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  • Lec 1 | MIT 18.01 Single Variable Calculus, Fall 2007

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    Watch at: 00:00 / 00:00:20The following content is provided under a CreativeCommons License.Your support will help MIT OpenCourseWare continue tooffer high quality educational resources for free.To make a donation or to view additional materials fromhundreds of MIT courses, visit MIT OpenCourseWareat ocw.mit.edu.Watch at: 00:20 / 00:40Professor: So, again welcome to 18.01.We're getting started today with what we're callingUnit One, a highly imaginative title.And it's differentiation.Watch at: 00:40 / 01:00So, let me first tell you, briefly, what's in store inthe next couple of weeks.The main topic today is what is a derivative.Watch at: 01:00 / 01:20And, we're going to look at this from several differentpoints of view, and the first one is thegeometric interpretation.That's what we'll spend most of today on.Watch at: 01:20 / 01:40And then, we'll also talk about a physical interpretationof what a derivative is.And then there's going to be something else which I guess isWatch at: 01:40 / 02:00maybe the reason why Calculus is so fundamental, and why wealways start with it in most science and engineeringschools, which is the importance of derivatives, ofWatch at: 02:00 / 02:20this, to all measurements.So that means pretty much every place.That means in science, in engineering, in economics,in political science, etc.Watch at: 02:20 / 02:40Polling, lots of commercial applications, justabout everything.Now, that's what we'll be getting started with, and thenthere's another thing that we're gonna do in this unit,which is we're going to explain how to differentiate anything.Watch at: 02:40 / 03:00So, how to differentiate any function you know.Watch at: 03:00 / 03:20And that's kind of a tall order, but let me justgive you an example.If you want to take the derivative - this we'll seetoday is the notation for the derivative of something - ofsome messy function like e ^ x arctanx.We'll work this out by the end of this unit.Watch at: 03:20 / 03:40All 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 geometricWatch at: 03:40 / 04:00interpretation.So let's begin with that.So here we go with the geometric interpretationof derivatives.Watch at: 04:00 / 04:20And, what we're going to do is just ask the geometric problemof finding the tangent line to some graph of someWatch at: 04:20 / 04:40function at some point.Which is to say (x0, y0).So that's the problem that we're addressing here.Watch at: 04:40 / 05:00Alright, so here's our problem, and now let meshow you the solution.So, well, let's graph the function.Here's it's graph.Watch at: 05:00 / 05:20Here'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, thiswas supposed to be an x0.That was some fixed place on the x-axis.Watch at: 05:20 / 05:40And now, in order to perform this mighty feat, I will useanother color of chalk.How about red?OK.So here it is.Watch at: 05:40 / 06:00There'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 interestingquestion, which unfortunately I can't solve for you inWatch at: 06:00 / 06:20this class, which is, how did I do that?That is, how physically did I manage to know what to doto 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 itWatch at: 06:20 / 06:40analytically, to do it in a way that a machine could just aswell as I did in drawing this tangent line.So, what did we learn in high school about whata tangent line is?Watch at: 06:40 / 07:00Well, a tangent line has an equation, and any line througha point has the equation y - y0 is equal to m theslope, times x - x0.So here's the equation for that line, and now there are twoWatch at: 07:00 / 07:20pieces of information that we're going to need to workout 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).Watch at: 07:20 / 07:40That's not a calculus problem, but anyway that's a veryimportant 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.Watch at: 07:40 / 08:00And 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 wehave to discuss now.So just to make that explicit here, I'm going to make aWatch at: 08:00 / 08:20definition, which is that f '(x0) , which is known as thederivative, of f, at x0, is the slope of the tangent line to yWatch at: 08:20 / 08:40= f (x) at the point, let's just call it p.Watch at: 08:40 / 09:00All right?So, that's what it is, but still I haven't made anyprogress in figuring out any better how I drew that line.Watch at: 09:00 / 09:20So I have to say something that's more concrete, becauseI 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, soI certainly am taking for granted that in sort ofnon-calculus part that I know what a line through a point is.Watch at: 09:20 / 09:40So I know this equation.But another possibility might be, this line here, how do Iknow - well, unfortunately, I didn't draw it quite straight,but there it is - how do I know that this orange line is not atangent line, but this other line is a tangent line?Watch at: 09:40 / 10:00Well, it's actually not so obvious, but I'm gonnadescribe it a little bit.It's not really