:58 --> 00:31:12
So this is the only bit of
calculus in this problem.
438
00:31:12 --> 00:31:15
But now we're not done.
439
00:31:15 --> 00:31:16
We have to finish it.
440
00:31:16 --> 00:31:19
We have to figure out all the
rest of these quantities so
441
00:31:19 --> 00:31:27
we can figure out the area.
442
00:31:27 --> 00:31:31
All right.
443
00:31:31 --> 00:31:40
So how do we do that?
444
00:31:40 --> 00:31:44
Well, to find this
point, this has a name.
445
00:31:44 --> 00:31:52
We're gonna find the so
called x-intercept.
446
00:31:52 --> 00:31:54
That's the first thing
we're going to do.
447
00:31:54 --> 00:31:58
So to do that, what we need
to do is to find where
448
00:31:58 --> 00:32:02
this horizontal line
meets that diagonal line.
449
00:32:02 --> 00:32:10
And the equation for the
x-intercept is y = 0.
450
00:32:10 --> 00:32:13
So we plug in y = 0, that's
this horizontal line,
451
00:32:13 --> 00:32:15
and we find this point.
452
00:32:15 --> 00:32:18
So let's do that into star.
453
00:32:18 --> 00:32:22
We get 0 minus, oh one other
thing we need to know.
454
00:32:22 --> 00:32:28
We know that y0 is f(x0)
, and f(x) is 1/x , so
455
00:32:28 --> 00:32:31
this thing is 1/x0.
456
00:32:31 --> 00:32:33
457
00:32:33 --> 00:32:36
And that's equal to -1/x0^2.
458
00:32:38 --> 00:32:41
And here's x, and here's x0.
459
00:32:41 --> 00:32:49
All right, so in order to find
this x value, I have to plug in
460
00:32:49 --> 00:32:53
one equation into the other.
461
00:32:53 --> 00:32:59
So this simplifies a bit.
462
00:32:59 --> 00:33:01
This is -x/x0^2.
463
00:33:03 --> 00:33:07
And this is plus 1/x0
because the x0 and
464
00:33:07 --> 00:33:10
x0^2 cancel somewhat.
465
00:33:10 --> 00:33:15
And so if I put this on
the other side, I get x /
466
00:33:15 --> 00:33:20
x0^2 is equal to 2 / x0.
467
00:33:20 --> 00:33:27
And if I then multiply through
- so that's what this implies -
468
00:33:27 --> 00:33:39
and if I multiply through
by x0^2 I get x = 2x0.
469
00:33:39 --> 00:33:51
OK, so I claim that this point
weve just calculated it's 2x0.
470
00:33:51 --> 00:33:57
Now, I'm almost done.
471
00:33:57 --> 00:34:00
I need to get the other one.
472
00:34:00 --> 00:34:03
I need to get this one up here.
473
00:34:03 --> 00:34:06
Now I'm gonna use a very
big shortcut to do that.
474
00:34:06 --> 00:34:27
So the shortcut to the
y-intercept is to use symmetry.
475
00:34:27 --> 00:34:30
All right, I claim I can stare
at this and I can look at that,
476
00:34:30 --> 00:34:33
and I know the formula
for the y-intercept.
477
00:34:33 --> 00:34:40
It's equal to 2y0.
478
00:34:40 --> 00:34:40
All right.
479
00:34:40 --> 00:34:42
That's what that one is.
480
00:34:42 --> 00:34:44
So this one is 2y0.
481
00:34:44 --> 00:34:48
And the reason I know this is
the following: so here's the
482
00:34:48 --> 00:34:52
symmetry of the situation,
which is not completely direct.
483
00:34:52 --> 00:34:56
It's a kind of mirror symmetry
around the diagonal.
484
00:34:56 --> 00:35:05
It involves the exchange of (x,
y) with (y, x); so trading
485
00:35:05 --> 00:35:06
the roles of x and y.
486
00:35:06 --> 00:35:09
So the symmetry that I'm using
is that any formula I get that
487
00:35:09 --> 00:35:13
involves x's and y's, if I
trade all the x's and replace
488
00:35:13 --> 00:35:16
them by y's and trade all the
y's and replace them by x's,
489
00:35:16 --> 00:35:18
then I'll have a correct
formula on the other ways.
490
00:35:18 --> 00:35:21
So if everywhere I see a y I
make it an x, and everywhere I
491
00:35:21 --> 00:35:24
see an x I make it a y, the
switch will take place.
492
00:35:24 --> 00:35:27
So why is that?
493
00:35:27 --> 00:35:30
That's just an accident
of this equation.
