The following is provided under
a Creative Commons License.
Your support will help
MIT OpenCourseWare
continue to offer high quality
educational resources for free.
To make a donation or to
view additional materials
from hundreds of MIT courses,
visit MIT OpenCourseWare
at ocw.mit.edu.
Professor: In today's
lecture I want
to develop several
more formulas that
will allow us to reach our goal
of differentiating everything.
So these are
derivative formulas,
and they come in two flavors.
The first kind is
specific, so some specific
function we're giving
the derivative of.
And that would be, for
example, x^n or (1/x) .
Those are the ones that we
did a couple of lectures ago.
And then there are
general formulas,
and the general
ones don't actually
give you a formula for
a specific function
but tell you something like,
if you take two functions
and add them together,
their derivative
is the sum of the derivatives.
Or if you multiply by a
constant, for example,
so c times u, the
derivative of that
is c times u' where
c is constant.
All right, so these
kinds of formulas
are very useful, both the
specific and the general kind.
For example, we need both
kinds for polynomials.
And more generally, pretty
much any set of formulas
that we give you, will
give you a few functions
to start out with
and then you'll
be able to generate lots more
by these general formulas.
So today, we wanna concentrate
on the trig functions,
and so we'll start out with
some specific formulas.
And they're going
to be the formulas
for the derivative
of the sine function
and the cosine function.
So that's what we'll spend the
first part of the lecture on,
and at the same time I hope to
get you very used to dealing
with trig functions,
although that's something
that you should think
of as a gradual process.
Alright, so in order
to calculate these,
I'm gonna start over here and
just start the calculation.
So here we go.
Let's check what happens
with the sine function.
So, I take sin (x + delta
x), I subtract sin x
and I divide by delta x.
Right, so this is the
difference quotient
and eventually I'm gonna have
to take the limit as delta
x goes to 0.
And there's really
only one thing
we can do with this to
simplify or change it,
and that is to use the sum
formula for the sine function.
So, that's this.
That's sin x cos delta x plus--
Oh, that's not what it is?
OK, so what is it?
sin x sin delta x.
OK, good.
Plus cosine.
No?
Oh, OK.
So which is it?
OK.
Alright, let's take a vote.
Is it sine, sine, or
is it sine, cosine?
Audience: [INAUDIBLE]
Professor: OK, so is this
going to be... cosine.
All right, you better remember
these formulas, alright?
OK, turns out that
it's sine, cosine.
All right.
Cosine, sine.
So here we go, no gotta
do x here, sin (delta x).
Alright, so now there's lots
of places to get confused here,
and you're gonna need to
make sure you get it right.
Alright, so we're gonna put
those in parentheses here.
sin (a + b) is sin a
cos b plus cos a sin b.
All right, now that's
what I did over here,
except the letter x was a,
and the letter b was delta x.
Now that's just the first part.
That's just this part
of the expression.
I still have to remember
the minus sin x.
That comes at the end.
Minus sin x.
And then, I have to remember the
denominator, which is delta x.
OK?
Alright, so now...
The next thing we're
gonna do is we're
gonna try to group the terms.
And the difficulty with all such
arguments is the following one:
any tricky limit is
basically 0 / 0 when
you set delta x equal to 0.
If I set delta x equal to
0, this is sin x - sin x.
So it's a 0 / 0 term.
Here we have
various things which
are 0 and various things
which are non-zero.
We must group the terms so
that a 0 stays over a 0.
Otherwise, we're
gonna have no hope.
If we get some 1 / 0
term, we'll get something
meaningless in the limit.
So I claim that the right
thing to do here is to notice,
and I'll just point
out this one thing.
When delta x goes to 0,
this cosine of 0 is 1.
So it doesn't cancel unless we
throw in this extra sine term
here.
So I'm going to use
this common factor,
and combine those terms.
So this is really
the only thing you're
gonna have to check in this
particular calculation.
So we have the
common factor of sin
x, and that multiplies
something that will cancel,
which is (cos delta
x - 1) / delta x.
That's the first term,
and now what's left,
well there's a cos
x that factors out,
and then the other factor is
(sin delta x) / (delta x).