the fact, this thing crosses at some otherplace, which is this point Q.Watch at: 10:00 / 10:20But it's not really the fact that the thing crosses at twoplace, because the line could be wiggly, the curve could bewiggly, 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'llfirst do it in language.Watch at: 10:20 / 10:40And it's the following idea: it's that if you take thisorange line, which is called a secant line, and you think ofthe point Q as getting closer and closer to P, then the slopeWatch at: 10:40 / 11:00of that line will get closer and closer to the slopeof the red line.And if we draw it close enough, then that's gonnabe the correct line.So that's really what I did, sort of in my brain whenI drew that first line.And so that's the way I'm going to articulate it first.Watch at: 11:00 / 11:20Now, so the tangent line is equal to the limit ofso called secant lines PQ, as Q tends to P.Watch at: 11:20 / 11:40And here we're thinking of P as being fixed and Q as variable.All right?Again, this is still the geometric discussion, but nowwe're gonna be able to put symbols and formulasWatch at: 11:40 / 12:00to 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 theseWatch at: 12:00 / 12:20points 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.Watch at: 12:20 / 12:40So this, gradually, we'll do this in two steps.And these steps will introduce us to the basic notations whichare used throughout calculus, including multi-variablecalculus, across the board.So the first notation that's used is you imagine here'sWatch at: 12:40 / 13:00the x-axis underneath, and here's the x0, the locationdirectly below the point P.And we're traveling here a horizontal distance whichis denoted by delta x.So that's delta x, so called.And we could also call it the change in x.Watch at: 13:00 / 13:20So that's one thing we want to measure in order to getthe 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.Watch at: 13:20 / 13:40And 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 thisWatch at: 13:40 / 14:00limit applies not just to the whole line itself, but also inparticular 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.Watch at: 14:00 / 14:20OK.Now, This is still a little general, and I want to work outWatch at: 14:20 / 14:40a more usable form here, a better formula for this.And in order to do that, I'm gonna write delta f, thenumerator more explicitly here.The change in f, so remember that the point P isWatch at: 14:40 / 15:00the 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 particularthe vertical distance here, I'm gonna have to get aformula for Q as well.Watch at: 15:00 / 15:20So if this horizontal distance is delta x, then thislocation is (x0delta x).And so the point above that point has aformula, which is (x0Watch at: 15:20 / 15:40delta 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,Watch at: 15:40 / 16:00which is the following: so this f'(x0) , which is the same asm, is going to be the limit as delta x goes to 0 of the changeWatch at: 16:00 / 16:19in f, well the change in f is the value of f at the upperpoint here, which is (x0delta x), and minus its value at the lower point P, which isf(x0), divided by delta x.Watch at: 16:19 / 16:40All right, so this is the formula.I'm going to put this in a little box, because this is byfar the most important formula today, which we use to derivepretty much everything else.And this is the way that we're going to be able tocompute these numbers.Watch at: 16:40 / 17:00So let's do an example.Watch at: 17:00 / 17:20This 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,Watch at: 17:20 / 17:40and sufficiently straightforward that we cancompute 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 formulaWatch at: 17:40 / 18:00here for that function.That's all we're going to do, and visually what we'reaccomplishing is somehow to take the hyperbola, and take apoint on the hyperbola, and figure out some tangent line.Watch at: 18:00 / 18:20That's what we're accomplishing when we do that.So we're accomplishing this geometrically but we'll bedoing it algebraically.So first, we consider this difference delta f / delta xand write out its formula.So I have to have a place.So I'm gonna make it again above this point x0, whichWatch at: 18:20 / 18:40is the general point.We'll make the general calculation.So the value of f at the top, when we move to the right byf(x), so I just read off from this, read off from here.The formula, the first thing I get here is 1 / x0Watch at: 18:40 / 19:00delta 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.Watch at: 19:00 / 19:20This thing is called a difference quotient.It's pretty complicated, because there's always adifference in the numerator.And in disguise, the denominator is a difference,because it's the difference between the value on theright side and the value on the left side here.Watch at: 19:20 / 19:40OK, 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 onthe next level here.Watch at: 19:40 / 20:00This 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 (x0delta x)x0.And so in the numerator for the first expressions Ihave x0, and for the second expression I have x0Watch