494
00:35:30 --> 00:35:46
That's because, so the symmetry
explained... is that the
495
00:35:46 --> 00:35:48
equation is y= 1 / x.
496
00:35:48 --> 00:35:53
But that's the same thing as xy
= 1, if I multiply through by
497
00:35:53 --> 00:35:58
x, which is the same
thing as x = 1/y.
498
00:35:58 --> 00:36:05
So here's where the x
and the y get reversed.
499
00:36:05 --> 00:36:13
OK now if you don't trust this
explanation, you can also get
500
00:36:13 --> 00:36:28
the y-intercept by plugging x
= 0 into the equation star.
501
00:36:28 --> 00:36:29
OK?
502
00:36:29 --> 00:36:34
We plugged y = 0 in and
we got the x value.
503
00:36:34 --> 00:36:43
And you can do the same thing
analogously the other way.
504
00:36:43 --> 00:36:47
All right so I'm almost done
with the geometry problem,
505
00:36:47 --> 00:36:58
and let's finish it off now.
506
00:36:58 --> 00:37:00
Well, let me hold off for one
second before I finish it off.
507
00:37:00 --> 00:37:05
What I'd like to say is just
make one more tiny remark.
508
00:37:05 --> 00:37:09
And this is the hardest part
of calculus in my opinion.
509
00:37:09 --> 00:37:15
So the hardest part of calculus
is that we call it one variable
510
00:37:15 --> 00:37:21
calculus, but we're perfectly
happy to deal with four
511
00:37:21 --> 00:37:25
variables at a time or
five, or any number.
512
00:37:25 --> 00:37:29
In this problem, I had an
x, a y, an x0 and a y0.
513
00:37:29 --> 00:37:32
That's already four different
things that have various
514
00:37:32 --> 00:37:35
relationships between them.
515
00:37:35 --> 00:37:37
Of course the manipulations we
do with them are algebraic, and
516
00:37:37 --> 00:37:41
when we're doing the
derivatives we just consider
517
00:37:41 --> 00:37:43
what's known as one
variable calculus.
518
00:37:43 --> 00:37:45
But really there are millions
of variable floating
519
00:37:45 --> 00:37:46
around potentially.
520
00:37:46 --> 00:37:49
So that's what makes things
complicated, and that's
521
00:37:49 --> 00:37:51
something that you
have to get used to.
522
00:37:51 --> 00:37:53
Now there's something else
which is more subtle, and that
523
00:37:53 --> 00:37:58
I think many people who teach
the subject or use the subject
524
00:37:58 --> 00:38:01
aren't aware, because they've
already entered into the
525
00:38:01 --> 00:38:04
language and they're so
comfortable with it that they
526
00:38:04 --> 00:38:06
don't even notice
this confusion.
527
00:38:06 --> 00:38:10
There's something deliberately
sloppy about the way we
528
00:38:10 --> 00:38:12
deal with these variables.
529
00:38:12 --> 00:38:14
The reason is very simple.
530
00:38:14 --> 00:38:16
There are already
four variables here.
531
00:38:16 --> 00:38:20
I don't wanna create six
names for variables or
532
00:38:20 --> 00:38:23
eight names for variables.
533
00:38:23 --> 00:38:26
But really in this problem
there were about eight.
534
00:38:26 --> 00:38:29
I just slipped them by you.
535
00:38:29 --> 00:38:30
So why is that?
536
00:38:30 --> 00:38:35
Well notice that the first time
that I got a formula for y0
537
00:38:35 --> 00:38:39
here, it was this point.
538
00:38:39 --> 00:38:44
And so the formula for y0,
which I plugged in right here,
539
00:38:44 --> 00:38:50
was from the equation of
the curve. y0 = 1 / x0.
540
00:38:50 --> 00:38:55
The second time I did it,
I did not use y = 1 / x.
541
00:38:55 --> 00:39:01
I used this equation here,
so this is not y = 1/x.
542
00:39:01 --> 00:39:03
That's the wrong thing to do.
543
00:39:03 --> 00:39:06
It's an easy mistake to make if
the formulas are all a blur to
544
00:39:06 --> 00:39:09
you and you're not paying
attention to where they
545
00:39:09 --> 00:39:11
are on the diagram.
546
00:39:11 --> 00:39:16
You see that x-intercept
calculation there involved
547
00:39:16 --> 00:39:21
where this horizontal line met
this diagonal line, and y = 0
548
00:39:21 --> 00:39:25
represented this line here.
549
00:39:25 --> 00:39:31
So the sloppines is that y
means two different things.
550
00:39:31 --> 00:39:34
And we do this constantly
because it's way, way more
551
00:39:34 --> 00:39:37
complicated not to do it.
