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

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MIT OpenCourseWare at ocw.mit.edu. 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=2×0. OK, so I claim that this point

weve just calculated it’s 2×0. 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 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 2×0 the height

is 2y0, and a half of that. So it’s 1/2( 2×0)(2y0) ,

which is (2×0)(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.

This is highschool pensum right? In Norway you learn this when ur 16, just wondering

dr wells 😀

Where can I find problems related only to this lecture?

This lecture was very helpful…It has cleared my all the doubts….Thank You so much!!!

I'm a high school senior and I don't get Algebra 1, so I don't know what I'm doing here.

just remember answer is 2 cause on mothership more logic im sure getting to another planet doesnt involve math better yet test color blind must pass red blue green test they like colors imao

I cant believe that I actualyl understand the class

My boy asked a very good question at the end!

A linguagem matemática é universal. Mesmo com um inglês (linguagem ) medíocre como o meu, consigo entender.

i love watching these videos but sometimes its hard to take the concepts into my classroom since all professors teach differently, for example my professor is teaching limits and has not even mentioned the idea of the secant line approaching the tangent line, explained this way i understand it , but my professor is only talking about velocity and time and uses no graphs , i like graphs tho i can see it and visualize it…

Can you please tell me what kind of chalks do MIT professors use?

The one this professor is using here? Is it possible for me to buy that? (The big white one)

Thanks beforehand, I would appreciate your help…

im trying to learn this in the sixth grade i need something simple that anyone can understand

I wish I can actually get a credit from watching this lecture series. I actually freakishly understand it!

this will be great for designing a pid motion loop control for a robot. thanks.

ow can i download this video?

Who is this professor

watching this at 3:30 AM why god why!

Thomas Calculus 11th ed. Textbook + Solution Manual ; http://turbobit.net/kxh9mj62g3yg.html

Yo professor, Area = (1/2)

(2X0)(2/X0) = 2 , symmetry ?we asians are smart because physics chemistry and Math with geometry were taught basically from seventh grade

I love how MIT courses start to teach from the very basics, the concepts are same as high school but it's crystal clear here in MIT

That wasn't how Newton wrote derivations, but it was Lagrange. Newton used a simple dot above x, while Leibniz used dy/dx, and Lagrange just did it as y' (or f').

Okay, morons who keep commenting "I am only 12 but I understand this" or "MIT students don't know that" need to chill the fuck down. It would be really inappropriate for professors to teach higher calculus right away. It is the 1st lecture and he needs to make an introduction to calculus and it's basis. By the 3rd lecture you won't even know what the fuck is he talking about, and not even to mention the Multivariable calculus. So you kids need to chill, because if it looks so easy to you, then damn, you are all going to be excellent engineers and mathematicians.

Coming from someone who is getting a degree in mathematics, such introductions in Calculus and Algebra are needed, and those 1st lectures would even my hairdresser understand.

not all profs at MIT are as good as Walter lewin

he should have derived (x^n)' = nx^(n-1)

Not trying to be rude but considering the fact that MIT is the best ( or one of the best ) college in the world, the course seems like a cakewalk ! And this is coming from someone who hasn't even started college yet !

I don't know why he said anything you write down, you can differentiate. Can you differentiate abs(x)? How about this function?

https://en.wikipedia.org/wiki/Weierstrass_function

Jesus crise! I'm from Spain and I did this even more complex in class but I'm 16…

Very good Lecture!

Who teaches differentiation before teaching continuity and differentiability ?

Can someone tell me why he claims 2xy/2=2?

Whats the Professor's name?

ISN'T calculs though in high school? how do they do physics without it?

This explanation was so much more complicated than it needed to be.

Its important that the profs begin with the basics. Had he had jumped to let's say multi variable Calculus well, the students would automatically feel denied and feel as if they are stupid.

Is this basically a semester long "Calculus 1" class? The name is vague

I'm fucked for college😨

so like for what age is this class?

@14:15 What has Napoleon got to do with calculus? Plus it should be 18:15 not 14:15

Am I the only Chinese to watching this video?

Are them mit students or high school students? lol

Who the fuck invented this shit??? No clue wtf is this!!!!!! This teacher might be an amazing quick inventing thinker

He's making this sound way more complicated then it really is

American failed education system so typical. he is just saying mathematical definitions not really explaining where they are coming from or what they are. I wanted that I would just buy a book and read the definitions instead of saying thousands of bucks.

He made this sound way more complicated then it really is

A very excellent class.. Thank you.

He made the differentiation example way too complicated and drawn out lol

Oh my god this comment section is so fucking ridiculous. Why dont you guys just appreciate that the best Engineering university of the planet is giving free math lectures and stop hating the MIT students or MIT in its own?

thanks for posting this. it is very exciting! /

I have a question: why is it that m is the lim of delta x -> 0 [in] delta f'/ delta x… why is it f' ?Thanks for an answer!

awesome, its very easy to learn this way

I love mathematics, but haven't studied any calculus. I left school at age 12, and it was never part of my curriculum. This is super helpful, thank you MIT for making this great information available.

("∃x" or "∃(x)") solve for μ when f= μ{x y f(x,y,) =>f(x) does this make sense when {x,y,z,}

F(x)= μ(h) solve for h

G = {(y, z) | ∃x ∈ R s.t. y = g(x)

Seems that h is the log or perhaps cos or tangent but possibly the hypotenuse in relation to some quantum mechanics of expansion???

