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Professor: So, again
welcome to 18.01.
We're getting started
today with what
we're calling Unit One, a
highly imaginative title.
And it's differentiation.
So, let me first
tell you, briefly,
what's in store in the
next couple of weeks.
The main topic today is
what is a derivative.
And, we're going to look at this
from several different points
of view, and the first one is
the geometric interpretation.
That's what we'll
spend most of today on.
And then, we'll also talk
about a physical interpretation
of what a derivative is.
And then there's going to
be something else which
I guess is maybe the reason
why Calculus is so fundamental,
and why we always start with it
in most science and engineering
schools, which is the importance
of derivatives, of this,
to all measurements.
So that means pretty
much every place.
That means in science,
in engineering,
in economics, in
political science, etc.
Polling, lots of
commercial applications,
just about everything.
Now, that's what we'll
be getting started with,
and then there's
another thing that we're
gonna do in this unit, which
is we're going to explain
how to differentiate anything.
So, how to differentiate
any function you know.
And that's kind of a
tall order, but let
me just give you an example.
If you want to
take the derivative
- this we'll see
today is the notation
for the derivative of something
- of some messy function like e
^ x arctan x.
We'll work this out by
the end of this unit.
All right?
Anything you can think of,
anything you can write down,
we can differentiate it.
All right, so that's what we're
gonna do, and today, as I said,
we're gonna spend
most of our time
on this geometric
interpretation.
So let's begin with that.
So here we go with the geometric
interpretation of derivatives.
And, what we're going to do is
just ask the geometric problem
of finding the tangent
line to some graph
of some function at some point.
Which is to say (x_0, y_0).
So that's the problem that
we're addressing here.
Alright, so here's
our problem, and now
let me show you the solution.
So, well, let's
graph the function.
Here's its graph.
Here's some point.
All right, maybe I should
draw it just a bit lower.
So here's a point P.
Maybe it's above the point
x_0. x_0, by the way, this
was supposed to be an x_0.
That was some fixed
place on the x-axis.
And now, in order to
perform this mighty feat,
I will use another
color of chalk.
How about red?
OK.
So here it is.
There's the tangent line,
well, not quite straight.
Close enough.
All right?
I did it.
That's the geometric problem.
I achieved what I
wanted to do, and it's
kind of an interesting
question, which unfortunately I
can't solve for you
in this class, which
is, how did I do that?
That is, how
physically did I manage
to know what to do to
draw this tangent line?
But that's what geometric
problems are like.
We visualize it.
We can figure it out
somewhere in our brains.
It happens.
And the task that we
have now is to figure out
how to do it analytically,
to do it in a way
that a machine could
just as well as I did
in drawing this tangent line.
So, what did we learn in high
school about what a tangent
line is?
Well, a tangent line
has an equation,
and any line through a
point has the equation y
- y_0 is equal to m, the
slope, times x - x_0.
So here's the equation
for that line,
and now there are two
pieces of information
that we're going to need to
work out what the line is.
The first one is the point.
That's that point P there.
And to specify P,
given x, we need
to know the level of y, which
is of course just f(x_0).
That's not a calculus
problem, but anyway that's
a very important
part of the process.
So that's the first
thing we need to know.
And the second thing we
need to know is the slope.
And that's this number m.
And in calculus we have
another name for it.
We call it f prime of x_0.
Namely, the derivative of f.
So that's the calculus part.
That's the tricky part,
and that's the part
that we have to discuss now.
So just to make
that explicit here,
I'm going to make a definition,
which is that f '(x_0) ,
which is known as the
derivative, of f, at x_0,
is the slope of the tangent
line to y = f(x) at the point,
let's just call it P.
All right?
So, that's what
it is, but still I
haven't made any progress in
figuring out any better how
I drew that line.
So I have to say
something that's
more concrete, because I
want to be able to cook up
what these numbers are.
I have to figure out
what this number m is.
And one way of thinking about
that, let me just try this,
so I certainly am
taking for granted that
in sort of non-calculus
part that I know
what a line through a point is.
So I know this equation.
But another possibility
might be, this line here,
how do I know - well,
unfortunately, I didn't draw
it quite straight,
but there it is -
how do I know that this orange
line is not a tangent line,
but this other line
is a tangent line?
Well, it's actually
not so obvious,
but I'm gonna describe
it a little bit.