552
00:39:37 --> 00:39:40
It's much more convenient for
us to allow ourselves the
553
00:39:40 --> 00:39:44
flexibility to change the role
that this letter plays in
554
00:39:44 --> 00:39:47
the middle of a computation.
555
00:39:47 --> 00:39:50
And similarly, later on, if I
had done this by this more
556
00:39:50 --> 00:39:54
straightforward method, for
the y-intercept, I would
557
00:39:54 --> 00:39:55
have set x equal to 0.
558
00:39:55 --> 00:39:59
That would have been this
vertical line, which is x = 0.
559
00:39:59 --> 00:40:03
But I didn't change the letter
x when I did that, because
560
00:40:03 --> 00:40:06
that would be a waste for us.
561
00:40:06 --> 00:40:09
So this is one of the main
confusions that happens.
562
00:40:09 --> 00:40:13
If you can keep yourself
straight, you're a lot better
563
00:40:13 --> 00:40:21
off, and as I say this is
one of the complexities.
564
00:40:21 --> 00:40:24
All right, so now let's
finish off the problem.
565
00:40:24 --> 00:40:30
Let me finally get
this area here.
566
00:40:30 --> 00:40:33
So, actually I'll just
finish it off right here.
567
00:40:33 --> 00:40:41
So the area of the triangle
is, well it's the base
568
00:40:41 --> 00:40:42
times the height.
569
00:40:42 --> 00:40:46
The base is 2x0 the height
is 2y0, and a half of that.
570
00:40:46 --> 00:40:54
So it's 1/2( 2x0)(2y0) ,
which is (2x0)(y0), which
571
00:40:54 --> 00:40:57
is, lo and behold, 2.
572
00:40:57 --> 00:41:00
So the amusing thing in this
case is that it actually didn't
573
00:41:00 --> 00:41:02
matter what x0 and y0 are.
574
00:41:02 --> 00:41:05
We get the same
answer every time.
575
00:41:05 --> 00:41:10
That's just an accident
of the function 1 / x.
576
00:41:10 --> 00:41:19
It happens to be the function
with that property.
577
00:41:19 --> 00:41:23
All right, so we have some
more business today,
578
00:41:23 --> 00:41:24
some serious business.
579
00:41:24 --> 00:41:30
So let me continue.
580
00:41:30 --> 00:41:41
So, first of all, I want to
give you a few more notations.
581
00:41:41 --> 00:41:50
And these are just other
notations that people use
582
00:41:50 --> 00:41:51
to refer to derivatives.
583
00:41:51 --> 00:41:54
And the first one is the
following: we already
584
00:41:54 --> 00:41:56
wrote y = f(x).
585
00:41:56 --> 00:42:00
And so when we write delta
y, that means the same
586
00:42:00 --> 00:42:01
thing as delta f.
587
00:42:01 --> 00:42:04
That's a typical notation.
588
00:42:04 --> 00:42:13
And previously we wrote f'
for the derivative, so
589
00:42:13 --> 00:42:20
this is Newton's notation
for the derivative.
590
00:42:20 --> 00:42:22
But there are other notations.
591
00:42:22 --> 00:42:28
And one of them is df/dx, and
another one is dy/ dx, meaning
592
00:42:28 --> 00:42:29
exactly the same thing.
593
00:42:29 --> 00:42:34
And sometimes we let the
function slip down below
594
00:42:34 --> 00:42:40
so that becomes d /
dx (f) and d/ dx(y) .
595
00:42:40 --> 00:42:43
So these are all notations
that are used for the
596
00:42:43 --> 00:42:49
derivative, and these were
initiated by Leibniz.
597
00:42:49 --> 00:42:55
And these notations are used
interchangeably, sometimes
598
00:42:55 --> 00:42:56
practically together.
599
00:42:56 --> 00:42:59
They both turn out to
be extremely useful.
600
00:42:59 --> 00:43:04
This one omits - notice that
this thing omits- the
601
00:43:04 --> 00:43:07
underlying base point, x0.
602
00:43:07 --> 00:43:09
That's one of the nuisances.
603
00:43:09 --> 00:43:11
It doesn't give you
all the information.
604
00:43:11 --> 00:43:17
But there are lots of
situations like that where
605
00:43:17 --> 00:43:20
people leave out some of the
important information, and
606
00:43:20 --> 00:43:23
you have to fill it
in from context.
607
00:43:23 --> 00:43:28
So that's another
couple of notations.
608
00:43:28 --> 00:43:33
So now I have one more
calculation for you today.