Where {0} Is an absolute value of the {x,y,z,}/1 where the vexation may exist as a prime factor representing vectrum as a constant changing variation due to continuum.???

Whereas the spherical wave equation?

Enquanto isso no Brasil —- "Que tiro foi esse ? Pá "

People in the comment section should be more encouraging. I graduated college with a degree in music, and I'm trying my best to learn calculus. Some of you are discouraging — this stuff isn't easy! I think it's amazing that MIT is posting their lectures online. Thank you!

hello sir could you please tell me which book is best for calculus or who is best author ??

excellent – nothing else

I am in Calculus I right now in college..my professor, book, and tutors explains "Calculus" very well, in the most simplest way.

The language, choice of symbols, and organization of his lecture is very confusing and all-over-the place!!!! Calculus pertaining to derivatives and limits is truly much simpler and smoother than this whole 51 minutes of bullshit out of the red bull's ass😂

If this was my 1st Calculus professor; I would have dropped this course!💯

Even other Calculus Youtubers explains these same topics BETTER

Can't wait to see the owners of these overly confidence filled comments driving intelligently coherent conversations on the third or further lecture's comment wall, oh wait, I'll have to wait, forever

has anyone solved for y = 2y?

GOT IT!!

Just wondering, is single variable calculus called that because the VARIATION occurs in a single variable at a time? (And not because we assume only a single variable will be involved)

3222 likes only?

Indians can teach better .

Basic problems are not solved

45:13 He saved(Actually wasted) blackboard energy.

45:17 He is running out of blackboards.

45:21 He is conserving chalk.

He is giving us a great example of sustainable development. 😀 LOL

Thank you MIT for giving the world, knowledge for free.

His job is to make things harder LOL

One minute silence for people who watching this without any knowledge of linear algebra

I'm really stressed out and can you guys please encourage me with some nice words

Calculus stands on the shoulders of Algebra, but is much easier…Keep at it, REALLY!

wow such a superficial introduction to Calculus.. Just deriving some formulas

Wow. I thought it would be interesting to review calculus over now that I have a degree in math and now studying electrical engineering. Trust me, this guy is awesome. Things like what he pointed out at 37:00 is priceless, and don't ever expect average intstructors to point stuff like that out.

That's really high-quality chalk that guy is using.

I realized that this course covers calc 1 and 2. So, until which video does the professor finish calculus 1 part?

I understand it but can't understand it, I just get brain dead, might aswell be in another language, makes me so angry, why can't I comprehend it Fkkkkkk

so that's where the chain rule comes from

So by definition the x-axis is a tangent line to y=x^3. But my high school teacher told me that's not the fact lol

It was really helpful to listen to this lecture. I like how he used a triangle formula. It is clever and simple.

Dumb questions around 24:00

This guy just spent 50 minutes on what my horrible high school teacher refused to explain and just jumped right into d/dx x^n = nx^n-1. She said "power to the front, leave base (variable raised to a power) the same, power minus one. Not only did he explain the fundamentals of calculus but he even used the binomial theorem to further explain it.

Why would you start with derivatives without going over what limits are????? That does not seem like a good pedagogical choice. Yeah it’s easy to say what the derivative is and briefly show and tell where the derivative comes from, but where’s the build up? Where’s having students build up intuition and understanding of limits and how they lead to what a derivative is? This is just front loading students information with an expectation of immediate understanding. “Sufficiently straight forward to compute the derivative”? Really??? It’s the first day…

Por aulas assim na UFPA rsrs

I am positively, absolutely, and undoubtedly unable to comprehend what the fuck I just watched. Tangents taste fucking delicious btw y'all.

Great handwritting, bold and clear.

Any Vietnamese here?

Thank u mit for a high quality education material for free

All this was my eleventh grade at school.

Thank you for making these materials available and accessible for free

i came from algebra, trigonometry, linear algebra i think is the right thing to do before going to Calculus

At 9:30, professor explains "…how do I know that this orange line is NOT a tangent line, but the other one is?"

He answers ".. its not really the fact that the thing crosses at two places, because the line could be wiggly, it can cross multiple times", instead, take the point Q and move slowly towards P, and when its close enough, that will be the tangent line. "limit of secant lines PQ as Q -> P, P is fixed"

I am not sure I get this definition of tangents – whats special about the point P, that it makes it a tangent? it seems like a circular definition. I would have understood the definition that it is a line that touches the curve only once, but he explicitly said that, thats not it.

what is the textbook used and who is the teacher'name ?

I dont even know why im watching this rn… like.. I do not have a reason to be here

The only thing I do not like too much about this lecture is the lack of motivation for WHY we need to be able to find a derivative. We need to be able to find a derivative because we want to be able to find out how quickly a function is increasing or decreasing at any given point on its graph, and this rate of increase/decrease is given by the slope of the tangent line to the graph at any given point where a [nonvertical] tangent line exists. WHY do we need to be able to find the rate of increase/decrease of a function at any point on its graph? A couple of simple examples from science or engineering would be informative here.

Será que estudar em Harvard é difícil

When it's year 2019 – and 2007 looks like it's 1999.

The guy overcomplicate a simple things. Probably this is why studying at MIT is so hard. 🙂