It's not really the
fact-- this thing
crosses at some
other place, which
is this point Q. But
it's not really the fact
that the thing
crosses at two place,
because the line
could be wiggly,
the curve could be wiggly,
and it could cross back
and forth a number of times.
That's not what distinguishes
the tangent line.
So I'm gonna have to
somehow grasp this,
and I'll first do
it in language.
And it's the
following idea: it's
that if you take this
orange line, which
is called a secant line,
and you think of the point Q
as getting closer and closer to
P, then the slope of that line
will get closer and closer
to the slope of the red line.
And if we draw it close
enough, then that's
gonna be the correct line.
So that's really what I did,
sort of in my brain when
I drew that first line.
And so that's the way I'm
going to articulate it first.
Now, so the tangent line is
equal to the limit of so called
secant lines PQ,
as Q tends to P.
And here we're thinking of P as
being fixed and Q as variable.
All right?
Again, this is still the
geometric discussion,
but now we're gonna be able
to put symbols and formulas
to this computation.
And we'll be able to work
out formulas in any example.
So let's do that.
So first of all, I'm gonna write
out these points P and Q again.
So maybe we'll put
P here and Q here.
And I'm thinking of
this line through them.
I guess it was orange, so
we'll leave it as orange.
All right.
And now I want to
compute its slope.
So this, gradually, we'll
do this in two steps.
And these steps
will introduce us
to the basic notations which
are used throughout calculus,
including multi-variable
calculus, across the board.
So the first
notation that's used
is you imagine here's
the x-axis underneath,
and here's the x_0, the location
directly below the point P.
And we're traveling here a
horizontal distance which
is denoted by delta x.
So that's delta x, so called.
And we could also call
it the change in x.
So that's one thing we want
to measure in order to get
the slope of this line PQ.
And the other thing
is this height.
So that's this distance here,
which we denote delta f,
which is the change in f.
And then, the slope is just
the ratio, delta f / delta x.
So this is the
slope of the secant.
And the process I just described
over here with this limit
applies not just to
the whole line itself,
but also in particular
to its slope.
And the way we write that is
the limit as delta x goes to 0.
And that's going
to be our slope.
So this is the slope
of the tangent line.
OK.
Now, This is still
a little general,
and I want to work out
a more usable form here,
a better formula for this.
And in order to
do that, I'm gonna
write delta f, the numerator
more explicitly here.
The change in f, so
remember that the point P
is the point (x_0, f(x_0)).
All right, that's what we got
for the formula for the point.
And in order to
compute these distances
and in particular the
vertical distance here,
I'm gonna have to get a
formula for Q as well.
So if this horizontal
distance is delta x,
then this location
is x_0 + delta x.
And so the point
above that point
has a formula, which
is x_0 plus delta
x, f of - and this is a
mouthful - x_0 plus delta x.
All right, so there's the
formula for the point Q.
Here's the formula
for the point P.
And now I can write a different
formula for the derivative,
which is the following:
so this f'(x_0) ,
which is the same as m, is going
to be the limit as delta x goes
to 0 of the change in f, well
the change in f is the value
of f at the upper point
here, which is x_0 + delta x,
and minus its value at the
lower point P, which is f(x_0),
divided by delta x.
All right, so this
is the formula.
I'm going to put
this in a little box,
because this is by far the
most important formula today,
which we use to derive
pretty much everything else.
And this is the way
that we're going to be
able to compute these numbers.
So let's do an example.
This example, we'll
call this example one.
We'll take the function
f(x) , which is 1/x .
That's sufficiently complicated
to have an interesting answer,
and sufficiently straightforward
that we can compute
the derivative fairly quickly.
So what is it that
we're gonna do here?
All we're going to do is we're
going to plug in this formula
here for that function.
That's all we're going
to do, and visually
what we're accomplishing is
somehow to take the hyperbola,
and take a point
on the hyperbola,
and figure out
some tangent line.
That's what we're
accomplishing when we do that.
So we're accomplishing
this geometrically
but we'll be doing
it algebraically.
So first, we consider this
difference delta f / delta x
and write out its formula.
So I have to have a place.
So I'm gonna make it again
above this point x_0, which
is the general point.
We'll make the
general calculation.
So the value of f at the top,
when we move to the right
by f(x), so I just read off
from this, read off from here.
The formula, the first
thing I get here is 1 /
(x_0 + delta x).
That's the left hand term.
Minus 1 / x_0, that's
the right hand term.
And then I have to
divide that by delta x.