at: 20:00 / 20:20delta 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.Watch at: 20:20 / 20:40The first one is that x0 and x0 cancel, so we have this.And then the second step is that these two expressionscancel, the numerator and the denominator.Watch at: 20:40 / 21:00Now we have a cancellation that we can make use of.So we'll write that under here.And this is equals -1 / (x0delta x)x0.And then the very last step is to take the limit as deltaWatch at: 21:00 / 21:20x 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, Ican pass to the limit.And all that happens is I set this delta x toWatch at: 21:20 / 21:400, and I get -1/x0^2.So that's the answer.All right, so in other words what I've shown - let me putit up here- is that f'(x0) = -1/x0^2.Watch at: 21:40 / 22:00Now, let's look at the graph just a little bit to check thisfor plausibility, all right?Watch at: 22:00 / 22:20What'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.Watch at: 22:20 / 22:40And the second thing that I would just like to point out isthat as x goes to infinity, that as we go farther to theright, it gets less and less steep.So as x0 goes to infinity, less and less steep.Watch at: 22:40 / 23:00So that's also consistent here, when x0 is very large, this isa smaller and smaller number in magnitude, althoughit's always negative.It's always sloping down.Watch at: 23:00 / 23:20All 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 againthis limiting process.Watch at: 23:20 / 23:40So the formula here is we have basically two numbers.So in other words, why is it that this expression, whendelta x tends to 0, is equal to -1/x0^2 ?Let me illustrate it by sticking in a number for x0to make it more explicit.All right, so for instance, let me stick in hereWatch at: 23:40 / 24:00for x0 the number 3.Then it's -1 / (3delta x)3.That's the situation that we've got.And now the question is what happens as this number getssmaller and smaller and smaller, and gets tobe practically 0?Watch at: 24:00 / 24:20Well, literally what we can do is just plug in 0there, and you get (30)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 thisextra number here.Watch at: 24:20 / 24:40Other questions?Yes.Student: [INAUDIBLE]Professor: So the question is what happened between thisWatch at: 24:40 / 25:00step 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 tobe the same as what's inWatch at: 25:00 / 25:20the numerator of this other expression.And then, at the same time as doing that, I put thatexpression, which is the difference of two fractions, Iexpressed it with a common denominator.So in the denominator here, you see the product ofthe denominators of the two fractions.And then I just figured out what the numerator hadto be without really...Watch at: 25:20 / 25:40Other questions?OK.So I claim that on the whole, calculus gets a bad rap, thatWatch at: 25:40 / 26:00it'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 calculusmade harder story here.So we have to make things harder, because that's our job.Watch at: 26:00 / 26:20And this is actually what most people do in calculus, and it'sthe reason why calculus has a bad reputation.So the secret is that when people ask problems incalculus, they generally ask them in context.And there are many, many other things going on.Watch at: 26:20 / 26:40And so the little piece of the problem which is calculus isactually fairly routine and has to be isolated andgotten through.But all the rest of it, relies on everything else you learnedin mathematics up to this stage, from grade schoolthrough high school.So that's the complication.So now we're going to do a little bit ofWatch at: 26:40 / 27:00calculus 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.Watch at: 27:00 / 27:20So 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 andWatch at: 27:20 / 27:40the tangent to y = 1/x.OK, so that's a geometry problem.Watch at: 27:40 / 28:00And 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.Watch at: 28:00 / 28:20And here's maybe one of our tangent lines, which iscoming 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 linein it, so we're gonna need to do some calculus toWatch at: 28:20 / 28:40answer 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 thatWatch at: 28:40 / 29:00the curve is y = 1/x.That's perfectly reasonable to do.And also, we're gonna calculate the areas of the triangles, andyou 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 domore than geometry.We're gonna have to do some of this analysis justas we've done before.Watch at: 29:00 / 29:20So I'm gonna pick a point and, consistent with the labelingwe've done before, I'm gonna to call it (x0, y0).So that's almost half the battle, having notations, x andy for the variables, and x0 and y0, for the specific point.Now, once you see that you have these labellings, I hope it'sWatch at: 29:20 / 29:40reasonable to do the following.So first of all, this is the point x0, and overhere 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'sWatch at: 29:40 / 30:00pretty clear that we should find the base, which is that weshould find this location here.And we should find the height, so we need tofind that