OK, now does anyone
remember from last time what
this thing goes to?
How many people say 1?
How many people say 0?
All right, it's 0.
That's my favorite
number, alright?
0.
It's the easiest
number to deal with.
So this goes to
0, and that's what
happens as delta x tends to 0.
How about this one?
This one goes to 1, my second
favorite number, almost as
easy to deal with as 0.
And these things are
picked for a reason.
They're the simplest
numbers to deal with.
So altogether, this thing as
delta x goes to 0 goes to what?
I want a single person to
answer, a brave volunteer.
Alright, back there.
Student: Cosine
Professor: Cosine,
because this factor is 0.
It cancels and this factor
has a 1, so it's cosine.
So it's cos x.
So our conclusion over here
- and I'll put it in orange -
is that the derivative of
the sine is the cosine.
OK, now I still wanna label
these very important limit
facts here.
This one we'll call
A, and this one we're
going to call B, because we
haven't checked them yet.
I promised you I would
do that, and I'll
have to do that this time.
So we're relying on
those things being true.
Now I'm gonna do the same
thing with the cosine function,
except in order to do it I'm
gonna have to remember the sum
rule for cosine.
So we're gonna do almost
the same calculation here.
We're gonna see that
that will work out,
but now you have to remember
that cos (a + b) = cos cos,
no it's not cosine^2, because
there are two different
quantities here.
It's cos a cos b - sin a sin b.
All right, so you'll have
to be willing to call those
forth at will right now.
So let's do the cosine now.
So that's cos (x + delta x)
- cos x divided by delta x.
OK, there's the
difference quotient
for the cosine function.
And now I'm gonna do the same
thing I did before except I'm
going to apply the
second rule, that
is the sum rule for cosine.
And that's gonna give me cos x
cos delta x - sin x sin delta
x.
And I have to remember again
to subtract the cosine divided
by this delta x.
And now I'm going to regroup
just the way I did before,
and I get the common factor of
cosine multiplying (cos delta x
- 1) / delta x.
And here I get the sin x
but actually it's -sin x.
And then I have (sin
delta x) / delta x.
All right?
The only difference
is this minus sign
which I stuck inside there.
Well that's not the
only difference,
but it's a crucial difference.
OK, again by A we get that this
is 0 as delta x tends to 0.
And this is 1.
Those are the properties
I called A and B.
And so the result here
as delta x tends to 0
is that we get negative sin x.
That's the factor.
So this guy is negative sin x.
I'll put a little
box around that too.
Alright, now these
formulas take a little bit
of getting used
to, but before I do
that I'm gonna explain to
you the proofs of A and B.
So we'll get ourselves
started by mentioning that.
Maybe before I do
that though, I want
to show you how A and B fit into
the proofs of these theorems.
So, let me just make
some remarks here.
So this is just
a remark but it's
meant to help you to frame
how these proofs worked.
So, first of all,
I want to point out
that if you take
the rate of change
of sin x, no let's
start with cosine
because a little
bit less obvious.
If I take the rate of change
of cos x, so in other words
this derivative at x =
0, then by definition
this is a certain limit
as delta x goes to 0.
So which one is it?
Well I have to evaluate
cosine at 0 + delta
x, but that's just delta x.
And I have to
subtract cosine at 0.
That's the base point,
but that's just 1.
And then I have to
divide by delta x.
And lo and behold you can see
that this is exactly the limit
that we had over there.
This is the one that we know is
0 by what we call property A.
And similarly, if I take the
derivative of sin x at x=0,
then that's going to be the
limit as delta x goes to 0
of sin delta x / delta x.
And that's because I should
be subtracting sine of 0
but sine of 0 is 0.
Right?
So this is going to be 1 by our
property B. And so the remark
that I want to make,
in addition to this,
is something about the
structure of these two proofs.
Which is the derivatives
of sine and cosine at x = 0
give all values of
d/dx sin x, d/dx cos x.
So that's really what this
argument is showing us,
is that we just need one
rate of change at one place
and then we work out
all the rest of them.
So that's really the
substance of this proof.