OK, so here's our expression.
And by the way this has a name.
This thing is called
a difference quotient.
It's pretty complicated,
because there's always
a difference in the numerator.
And in disguise, the
denominator is a difference,
because it's the difference
between the value
on the right side and the
value on the left side here.
OK, so now we're going to
simplify it by some algebra.
So let's just take a look.
So this is equal to, let's
continue on the next level
here.
This is equal to 1
/ delta x times...
All I'm going to do is put
it over a common denominator.
So the common denominator
is (x_0 + delta x) * x_0.
And so in the numerator for the
first expressions I have x_0,
and for the second expression
I have x_0 + delta x.
So this is the same thing as
I had in the numerator before,
factoring out this denominator.
And here I put that numerator
into this more amenable form.
And now there are two
basic cancellations.
The first one is that x_0 and
x_0 cancel, so we have this.
And then the second step is that
these two expressions cancel,
the numerator and
the denominator.
Now we have a cancellation
that we can make use of.
So we'll write that under here.
And this is equals -1 over
x_0 plus delta x times x_0.
And then the very last step
is to take the limit as delta
x tends to 0, and
now we can do it.
Before we couldn't do it.
Why?
Because the numerator and the
denominator gave us 0 / 0.
But now that I've made
this cancellation,
I can pass to the limit.
And all that happens is
I set this delta x to 0,
and I get -1/x_0^2.
So that's the answer.
All right, so in other
words what I've shown -
let me put it up here - is
that f'(x_0) = -1/x_0^2.
Now, let's look at the
graph just a little
bit to check this for
plausibility, all right?
What's happening here is,
first of all it's negative.
It's less than 0,
which is a good thing.
You see that slope
there is negative.
That's the simplest check
that you could make.
And the second thing that I
would just like to point out
is that as x goes to infinity,
that as we go farther
to the right, it gets
less and less steep.
So as x_0 goes to infinity,
less and less steep.
So that's also
consistent here, when
x_0 is very large, this is
a smaller and smaller number
in magnitude, although
it's always negative.
It's always sloping down.
All right, so I've managed
to fill the boards.
So maybe I should stop
for a question or two.
Yes?
Student: [INAUDIBLE]
Professor: So the question
is to explain again
this limiting process.
So the formula here is we
have basically two numbers.
So in other words, why is
it that this expression,
when delta x tends to 0,
is equal to -1 / x_0^2 ?
Let me illustrate it by
sticking in a number for x_0
to make it more explicit.
All right, so for
instance, let me stick
in here for x_0 the number 3.
Then it's -1 over 3
plus delta x times 3.
That's the situation
that we've got.
And now the question
is what happens
as this number gets smaller
and smaller and smaller,
and gets to be practically 0?
Well, literally what we can
do is just plug in 0 there,
and you get 3 plus 0 times
3 in the denominator.
-1 in the numerator.
So this tends to
-1/9 (over 3^2).
And that's what I'm saying in
general with this extra number
here.
Other questions?
Yes.
Student: [INAUDIBLE]
Professor: So the
question is what
happened between this
step and this step, right?
Explain this step here.
Alright, so there were
two parts to that.
The first is this delta x which
is sitting in the denominator,
I factored all
the way out front.
And so what's in
the parentheses is
supposed to be
the same as what's
in the numerator of
this other expression.
And then, at the
same time as doing
that, I put that
expression, which
is the difference
of two fractions,
I expressed it with
a common denominator.
So in the denominator
here, you see
the product of the denominators
of the two fractions.
And then I just figured out what
the numerator had to be without
really...
Other questions?
OK.
So I claim that on
the whole, calculus
gets a bad rap,
that it's actually
easier than most things.
But there's a perception
that it's harder.
And so I really have a duty
to give you the calculus made
harder story here.
So we have to make things
harder, because that's our job.
And this is actually what
most people do in calculus,
and it's the reason why
calculus has a bad reputation.
So the secret is
that when people
ask problems in calculus, they
generally ask them in context.
And there are many, many
other things going on.
And so the little piece of
the problem which is calculus
is actually fairly routine and
has to be isolated and gotten
through.
But all the rest of it,
relies on everything else
you learned in mathematics up
to this stage, from grade school
through high school.
So that's the complication.
So now we're going to do
a little bit of calculus
made hard.
By talking about a word problem.
We only have one sort of word
problem that we can pose,
because all we've talked about
is this geometry point of view.
