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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.

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Lec 1 | MIT 18.01 Single Variable Calculus, Fall 2007

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    yukang920108 に公開 2022 年 08 月 13 日
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