That of course really then
shows that it boils down
to showing what this rate of
change is in these two cases.
So now there's enough
suspense that we
want to make sure that we know
that those answers are correct.
OK, so let's demonstrate
both of them.
I'll start with B. I need to
figure out property B. Now,
we only have one alternative
as to a type of proof
that we can give of
this kind of result,
and that's because we only
have one way of describing
sine and cosine functions,
that is geometrically.
So we have to give
a geometric proof.
And to write down
a geometric proof
we are going to have
to draw a picture.
And the first step
in the proof, really,
is to replace this
variable delta
x which is going to 0
with another name which
is suggestive of what we're
gonna do which is the letter
theta for an angle.
OK, so let's draw
a picture of what
it is that we're going to do.
Here is the circle.
And here is the origin.
And here's some little
angle, well I'll
draw it a little
larger so it's visible.
Here's theta, alright?
And this is the unit circle.
I won't write that down on here
but that's the unit circle.
And now sin theta is this
vertical distance here.
Maybe, I'll draw it
in a different color
so that we can see it all.
OK so here's this distance.
This distance is sin theta.
OK?
Now almost the
only other thing we
have to write down in this
picture to have it work out
is that we have to recognize
that when theta is the angle,
that's also the arc length
of this piece of the circle
when measured in radians.
So this length here is
also arc length theta.
That little piece in there.
So maybe I'll use a different
color for that to indicate it.
So that's orange and that's
this little chunk there.
So those are the two pieces.
Now in order to persuade
you now that the limit is
what it's supposed
to be, I'm going
to extend the picture
just a little bit.
I'm going to double it, just
for my own linguistic sake
and so that I can
tell you a story.
Alright, so that
you'll remember this.
So I'm going to take
a theta angle below
and I'll have another copy
of sin theta down here.
And now the total
picture is really
like a bow and its
bow string there.
Alright?
So what we have here is
a length of 2 sin theta.
So maybe I'll write it
this way, 2 sin theta.
I just doubled it.
And here I have underneath,
whoops, I got it backwards.
Sorry about that.
Trying to be fancy
with the colored chalk
and I have it reversed here.
So this is not 2 sin theta.
2 sin theta is the vertical.
That's the green.
So let's try that again.
This is 2 sin theta, alright?
And then in the denominator
I have the arc length
which is theta is the first half
and so double it is 2 theta.
Alright?
So if you like, this is
the bow and up here we
have the bow string.
And of course we
can cancel the 2's.
That's equal to
sin theta / theta.
And so now why does this
tend to 1 as theta goes to 0?
Well, it's because as the
angle theta gets very small,
this curved piece looks more
and more like a straight one.
Alright?
And if you get very, very
close here the green segment
and the orange segment
would just merge.
They would be practically
on top of each other.
And they have closer and closer
and closer to the same length.
So that's why this is true.
I guess I'll articulate that
by saying that short curves are
nearly straight.
Alright, so that's the
principle that we're using.
Or short pieces of curves, if
you like, are nearly straight.
So if you like, this
is the principle.
So short pieces of curves.
Alright?
So now I also need to
give you a proof of A.
And that has to do with
this cosine function here.
This is the property
A. So I'm going
to do this by
flipping it around,
because it turns out that this
numerator is a negative number.
If I want to interpret
it as a length,
I'm gonna want a
positive quantity.
So I'm gonna write
down 1 - cos theta here
and then I'm gonna
divide by theta there.
Again I'm gonna make some
kind of interpretation.
Now this time I'm going to draw
the same sort of bow and arrow
arrangement, but maybe I'll
exaggerate it a little bit.
So here's the vertex
of the sector,
but we'll maybe make
it a little longer.
Alright, so here it is, and
here was that middle line
which was the unit-- Whoops.
OK, I think I'm going
to have to tilt it up.
OK, let's try from here.
Alright, well you know
on your pencil and paper
it will look better than
it does on my blackboard.
OK, so here we are.
Here's this shape.
Now this angle is
supposed to be theta
and this angle is another theta.
So here we have a length
which is again theta
and another length which
is theta over here.
That's the same as
in the other picture,
except we've
exaggerated a bit here.