So far those are the only kinds
of word problems we can pose.
So what we're gonna do is
just pose such a problem.
So find the areas of
triangles, enclosed
by the axes and the
tangent to y = 1/x.
OK, so that's a
geometry problem.
And let me draw a picture of it.
It's practically the same as
the picture for example one.
We only consider
the first quadrant.
Here's our shape.
All right, it's the hyperbola.
And here's maybe one
of our tangent lines,
which is coming in like this.
And then we're trying
to find this area here.
Right, so there's our problem.
So why does it have
to do with calculus?
It has to do with
calculus because there's
a tangent line in
it, so we're gonna
need to do some calculus
to answer this question.
But as you'll see, the
calculus is the easy part.
So let's get started
with this problem.
First of all, I'm gonna
label a few things.
And one important thing
to remember of course,
is that the curve is y = 1/x.
That's perfectly
reasonable to do.
And also, we're gonna calculate
the areas of the triangles,
and you could ask
yourself, in terms of what?
Well, we're gonna have to pick
a point and give it a name.
And since we need
a number, we're
gonna have to do
more than geometry.
We're gonna have to do
some of this analysis
just as we've done before.
So I'm gonna pick a point and,
consistent with the labeling
we've done before, I'm
gonna to call it (x_0, y_0).
So that's almost half the
battle, having notations, x
and y for the variables,
and x_0 and y_0,
for the specific point.
Now, once you see that
you have these labelings,
I hope it's reasonable
to do the following.
So first of all, this
is the point x_0,
and over here is the point y_0.
That's something that
we're used to in graphs.
And in order to figure out
the area of this triangle,
it's pretty clear
that we should find
the base, which is that we
should find this location here.
And we should find
the height, so we
need to find that value there.
Let's go ahead and do it.
So how are we going to do this?
Well, so let's just take a look.
So what is it that
we need to do?
I claim that there's
only one calculus step,
and I'm gonna put a star
here for this tangent line.
I have to understand
what the tangent line is.
Once I've figured out
what the tangent line is,
the rest of the problem
is no longer calculus.
It's just that
slope that we need.
So what's the formula
for the tangent line?
Put that over here. it's going
to be y - y_0 is equal to,
and here's the magic number,
we already calculated it.
It's in the box over there.
It's -1/x_0^2 ( x - x_0).
So this is the only bit of
calculus in this problem.
But now we're not done.
We have to finish it.
We have to figure out all
the rest of these quantities
so we can figure out the area.
All right.
So how do we do that?
Well, to find this
point, this has a name.
We're gonna find the
so called x-intercept.
That's the first thing
we're going to do.
So to do that,
what we need to do
is to find where this horizontal
line meets that diagonal line.
And the equation for the
x-intercept is y = 0.
So we plug in y = 0, that's
this horizontal line,
and we find this point.
So let's do that into star.
We get 0 minus, oh one
other thing we need to know.
We know that y0 is f(x_0)
, and f(x) is 1/x ,
so this thing is 1/x_0.
And that's equal to -1/x_0^2.
And here's x, and here's x_0.
All right, so in order
to find this x value,
I have to plug in one
equation into the other.
So this simplifies a bit.
This is -x/x_0^2.
And this is plus 1/x_0 because
the x_0 and x0^2 cancel
somewhat.
And so if I put this
on the other side,
I get x / x_0^2 is
equal to 2 / x_0.
And if I then multiply through
- so that's what this implies -
and if I multiply through
by x_0^2 I get x = 2x_0.
OK, so I claim that this
point we've just calculated,
it's 2x_0.
Now, I'm almost done.
I need to get the other one.
I need to get this one up here.
Now I'm gonna use a very
big shortcut to do that.
So the shortcut to the
y-intercept is to use symmetry.
All right, I claim I can stare
at this and I can look at that,
and I know the formula
for the y-intercept.
It's equal to 2y_0.
All right.
That's what that one is.
So this one is 2y_0.
And the reason I know this
is the following: so here's
the symmetry of the situation,
which is not completely direct.
It's a kind of mirror
symmetry around the diagonal.
It involves the exchange
of (x, y) with (y, x);
so trading the roles of x and y.
So the symmetry
that I'm using is
that any formula I get that
involves x's and y's, if I
trade all the x's and
replace them by y's and trade
all the y's and replace
them by x's, then
I'll have a correct
formula on the other way.