609
00:43:33 --> 00:43:35
I carried out this calculation
of the derivative of
610
00:43:35 --> 00:43:45
the function 1 / x.
611
00:43:45 --> 00:43:48
I wanna take care of
some other powers.
612
00:43:48 --> 00:43:59
So let's do that.
613
00:43:59 --> 00:44:07
So Example 2 is going to be
the function f(x) = x^n.
614
00:44:08 --> 00:44:14
n = 1, 2, 3; one of these guys.
615
00:44:14 --> 00:44:18
And now what we're trying to
figure out is the derivative
616
00:44:18 --> 00:44:21
with respect to x of x^n in
our new notation, what
617
00:44:21 --> 00:44:27
this is equal to.
618
00:44:27 --> 00:44:32
So again, we're going to
form this expression,
619
00:44:32 --> 00:44:35
delta f / delta x.
620
00:44:35 --> 00:44:38
And we're going to make some
algebraic simplification.
621
00:44:38 --> 00:44:41
So what we plug in
for delta f is ((x
622
00:44:41 --> 00:44:46
delta x)^n - x^n)/delta x.
623
00:44:48 --> 00:44:50
Now before, let me just
stick this in then
624
00:44:50 --> 00:44:52
I'm gonna erase it.
625
00:44:52 --> 00:44:56
Before, I wrote x0
here and x0 there.
626
00:44:56 --> 00:45:00
But now I'm going to get rid of
it, because in this particular
627
00:45:00 --> 00:45:01
calculation, it's a nuisance.
628
00:45:01 --> 00:45:04
I don't have an x floating
around, which means something
629
00:45:04 --> 00:45:06
different from the x0.
630
00:45:06 --> 00:45:08
And I just don't wanna
have to keep on writing
631
00:45:08 --> 00:45:10
all those symbols.
632
00:45:10 --> 00:45:13
It's a waste of
blackboard energy.
633
00:45:13 --> 00:45:17
There's a total amount of
energy, and I've already filled
634
00:45:17 --> 00:45:21
up so many blackboards that,
there's just a limited amount.
635
00:45:21 --> 00:45:23
Plus, I'm trying to
conserve chalk.
636
00:45:23 --> 00:45:25
Anyway, no 0's.
637
00:45:25 --> 00:45:28
So think of x as fixed.
638
00:45:28 --> 00:45:40
In this case, delta x moves and
x is fixed in this calculation.
639
00:45:40 --> 00:45:42
All right now, in order to
simplify this, in order to
640
00:45:42 --> 00:45:45
understand algebraically what's
going on, I need to understand
641
00:45:45 --> 00:45:48
what the nth power of a sum is.
642
00:45:48 --> 00:45:50
And that's a famous formula.
643
00:45:50 --> 00:45:52
We only need a little tiny
bit of it, called the
644
00:45:52 --> 00:45:56
binomial theorem.
645
00:45:56 --> 00:46:08
So, the binomial theorem which
is in your text and explained
646
00:46:08 --> 00:46:15
in an appendix, says that if
you take the sum of two guys
647
00:46:15 --> 00:46:18
and you take them to the nth
power, that of course is (x
648
00:46:18 --> 00:46:24
delta x) multiplied
by itself n times.
649
00:46:24 --> 00:46:29
And so the first term is
x^n, that's when all of
650
00:46:29 --> 00:46:31
the n factors come in.
651
00:46:31 --> 00:46:35
And then, you could have
this factor of delta x
652
00:46:35 --> 00:46:36
and all the rest x's.
653
00:46:36 --> 00:46:39
So at least one term of the
form (x^(n-1))delta x.
654
00:46:39 --> 00:46:41
655
00:46:41 --> 00:46:43
And how many times
does that happen?
656
00:46:43 --> 00:46:46
Well, it happens when there's a
factor from here, from the next
657
00:46:46 --> 00:46:48
factor, and so on, and
so on, and so on.
658
00:46:48 --> 00:46:54
There's a total of n possible
times that that happens.
659
00:46:54 --> 00:46:59
And now the great thing is
that, with this alone, all the
660
00:46:59 --> 00:47:06
rest of the terms are junk that
we won't have to worry about.
661
00:47:06 --> 00:47:11
So to be more specific,
there's a very careful
662
00:47:11 --> 00:47:12
notation for the junk.
663
00:47:12 --> 00:47:14
The junk is what's called
big O((delta x)^2).
664
00:47:14 --> 00:47:18
665
00:47:18 --> 00:47:27
What that means is that these
are terms of order, so with
666
00:47:27 --> 00:47:33
(delta x)^2, (delta
x)^3 or higher.