value there.Let's go ahead and do it.So how are we going to do this?Watch at: 30:00 / 30:20Well, 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 gonnaWatch at: 30:20 / 30:40put 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 ofthe 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,Watch at: 30:40 / 31:00and 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.Watch at: 31:00 / 31:20But now we're not done.We have to finish it.We have to figure out all the rest of these quantities sowe can figure out the area.Watch at: 31:20 / 31:40All right.So how do we do that?Watch at: 31:40 / 32:00Well, 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 wherethis horizontal line meets that diagonal line.Watch at: 32:00 / 32:20And 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.Watch at: 32:20 / 32:39We know that y0 is f(x0) , and f(x) is 1/x , sothis thing is 1/x0.And that's equal to -1/x0^2.And here's x, and here's x0.Watch at: 32:39 / 33:00All right, so in order to find this x value, I have to plug inone equation into the other.So this simplifies a bit.This is -x/x0^2.Watch at: 33:00 / 33:20And this is plus 1/x0 because the x0 andx0^2 cancel somewhat.And so if I put this on the other side, I get x /x0^2 is equal to 2 / x0.Watch at: 33:20 / 33:39And 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.Watch at: 33:39 / 34:00Now, I'm almost done.I need to get the other one.Watch at: 34:00 / 34:20I 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.Watch at: 34:20 / 34:40All 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.Watch at: 34:40 / 35:00All right.That's what that one is.So this one is 2y0.And the reason I know this is the following: so here's thesymmetry 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 tradingWatch at: 35:00 / 35:20the roles of x and y.So the symmetry that I'm using is that any formula I get thatinvolves x's and y's, if I trade all the x's and replacethem 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 IWatch at: 35:20 / 35:40see 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 theWatch at: 35:40 / 36:00equation is y= 1 / x.But that's the same thing as xy = 1, if I multiply through byx, which is the same thing as x = 1/y.So here's where the x and the y get reversed.Watch at: 36:00 / 36:20OK now if you don't trust this explanation, you can also getthe y-intercept by plugging x = 0 into the equation star.Watch at: 36:20 / 36:40OK?We plugged y = 0 in and we got the x value.And you can do the same thing analogously the other way.Watch at: 36:40 / 37:00All 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.Watch at: 37:00 / 37:20What 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 variablecalculus, but we're perfectly happy to deal with fourWatch at: 37:20 / 37:40variables 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 variousrelationships between them.Of course the manipulations we do with them are algebraic, andwhen we're doing the derivatives we just considerWatch at: 37:40 / 38:00what's known as one variable calculus.But really there are millions of variable floatingaround potentially.So that's what makes things complicated, and that'ssomething that you have to get used to.Now there's something else which is more subtle, and thatI think many people who teach the subject or use the subjectaren't aware, because they've already entered into theWatch at: 38:00 / 38:20language and they're so comfortable with it that theydon't even notice this confusion.There's something deliberately sloppy about the way wedeal with these variables.The reason is very simple.There are already four variables here.I don't wanna create six names for variables orWatch at: 38:20 / 38:40eight 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 y0here, it was this point.And so the formula for y0, which I plugged in right here,Watch at: 38:40 / 39:00was 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.Watch at: 39:00 / 39:20That's the wrong thing to do.It's an easy mistake to make if the formulas are all a blur toyou and you're not paying attention to where theyare on the diagram.You see that x-intercept calculation there involvedwhere this horizontal line met this diagonal line, and y = 0Watch at: 39:20 / 39:40represented this line here.So the sloppines is that y means two different things.And we do this constantly because it's way, way morecomplicated not to do it.It's much more convenient for us to allow ourselves theWatch at: 39:40 / 40:00flexibility to change the role that this letter plays inthe middle of a computation.And similarly, later on, if I had done this by this morestraightforward method, for the y-intercept, I wouldhave 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, becauseWatch at: 40:00 / 40:20that 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 betteroff, and as I say this is one of the complexities.Watch at: 40:20 / 40:40All 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 baseWatch at: 40:40 / 41:00times 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), whichis, lo and behold, 2.So the amusing thing in this case is that it actually didn'tWatch at: 41:00 / 41:20matter 