And now we have
this vertical line,
which again I'm gonna draw
in green, the bow string.
But notice that as the vertex
gets farther and farther away,
the curved line gets
closer and closer
to being a vertical line.
That's sort of the flip
side, by expansion,
of the zoom in principle.
The principle that
curves are nearly
straight when you zoom in.
If you zoom out that would
mean sending this vertex way,
way out somewhere.
The curved line, the
piece of the circle,
gets more and more straight.
And now let me show you
where this numerator 1 - cos
theta is on this picture.
So where is it?
Well, this whole distance is 1.
But the distance from
the vertex to the green
is cosine of theta.
Right, because this is
theta, so dropping down
the perpendicular this
distance back to the origin
is cos theta.
So this little
tiny, bitty segment
here is basically the
gap between the curve
and the vertical segment.
So the gap is equal
to 1 - cos theta.
So now you can see that as
this point gets farther away,
if this got sent off
to the Stata Center,
you would hardly be able
to tell the difference.
The bow string would
coincide with the bow
and this little gap
between the bow string
and the bow would
be tending to 0.
And that's the
statement that this
tends to 0 as theta tends to 0.
The scaled version of that.
Yeah, question down here.
Student: Doesn't the denominator
also tend to 0 though?
Professor: Ah, the question is
"doesn't the denominator also
tend to 0?"
And the answer is yes.
In my strange analogy with
zooming in, what I did
was I zoomed out the picture.
So in other words, if you
imagine you're taking this
and you're putting it under
a microscope over here
and you're looking
at something where
theta is getting smaller
and smaller and smaller
and smaller.
Alright?
But now because I want my
picture, I expanded my picture.
So the ratio is the
thing that's preserved.
So if I make it so that
this gap is tiny...
Let me say this one more time.
I'm afraid I've made life
complicated for myself.
If I simply let this
theta tend in to 0,
that would be the same
effect as making this
closer and closer in and then
the vertical would approach.
But I want to keep on
blowing up the picture
so that I can see the
difference between the vertical
and the curve.
So that's very much like if
you are on a video screen
and you zoom in, zoom
in, zoom in, and zoom in.
So the question is what
would that look like?
That has the same effect
as sending this point out
farther and farther in that
direction, to the left.
And so I'm just
trying to visualize it
for you by leaving the
theta at this scale,
but actually the
scale of the picture
is then changing when I do that.
So theta is going
to 0, but I I'm
rescaling so that it's of a
size that we can look at it,
And then imagine
what's happening to it.
OK, does that answer
your question?
Student: My question
then is that seems
to prove that that
limit is equal to 0/0.
Professor: It proves more
than it is equal to 0/0.
It's the ratio of this little
short thing to this longer
thing.
And this is getting much, much
shorter than this total length.
You're absolutely
right that we're
comparing two quantities which
are going to 0, but one of them
is much smaller than the other.
In the other case we
compared two quantities
which were both going to
0 and they both end up
being about equal in length.
Here the previous one
was this green one.
Here it's this
little tiny bit here
and it's way shorter than
the 2 theta distance.
Yeah, another question.
Student: cos theta - 1 over
cos theta is the same as 1-
cos theta over theta?
Professor: cos theta - 1 over...
Student: [INAUDIBLE]
Professor: So here, what I
wrote is (cos delta x - 1)
/ delta x, OK, and I
claimed that it goes to 0.
Here, I wrote minus that, that
is I replaced delta x by theta.
But then I wrote this thing.
So (cos theta - 1) minus
1 is the negative of this.
Alright?
And if I show that
this goes to 0,
it's the same as showing
the other one goes to 0.
Another question?
Student: [INAUDIBLE]
Professor: So the question
is, what about this business
about arc length.
So the word arc length,
that orange shape is an arc.
And we're just talking about
the length of that arc,
and so we're calling
it arc length.
That's what the word
arc length means,
it just means the
length of the arc.
Student: [INAUDIBLE]
Professor: Why is
this length theta?
Ah, OK so this is a
very important point,
and in fact it's the very next
point that I wanted to make.