So if everywhere I see
a y I make it an x,
and everywhere I see
an x I make it a y,
the switch will take place.
So why is that?
That's just an accident
of this equation.
That's because, so the
symmetry explained...
is that the equation is y = 1/x.
But that's the same
thing as xy = 1,
if I multiply
through by x, which
is the same thing as x = 1/y.
So here's where the x
and the y get reversed.
OK now if you don't
trust this explanation,
you can also get the
y-intercept by plugging x = 0
into the equation star.
OK?
We plugged y = 0 in
and we got the x-value.
And you can do the same thing
analogously the other way.
All right so I'm almost done
with the geometry problem,
and let's finish it off now.
Well, let me hold off for one
second before I finish it off.
What I'd like to say is just
make one more tiny remark.
And this is the hardest part
of calculus in my opinion.
So the hardest
part of calculus is
that we call it one
variable calculus,
but we're perfectly
happy to deal
with four variables at a
time or five, or any number.
In this problem, I had an
x, a y, an x_0 and a y_0.
That's already four
different things
that have various
relationships between them.
Of course the manipulations
we do with them are algebraic,
and when we're doing
the derivatives
we just consider what's known
as one variable calculus.
But really there are millions
of variable floating around
potentially.
So that's what makes
things complicated,
and that's something that
you have to get used to.
Now there's something
else which is more subtle,
and that I think many
people who teach the subject
or use the subject aren't
aware, because they've already
entered into the language and
they're so comfortable with it
that they don't even
notice this confusion.
There's something deliberately
sloppy about the way
we deal with these variables.
The reason is very simple.
There are already
four variables here.
I don't wanna create six names
for variables or eight names
for variables.
But really in this problem
there were about eight.
I just slipped them by you.
So why is that?
Well notice that the first time
that I got a formula for y_0
here, it was this point.
And so the formula for y_0,
which I plugged in right here,
was from the equation of
the curve. y_0 = 1 / x_0.
The second time I did it,
I did not use y = 1/x.
I used this equation here,
so this is not y = 1/x.
That's the wrong thing to do.
It's an easy mistake to
make if the formulas are
all a blur to you and
you're not paying attention
to where they are
on the diagram.
You see that x-intercept
calculation there involved
where this horizontal line met
this diagonal line, and y = 0
represented this line here.
So the sloppiness is that y
means two different things.
And we do this constantly
because it's way, way more
complicated not to do it.
It's much more convenient
for us to allow ourselves
the flexibility
to change the role
that this letter plays in
the middle of a computation.
And similarly, later
on, if I had done this
by this more straightforward
method, for the y-intercept,
I would have set x equal to 0.
That would have been this
vertical line, which is x = 0.
But I didn't change the letter
x when I did that, because that
would be a waste for us.
So this is one of the main
confusions that happens.
If you can keep
yourself straight,
you're a lot better
off, and as I
say this is one of
the complexities.
All right, so now let's
finish off the problem.
Let me finally get
this area here.
So, actually I'll just
finish it off right here.
So the area of the
triangle is, well
it's the base times the height.
The base is 2x_0, the height
is 2y_0, and a half of that.
So it's 1/2 (2x_0) * (2y_0) ,
which is 2x_0 y_0, which is,
lo and behold, 2.
So the amusing
thing in this case
is that it actually didn't
matter what x_0 and y_0 are.
We get the same
answer every time.
That's just an accident
of the function 1 / x.
It happens to be the
function with that property.
All right, so we have
some more business today,
some serious business.
So let me continue.
So, first of all, I want to
give you a few more notations.
And these are just
other notations
that people use to
refer to derivatives.
And the first one
is the following:
we already wrote y = f(x).
And so when we write
delta y, that means
the same thing as delta f.
That's a typical notation.
And previously we wrote f
prime for the derivative,
so this is Newton's
notation for the derivative.
But there are other notations.
And one of them is df/dx,
and another one is dy/dx,
meaning exactly the same thing.
And sometimes we
let the function
slip down below so that becomes
d/dx of f and d/dx of y.
So these are all notations that
are used for the derivative,
and these were
initiated by Leibniz.
And these notations are used
interchangeably, sometimes
practically together.
They both turn out to
be extremely useful.
This one omits - notice
that this thing omits
- the underlying
base point, x_0.
That's one of the nuisances.
It doesn't give you
all the information.
But there are lots of situations
like that where people leave
out some of the
important information,
and you have to fill
it in from context.