667
00:47:34 --> 00:47:38
All right, that's how.
668
00:47:38 --> 00:47:42
Very exciting,
higher order terms.
669
00:47:42 --> 00:47:47
OK, so this is the only algebra
that we need to do, and now
670
00:47:47 --> 00:47:50
we just need to combine it
together to get our result.
671
00:47:50 --> 00:47:54
So, now I'm going to just
carry out the cancellations
672
00:47:54 --> 00:48:02
that we need.
673
00:48:02 --> 00:48:03
So here we go.
674
00:48:03 --> 00:48:10
We have delta f / delta x,
which remember was 1 / delta
675
00:48:10 --> 00:48:22
x times this, which is this
times, now this is (x^n
676
00:48:22 --> 00:48:27
nx^(n-1) delta x
677
00:48:27 --> 00:48:35
this junk term) - x^n.
678
00:48:35 --> 00:48:38
So that's what we have
so far based on our
679
00:48:38 --> 00:48:41
previous calculations.
680
00:48:41 --> 00:48:48
Now, I'm going to do the main
cancellation, which is this.
681
00:48:48 --> 00:48:49
All right.
682
00:48:49 --> 00:48:55
So, that's 1/delta x(
nx^(n-1) delta x
683
00:48:55 --> 00:49:01
this term here).
684
00:49:01 --> 00:49:05
And now I can divide
in by delta x.
685
00:49:05 --> 00:49:08
So I get nx^(n-1)
686
00:49:08 --> 00:49:09
now it's O(delta x).
687
00:49:10 --> 00:49:13
There's at least one factor of
delta x not two factors of
688
00:49:13 --> 00:49:17
delta x, because I have
to cancel one of them.
689
00:49:17 --> 00:49:19
And now I can just
take the limit.
690
00:49:19 --> 00:49:22
And the limit this
term is gonna be 0.
691
00:49:22 --> 00:49:25
That's why I called
it junk originally,
692
00:49:25 --> 00:49:26
because it disappears.
693
00:49:26 --> 00:49:31
And in math, junk is
something that goes away.
694
00:49:31 --> 00:49:36
So this tends to, as delta
x goes to 0, nx ^ (n-1).
695
00:49:37 --> 00:49:43
And so what I've shown you is
that d/dx of x to the n minus -
696
00:49:43 --> 00:49:47
sorry -n, is equal to nx^(n-1).
697
00:49:47 --> 00:49:51
698
00:49:51 --> 00:49:54
So now this is gonna be super
important to you right on your
699
00:49:54 --> 00:49:57
problem set in every possible
way, and I want to tell you one
700
00:49:57 --> 00:50:00
thing, one way in which
it's very important.
701
00:50:00 --> 00:50:02
One way that extends
it immediately.
702
00:50:02 --> 00:50:10
So this thing extends
to polynomials.
703
00:50:10 --> 00:50:13
We get quite a lot out of
this one calculation.
704
00:50:13 --> 00:50:18
Namely, if I take d / dx
of something like (x^3
705
00:50:18 --> 00:50:25
5x^10) that's gonna be equal
to 3x^2, that's applying
706
00:50:25 --> 00:50:27
this rule to x^3.
707
00:50:27 --> 00:50:33
And then here, I'll
get 5*10 so 50x^9.
708
00:50:35 --> 00:50:38
So this is the type of thing
that we get out of it, and
709
00:50:38 --> 00:50:50
we're gonna make more hay
with that next time.
710
00:50:50 --> 00:50:50
Question.
711
00:50:50 --> 00:50:50
Yes.
712
00:50:50 --> 00:50:51
I turned myself off.
713
00:50:51 --> 00:50:52
Yes?
714
00:50:52 --> 00:50:56
Student: [INAUDIBLE]
715
00:50:56 --> 00:51:01
Professor: The question was the
binomial theorem only works
716
00:51:01 --> 00:51:04
when delta x goes to 0.
717
00:51:04 --> 00:51:07
No, the binomial theorem is a
general formula which also
718
00:51:07 --> 00:51:10
specifies exactly
what the junk is.
719
00:51:10 --> 00:51:11
It's very much more detailed.
720
00:51:11 --> 00:51:13
But we only needed this part.
721
00:51:13 --> 00:51:18
We didn't care what all
these crazy terms were.
722
00:51:18 --> 00:51:24
It's junk for our purposes now,
because we don't happen to need
723
00:51:24 --> 00:51:27
any more than those
first two terms.
724
00:51:27 --> 00:51:29
Yes, because delta x goes to 0.
725
00:51:29 --> 00:51:32
OK, see you next time.