Lec 1 | MIT 18.01 Single Variable Calculus, Fall 2007

Lec 1 | MIT 18.01 Single Variable Calculus, Fall 2007

August 26, 2019 100 By Stanley Isaacs


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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.