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The main topic today is
what is
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.
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
. Anything you can think of,
anything you
can write down, we can
differentiate it.
All right, so that
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.
So
that
’
s the problem that
we
’
re addressing here.
All right, so
here
’
s our problem, and now
let me show you the solution.
Let
’
s graph the
function. Here is the graph.
Here
’
s some point
P
. It
’
s above the
point x0.x0
was some fixed place on the
x-axis. So here the tangent line
is.
That
’
s the geometric
problem. I
achieved what I wanted to
do. And the task that we have now is to figure out
how to do it.
So, what did we learn
about what a tangent line is? 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
is
the
equation
for that line, and now there are two pieces of
information that we
’
re going
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). 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. In
Calculus we
have another name for it.
We call it prime of x0. Namely, the derivative of
f. 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 know 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
.
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 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
that
this
orange
line
is
not
a
tangent
line,
but
this
other
line
is
a
tangent line? In fact,
this line crosses at some other point, which is
this point Q. But it
’
s not
really
the fact that the line crosses
at two points, 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
（割线）
, 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 closer enough,
then that
’
s
gonna
be the correct line.
Now,
so the tangent line is equal to the limit of so
called secant line PQ, as Q tends to P
.
And
here we
’
re
thinking of P as being
fixed
（不变）
and Q
as variable
（变化）
.
Then we
’
re gonna
be able to put symbols and
formulas
（符号和公式）
to this computation.
And
we
’
ll be able to work out
formulas in any example. So
let
’
s do that.
First
of
all,
I
’
m
gonna write
out these
points
P
and
Q
again.
And I
’
m
thinking
of
this
line
through them. 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.
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
Δ
x. So
that
’
s delta x, so called.
And we could also call it the change in x.
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
Δ
f, which is
the change in f.
And then,
the slope is just the ratio,
Δ
f/
Δ
x
. This is
the slope of the secant
.
And the way we write that
is
the limit as
Δ
x goes to 0
.
That
’
s going to be our
slope. So this
is
slope of
the tangent line
.
I want to
fine a more usable formula for this. In order to
do that, I
’
m going to write
Δ
f more
explicitly. The change in f, so
remember that the point P is the point (x0,
f(x0)). 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
Δ
x,
then this location is
(x0+
Δ
x). And so the point
above that point has a formula, which is
(x0+
Δ
x,
f(x0+
Δ
x)).
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
Δ
x
goes to o of
the
change in f, well the change in f
is the value of f at the upper point here, which
is (x0+
Δ
x
) and
minus
its
value
at
the
lower
point
P
, which
if
f(x0),
divided
by
Δ
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.
W
e
’
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 that
we
’
re gonna do here? All
we
’
re going
to is
we
’
re going to plug in this
formula here 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 its
algebraically. So first,
we consider
this difference
Δ
f/
Δ
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+
Δ
x).
That
’
s the left
hand term. Minus 1/x0,
that
’
s the right hand term.
And then I have to divide
that
by
Δ
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/
Δ
x
times
……
All
I
’
m going
to do
is put it over a common denominator
. So
the common denominator is
(x0+
Δ
x)*x0. And
so in the numerator for the first
expression I, have x0, and for the second
expression I have
x0+
Δ
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+
Δ
x)x0. And then the
very last step is to take the limit as
Δ
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
2
this
Δ
x to 0, and
I
get -1/
x0
. So
that
’
s the answer. All
right, so in order 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.
Y
ou
see that slope there 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(not zero), 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.
Y
es.
学生提问
. 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
Δ
x tends to 0, is equal to
-1/x0
? 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+
Δ
x)3.
That
’
s
the
situation
that
we
’
ve
got.
And
now
the
question
is
what
happens this number
gets smaller and smaller and smaller, and gets to
be practically 0? W
ell,
2
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*3). And
that
’
s what
I’
m saying in general
with this extra number here.
Other questions? 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
Δ
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.
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 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
there
’
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 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
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
labeling, 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
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