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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 arctanx.

  • 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 (x0, y0).

  • 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 it's 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 x0. x0, by the way, this

  • was supposed to be an x0.

  • 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 - y0 is equal to m the

  • slope, times x - x0.

  • 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(x0).

  • 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 x0.

  • 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 '(x0) , which is known as the

  • derivative, of f, at x0, 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 x0, 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 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 (x0, f(x0)).

  • 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 (x0

  • delta x).

  • And so the point above that point has a

  • formula, which is (x0

  • delta x, f(x0 and this is a mouthful,

  • 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'(x0) , 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 (x0

  • delta x), and minus its value at the lower point P, which is

  • f(x0), 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 x0, 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 / x0

  • delta x.

  • That's the left hand term.

  • Minus 1 / x0, 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 (x0

  • delta x)x0.

  • And so in the numerator for the first expressions I

  • have x0, and for the second expression I have x0

  • 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 x0 and x0 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 / (x0

  • delta x)x0.

  • 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/x0^2.

  • So that's the answer.

  • All right, so in other words what I've shown - let me put

  • it up here- is that f'(x0) = -1/x0^2.

  • 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 x0 goes to infinity, less and less steep.

  • So that's also consistent here, when x0 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/x0^2 ?

  • Let me illustrate it by sticking in a number for x0

  • to make it more explicit.

  • All right, so for instance, let me stick in here

  • for x0 the number 3.

  • Then it's -1 / (3

  • delta x)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

  • 0)3 in the denominator.

  • Minus one 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 (x0, y0).

  • So that's almost half the battle, having notations, x and

  • y for the variables, and x0 and y0, for the specific point.

  • Now, once you see that you have these labellings, I hope it's

  • reasonable to do the following.

  • So first of all, this is the point x0, and over

  • here is the point y0.

  • 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 - y0 is equal to,

  • and here's the magic number, we already calculated it.

  • It's in the box over there.

  • It's -1/x0^2 ( x - x0).

  • 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(x0) , and f(x) is 1/x , so

  • this thing is 1/x0.

  • And that's equal to -1/x0^2.

  • And here's x, and here's x0.

  • 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/x0^2.

  • And this is plus 1/x0 because the x0 and

  • x0^2 cancel somewhat.

  • And so if I put this on the other side, I get x /

  • x0^2 is equal to 2 / x0.

  • And if I then multiply through - so that's what this implies -

  • and if I multiply through by x0^2 I get x = 2x0.

  • OK, so I claim that this point weve just calculated it's 2x0.

  • 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 2y0.

  • All right.

  • That's what that one is.

  • So this one is 2y0.

  • 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 ways.

  • 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 x0 and a y0.

  • 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 y0

  • here, it was this point.

  • And so the formula for y0, which I plugged in right here,

  • was from the equation of the curve. y0 = 1 / x0.

  • 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 sloppines 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 2x0 the height is 2y0, and a half of that.

  • So it's 1/2( 2x0)(2y0) , which is (2x0)(y0), which

  • is, lo and behold, 2.

  • So the amusing thing in this case is that it actually didn't

  • matter what x0 and y0 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' 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 (f) and d/ dx(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, x0.

  • 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 x0 here and x0 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 x0.

  • 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))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((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

  • nx^(n-1) delta x

  • this junk term) - 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( nx^(n-1) delta x

  • this term here).

  • And now I can divide in by delta x.

  • So I get nx^(n-1)

  • 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.

  • And 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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