So that's another
couple of notations.
So now I have one more
calculation for you today.
I carried out this
calculation of the derivative
of the function 1 / x.
I wanna take care of
some other powers.
So let's do that.
So Example 2 is going to
be the function f(x) = x^n.
n = 1, 2, 3; one of these guys.
And now what we're trying to
figure out is the derivative
with respect to x of
x^n in our new notation,
what this is equal to.
So again, we're going to form
this expression, delta f /
delta x.
And we're going to make some
algebraic simplification.
So what we plug in for
delta f is ((x delta x)^n -
x^n)/delta x.
Now before, let
me just stick this
in then I'm gonna erase it.
Before, I wrote x_0
here and x_0 there.
But now I'm going
to get rid of it,
because in this particular
calculation, it's a nuisance.
I don't have an x
floating around,
which means something
different from the x_0.
And I just don't
wanna have to keep
on writing all those symbols.
It's a waste of
blackboard energy.
There's a total
amount of energy,
and I've already filled
up so many blackboards
that, there's just
a limited amount.
Plus, I'm trying
to conserve chalk.
Anyway, no 0's.
So think of x as fixed.
In this case, delta x moves and
x is fixed in this calculation.
All right now, in order
to simplify this, in order
to understand algebraically
what's going on,
I need to understand what
the nth power of a sum is.
And that's a famous formula.
We only need a little
tiny bit of it,
called the binomial theorem.
So, the binomial theorem
which is in your text
and explained in
an appendix, says
that if you take
the sum of two guys
and you take them to the
nth power, that of course
is (x + delta x) multiplied
by itself n times.
And so the first term is
x^n, that's when all of the n
factors come in.
And then, you could have this
factor of delta x and all
the rest x's.
So at least one term of the
form (x^(n-1)) times delta x.
And how many times
does that happen?
Well, it happens when
there's a factor from here,
from the next factor, and
so on, and so on, and so on.
There's a total of n possible
times that that happens.
And now the great thing
is that, with this alone,
all the rest of the
terms are junk that we
won't have to worry about.
So to be more specific,
there's a very careful notation
for the junk.
The junk is what's called
big O of (delta x)^2.
What that means is that
these are terms of order,
so with (delta x)^2,
(delta x)^3 or higher.
All right, that's how.
Very exciting,
higher order terms.
OK, so this is the only
algebra that we need to do,
and now we just need to combine
it together to get our result.
So, now I'm going to just
carry out the cancellations
that we need.
So here we go.
We have delta f / delta x, which
remember was 1 / delta x times
this, which is this times, now
this is x^n plus nx^(n-1) delta
x plus this junk
term, minus x^n.
So that's what we
have so far based
on our previous calculations.
Now, I'm going to do the main
cancellation, which is this.
All right.
So, that's 1/delta x times
nx^(n-1) delta x plus this term
here.
And now I can divide
in by delta x.
So I get nx^(n-1) plus,
now it's O(delta x).
There's at least
one factor of delta
x not two factors of
delta x, because I
have to cancel one of them.
And now I can just
take the limit.
In the limit this
term is gonna be 0.
That's why I called
it junk originally,
because it disappears.
And in math, junk is
something that goes away.
So this tends to, as delta
x goes to 0, nx^(n-1).
And so what I've shown you is
that d/dx of x to the n minus--
sorry, n, is equal to nx^(n-1).
So now this is gonna be
super important to you
right on your problem set
in every possible way,
and I want to tell you one
thing, one way in which it's
very important.
One way that extends
it immediately.
So this thing extends
to polynomials.
We get quite a lot out
of this one calculation.
Namely, if I take d/dx of
something like (x^3 + 5x^10)
that's gonna be equal to 3x^2,
that's applying this rule
to x^3.
And then here, I'll
get 5*10 so 50x^9.
So this is the type of
thing that we get out of it,
and we're gonna make more
hay with that next time.
Question.
Yes.
I turned myself off.
Yes?
Student: [INAUDIBLE]
Professor: The question
was the binomial theorem
only works when
delta x goes to 0.
No, the binomial theorem
is a general formula
which also specifies
exactly what the junk is.
It's very much more detailed.
But we only needed this part.
We didn't care what all
these crazy terms were.
It's junk for our
purposes now, because we
don't happen to need any more
than those first two terms.
Yes, because delta x goes to 0.
OK, see you next time.