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,Watch at: 41:20 / 41:40some serious business.So let me continue.So, first of all, I want to give you a few more notations.Watch at: 41:40 / 42:00And these are just other notations that people useto refer to derivatives.And the first one is the following: we alreadywrote y = f(x).And so when we write delta y, that means the sameWatch at: 42:00 / 42:20thing as delta f.That's a typical notation.And previously we wrote f' for the derivative, sothis is Newton's notation for the derivative.Watch at: 42:20 / 42:40But there are other notations.And one of them is df/dx, and another one is dy/ dx, meaningexactly the same thing.And sometimes we let the function slip down belowso that becomes d / dx (f) and d/ dx(y) .Watch at: 42:40 / 43:00So these are all notations that are used for thederivative, and these were initiated by Leibniz.And these notations are used interchangeably, sometimespractically together.They both turn out to be extremely useful.This one omits - notice that this thing omits- theWatch at: 43:00 / 43:20underlying 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 wherepeople leave out some of the important information, andWatch at: 43:20 / 43:40you 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 ofthe function 1 / x.Watch at: 43:40 / 44:00I 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.Watch at: 44:00 / 44:20n = 1, 2, 3; one of these guys.And now what we're trying to figure out is the derivativewith respect to x of x^n in our new notation, whatWatch at: 44:20 / 44:40this 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 ((xWatch at: 44:40 / 45:00delta x)^n - x^n)/delta x.Now before, let me just stick this in thenI'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 particularWatch at: 45:00 / 45:20calculation, it's a nuisance.I don't have an x floating around, which means somethingdifferent from the x0.And I just don't wanna have to keep on writingall those symbols.It's a waste of blackboard energy.There's a total amount of energy, and I've already filledup so many blackboards that, there's just a limited amount.Watch at: 45:20 / 45:40Plus, 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.Watch at: 45:40 / 46:00All right now, in order to simplify this, in order tounderstand algebraically what's going on, I need to understandwhat the nth power of a sum is.And that's a famous formula.We only need a little tiny bit of it, called thebinomial theorem.So, the binomial theorem which is in your text and explainedWatch at: 46:00 / 46:20in an appendix, says that if you take the sum of two guysand you take them to the nth power, that of course is (xdelta x) multiplied by itself n times.Watch at: 46:20 / 46:40And so the first term is x^n, that's when all ofthe n factors come in.And then, you could have this factor of delta xand all the rest x's.So at least one term of the form (x^(n-1))delta x.Watch at: 46:40 / 47:00And how many times does that happen?Well, it happens when there's a factor from here, from the nextfactor, 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 therest of the terms are junk that we won't have to worry about.Watch at: 47:00 / 47:20So to be more specific, there's a very carefulnotation 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 withWatch at: 47:20 / 47:40(delta x)^2, (delta x)^3 or higher.All right, that's how.Very exciting, higher order terms.Watch at: 47:40 / 48:00OK, so this is the only algebra that we need to do, and nowwe just need to combine it together to get our result.So, now I'm going to just carry out the cancellationsthat we need.Watch at: 48:00 / 48:20So here we go.We have delta f / delta x, which remember was 1 / deltax times this, which is this times, now this is (x^nWatch at: 48:20 / 48:40nx^(n-1) delta xthis junk term) - x^n.So that's what we have so far based on ourprevious calculations.Watch at: 48:40 / 49:00Now, I'm going to do the main cancellation, which is this.All right.So, that's 1/delta x( nx^(n-1) delta xthis term here).Watch at: 49:00 / 49:20And 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 ofdelta 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.Watch at: 49:20 / 49:40That'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 -Watch at: 49:40 / 50:00sorry -n, is equal to nx^(n-1).So now this is gonna be super important to you right on yourproblem set in every possible way, and I want to tell you onething, one way in which it's very important.Watch at: 50:00 / 50:20One 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^35x^10) that's gonna be equal to 3x^2, that's applyingWatch at: 50:20 / 50:40this 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, andwe're gonna make more hay with that next time.Watch at: 50:40 / 51:00Question.Yes.I turned myself off.Yes?Student: [INAUDIBLE]Professor: The question was the binomial theorem only worksWatch at: 51:00 / 51:20when delta x goes to 0.No, the binomial theorem is a general formula which alsospecifies 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 needWatch at: 51:20 / 51:40any more than those first two terms.Yes, because delta x goes to 0.OK, see you next time.