Namely, notice that
in this calculation
it was very important
that we used length.
And that means that the way
that we're measuring theta,
is in what are known as radians.
Right, so that applies to both B
and A, it's a scale change in A
and doesn't really matter
but in B it's very important.
The only way that
this orange length
is comparable to
this green length,
the vertical is
comparable to the arc,
is if we measure them in terms
of the same notion of length.
If we measure them in
degrees, for example,
it would be completely wrong.
We divide up the angles
into 360 degrees,
and that's the wrong
unit of measure.
The correct measure is the
length along the unit circle,
which is what radians are.
And so this is only
true if we use radians.
So again, a little warning
here, that this is in radians.
Now here x is in radians.
The formulas are just wrong
if you use other units.
Ah yeah?
Student: [INAUDIBLE].
Professor: OK so the
second question is why
is this crazy length here 1.
And the reason is
that the relationship
between this picture up here
and this picture down here,
is that I'm drawing
a different shape.
Namely, what I'm
really imagining here
is a much, much smaller theta.
OK?
And then I'm blowing
that up in scale.
So this scale of this
picture down here
is very different from the
scale of the picture up there.
And if the angle is
very, very, very small
then one has to be
very, very long in order
for me to finish the circle.
So, in other words,
this length is 1
because that's what
I'm insisting on.
So, I'm claiming that that's
how I define this circle,
to be of unit radius.
Another question?
Student: [INAUDIBLE]
the ratio between 1 -
cos theta and theta will
get closer and closer to 1.
I don't understand [INAUDIBLE].
Professor: OK, so
the question is it's
hard to visualize
this fact here.
So let me, let me take you
through a couple of steps,
because I think probably other
people are also having trouble
with this visualization.
The first part of
the visualization I'm
gonna try to demonstrate
on this picture up here.
The first part of
the visualization
is that I should think
of a beak of a bird
closing down, getting
narrower and narrower.
So in other words,
the angle theta
has to be getting smaller
and smaller and smaller.
OK, that's the first step.
So that's the process
that we're talking about.
Now, in order to draw that, once
theta gets incredibly narrow,
in order to depict that I have
to blow the whole picture back
up in order be able to see it.
Otherwise it just
disappears on me.
In fact in the limit theta
= 0, it's meaningless.
It's just a flat line.
That's the whole problem
with these tricky limits.
They're meaningless right
at the zero-zero level.
It's only just a little away
that they're actually useful,
that you get useful geometric
information out of them.
So we're just a little away.
So that's what this picture down
below in part A is meant to be.
It's supposed to be that theta
is open a tiny crack, just
a little bit.
And the smallest I can draw
it on the board for you
to visualize it is using the
whole length of the blackboard
here for that.
So I've opened a little
tiny bit and by the time
we get to the other
end of the blackboard,
of course it's fairly wide.
But this angle theta
is a very small angle.
Alright?
So I'm trying to imagine what
happens as this collapses.
Now, when I imagine
that I have to imagine
a geometric interpretation
of both the numerator
and the denominator
of this quantity here.
And just see what happens.
Now I claimed the numerator is
this little tiny bit over here
and the denominator is actually
half of this whole length here.
But the factor of
2 doesn't matter
when you're seeing whether
something tends to 0 or not.
Alright?
And I claimed that
if you stare at this,
it's clear that
this is much shorter
than that vertical curve there.
And I'm claiming, so this
is what you have to imagine,
is this as it gets smaller
and smaller and smaller still
that has the same effect of
this thing going way, way way,
farther away and this vertical
curve getting closer and closer
and closer to the green.
And so that the gap between
them gets tiny and goes to 0.
Alright?
So not only does it go to
0, that's not enough for us,
but it also goes to 0 faster
than this theta goes to 0.
And I hope the evidence
is pretty strong here
because it's so tiny
already at this stage.
Alright.
We are going to move
forward and you'll
have to ponder these
things some other time.
So I'm gonna give you
an even harder thing
to visualize now so be prepared.
OK, so now, the next
thing that I'd like to do
is to give you a second proof.
Because it really
is important, I
think, to understand this
particular fact more thoroughly
and also to get
a lot of practice
with sines and cosines.
So I'm gonna give
you a geometric proof
of the formula for sine here,
for the derivative of sine.
So here we go.
This is a geometric
proof of this fact.
This is for all theta.
So far we only did
it for theta = 0
and now we're going to
do it for all theta.
So this is a different
proof, but it uses
exactly the same principles.
Right?
So, I want do this by
drawing another picture,
and the picture is
going to describe
y, which is sin
theta, which is if you
like the vertical position
of some circular motion.
So I'm imagining that something
is going around in a circle.
Some particle is going
around in a circle.
And so here's the
circle, here the origin.
This is the unit distance.
And right now it happens to
be at this location P. Maybe
we'll put P a little over here.
And here's the angle theta.
And now we're going to move it.
We're going to vary
theta and we're
interested in the
rate of change of y.
So y is the height
of P but we're
gonna move it to
another location.
We'll move it along
the circle to Q. Right?
So here it is.
Here's the thing.
So how far did we move it?
Well we moved it by
an angle delta theta.
So we started theta,
theta is going
to be fixed in this
argument, and we're
going to move a little
bit delta theta.
And now we're just
gonna try to figure out
how far the thing moved.
Well, in order to
do that we've got
to keep track of the height,
the vertical displacement here.
So we're going to draw
this right angle here, this
is the position R. And
then this distance here
is the change in y.
Alright?
So the picture is
we have something
moving around a unit circle.
A point moving
around a unit circle.
It starts at P, it moves to
Q. It moves from angle theta
to angle theta plus delta theta.
And the issue is how
much does y move?
And the formula
for y is sin theta.
So that's telling us the
rate of change of sin theta.
Alright, well so let's just
try to think a little bit
about what this is.
So, first of all,
I've already said this
and I'm going to repeat it here.
Delta y is PR.
It's going from P and
going straight up to R.
That's how far y moves.
That's the change in y
That's what I said up
in the right hand corner there.
Oops.
I said PR but I wrote PQ.
Alright, that's not a good idea.
Alright.
So delta Y is PR.
And now I want to draw the
diagram again one time.
So here's Q, here's R,
and here's P, and here's
my triangle.
And now what I'd like to
do is draw this curve here
which is a piece of
the arc of the circle.
But really what I
want to keep in mind
is something that I did also
in all these other arguments.
Which is, maybe I
should have called
this orange, that I'm gonna
think of the straight line
between.
So it's the straight line
approximation to the curve
that we're always interested in.
So the straight line
is much simpler,
because then we just
have a triangle here.
And in fact it's
a right triangle.
Right, so we have the geometry
of a right triangle which
is going to now let us do
all of our calculations.
OK, so now the key step
is this same principle
that we already used which is
that short pieces of curves
are nearly straight.
So that means that this piece
of the circular arc here from P
to Q is practically the same
as the straight segment from P
to Q. So, that's this
principle that - well,
let's put it over
here - Is that PQ
is practically the same as the
straight segment from P to Q.
So how are we going to use that?
We want to use
that quantitatively
in the following way.
What we want to notice is
that the distance from P to Q
is approximately delta theta.
Right?
Because the arc length
along that curve,
the length of the
curve is delta theta.
So the length of the green which
is PQ is almost delta theta.
So this is essentially delta
theta, this distance here.
Now the second step, which
is a little trickier,
is that we have to work
out what this angle is.
So our goal, and I'm gonna put
it one step below because I'm
gonna put the geometric
reasoning in between,
is I need to figure out
what the angle QPR is.
If I can figure out
what this angle is,
then I'll be able to figure out
what this vertical distance is
because I'll know the
hypotenuse and I'll
know the angle so I'll be
able to figure out what
the side of the triangle is.
So now let me show you
why that's possible to do.
So in order to do that first of
all I'm gonna trade the boards
and show you where
the line PQ is.
So the line PQ is here.
That's the whole thing.
And the key point about this
line that I need you to realize
is that it's practically
perpendicular,
it's almost perpendicular,
to this ray here.
Alright?
It's not quite because the
distance between P to Q
is non-zero.
So it isn't quite,
but in the limit
it's going to be perpendicular.
Exactly perpendicular.
The tangent line to the circle.
So the key thing
that I'm going to use
is that PQ is almost
perpendicular to OP.
Alright?
The ray from the
origin is basically
perpendicular to
that green line.
And then the second
thing I'm going to use
is something that's obvious
which is that PR is vertical.
OK?
So those are the two pieces of
geometry that I need to see.
And now notice what's happening
upstairs on the picture here
in the upper right.
What I have is the
angle theta is the angle
between the horizontal and OP.
That's angle theta.
If I rotate it by ninety
degree, the horizontal
becomes vertical.
It becomes PR and
the other thing
rotated by 90 degrees
becomes the green line.
So the angle that I'm talking
about I get by taking this guy
and rotating it by 90 degrees.
It's the same angle.
So that means that this angle
here is essentially theta.
That's what this angle is.
Let me repeat that
one more time.
We started out
with an angle that
looks like this, which is
the horizontal-- that's
the origin straight
out horizontally.
That's the thing labeled 1.
That distance there.
That's my right arm
which is down here.
My left arm is pointing up
and it's going from the origin
to the point P. So
here's the horizontal
and the angle between
them is theta.
And now, what I claim is that
if I rotate by 90 degrees up,
like this, without
changing anything -
so that was what I
did - the horizontal
will become a vertical.
That's PR.
That's going up, PR.
And if I rotate OP 90
degrees, that's exactly PQ.
OK?
So let me draw it
on there one time.
Let's do it with
some arrows here.
So I started out with
this and then, we'll
label this as orange,
OK so red to orange,
and then I rotate by
90 degrees and the red
becomes this starting from P
and the orange rotates around 90
degrees and becomes
this thing here.
Alright?
So this angle here is
the same as the other one
which I've just drawn.
Different vertices
for the angles.
OK?
Well I didn't say
that all arguments
were supposed to be easy.
Alright, so I claim
that the conclusion
is that this angle is
approximately theta.
And now we can finish
our calculation,
because we have something with
the hypotenuse being delta
theta and the angle being theta
and so this segment here PR is
approximately the
hypotenuse length
times the cosine of the angle.
And that is exactly
what we wanted.
If we divide, we divide by
delta theta, we get (delta y)
/ (delta theta) is
approximately cos theta.
And that's the same thing as...
So what we want in
the limit is exactly
the delta theta going to 0
of (delta y) / (delta theta)
is equal to cos theta.
So we get an approximation on
a scale that we can visualize
and in the limit the
formula is exact.
OK, so that is a geometric
argument for the same result.
Namely that the derivative
of sine is cosine.
Yeah?
Student: [INAUDIBLE].
Professor: You will have to do
some kind of geometric proofs
sometimes.
When you'll really need
this is probably in 18.02.
So you'll need to make
reasoning like this.
This is, for example, the
way that you actually develop
the theory of arc length.
Dealing with delta x's and
delta y's is a common tool.
Alright, I have one
more thing that I
want to talk about today,
which is some general rules.
We took a little bit more time
than I expected with this.
So what I'm gonna do is
just tell you the rules
and we'll discuss
them in a few days.
So let me tell you
the general rules.
So these were the specific ones
and here are some general ones.
So the first one is
called the product rule.
And what it says is that if
you take the product of two
functions and
differentiate them,
you get the derivative of
one times the other plus
the other times the
derivative of the one.
Now the way that you
should remember this,
and the way that I'll
carry out the proof,
is that you should think of it
is you change one at a time.
And this is a very useful way of
thinking about differentiation
when you have
things which depend
on more than one function.
So this is a general procedure.
The second formula that
I wanted to mention
is called the quotient rule
and that says the following.
That (u / v) prime
has a formula as well.
And the formula is
(u'v - uv' ) / v^2.
So this is our second formula.
Let me just mention, both of
them are extremely valuable
and you'll use
them all the time.
This one of course only
works when v is not 0.
Alright, so because
we're out of time
we're not gonna
prove these today
but we'll prove these next
time and you're definitely
going to be responsible
for these kinds of proofs.
