PROFESSOR: Hi.
Well, today's the chain rule.
Very, very useful rule, and it's
kind of neat, natural.
Can I explain what a chain
of functions is?
There is a chain of functions.
And then we want to know the
slope, the derivative.
So how does the chain work?
So there x is the input.
It goes into a function
g of x.
We could call that inside
function y.
So the first step
is y is g of x.
So we get an output
from g, call it y.
That's halfway, because that y
then becomes the input to f.
That completes the chain.
It starts with x, produces
y, which is the inside
function g of x.
And then let me call
it z is f of y.
And what I want to know
is how quickly does
z change as x changes?
That's what the chain
rule asks.
It's the slope of that chain.
Can I maybe just tell
you the chain rule?
And then we'll try it
on some examples.
You'll see how it works.
OK, here it is.
The derivative, the slope of
this chain dz dx, notice I
want the change in the whole
thing when I change the
original input.
Then the formula is that
I take-- it's nice.
You take dz dy times dy dx.
So the derivative that we're
looking for, the slope, the
speed, is a product of two
simpler derivatives that we
probably know.
And when we put the chain
together, we multiply those
derivatives.
But there's one catch
that I'll explain.
I can give you a hint
right away.
dz dy, this first factor,
depends on y.
But we're looking for the change
due to the original
change in x.
When I find dz dy, I'm going
to have to get back to x.
Let me just do an example
with a picture.
You'll see why I
have to do it.
So let the chain be cosine
of-- oh, sine.
Why not?
Sine of 3x.
Let me take sine of 3x.
So that's my sine of 3x.
I would like to know if that's
my function, and I can draw it
and will draw it, what
is the slope?
OK, so what's the
inside function?
What's y here?
Well, it's sitting there
in parentheses.
Often it's in parentheses so
we identify it right away.
y is 3x.
That's the inside function.
And then the outside function
is the sine of y.
So what's the derivative
by the chain rule?
I'm ready to use the chain rule,
because these are such
simple functions, I know their
separate derivative.
So if this whole thing
is z, the chain rule
says that dz dx is--
I'm using this rule.
I first name dz dy, the
derivative of z with respect
to y, which is cosine of y.
And then the second factor is
dy dx, and that's just a
straight line with slope
3, so dy dx is 3.
Good.
Good, but not finished, because
I'm getting an answer
that's still in terms of y, and
I have to get back to x,
and no problem to do it.
I know the link from y to x.
So here's the 3.
I can usually write it out here,
and then I wouldn't need
parentheses.
That's just that 3.
Now the part I'm caring
about: cosine of 3x.
Not cosine x, even though
this was just sine.
But it was sine of y, and
therefore, we need cosine 3x.
Let me draw a picture of this
function, and you'll see
what's going on.
If I draw a picture of--
I'll start with a picture of
sine x, maybe out to 180
degrees pi.
This direction is now x.
And this direction is going to
be-- well, there is the sine
of x, but that's not
my function.
My function is sine of 3x, and
it's worth realizing what's
the difference.
How does the graph change if
I have 3x instead of x?
Well, things come sooner.
Things are speeded up.
Here at x equal pi, 180 degrees,
is when the sine
goes back to 0.
But for 3x, it'll be back to
0 already when x is 60
degrees, pi over 3.
So 1/3 of the way along, right
there, my sine 3x is this one.
It's just like the sine
curve but faster.
That was pi over 3 there,
60 degrees.
So this is my z of x curve, and
you can see that the slope
is steeper at the beginning.
You can see that the slope--
things are happening
three times faster.
Things are compressed by 3.
This sine curve is
compressed by 3.
That makes it speed up so the
slope is 3 at the start, and I
claim that it's 3
cosine of 3x.
Oh, let's draw the slope.
All right, draw the slope.
All right, let me start with
the slope of sine-- so this
was just old sine x.
So its slope is just
cosine x along to--
right?
That's the slope starts at 1.
This is now cosine x.
But that's going out
to pi again.
That's the slope of the original
one, not the slope of
our function, of our chain.
So the slope of our
chain will be--
I mean, it doesn't
go out so far.
It's all between here and
pi over 3, right?
Our function, the one
we're looking at, is
just on this part.
And the slope starts out at 3,
and it's three times bigger,
so it's going to be--
well, I'll just
about get it on there.
It's going to go down.
I don't know if that's great,
but it maybe makes the point
that I started up here at 3, and
I ended down here at minus
3 when x was 60 degrees
because then--
you see, this is a picture
of 3 cosine of 3x.
I had to replace y by
3x at this point.
OK, let me do two or three
more examples,
just so you see it.
Let's take an easy one.
Suppose z is x cubed squared.
All right, here is the
inside function.
y is x cubed, and z is--
do you see what z is?
z is x cubed squared.
So x cubed is the
inside function.
What's the outside function?
It's a function of y.
I'm not going to write--
it's going to be the
squaring function.
That's what we do outside.
I'm not going to write
x squared.
It's y squared.
This is y.
It's y squared that
gets squared.
Then the derivative dz dx by
the chain rule is dz dy.
Shall I remember
the chain rule?
dz dy, dy dx.
Easy to remember because in the
mind of everybody, these
dy's, you see that they're
sort of canceling.
So what's dz dy?
z is y squared, so this
is 2y, that factor.
What's dy dx?
y is x cubed.
We know the derivative
of x cubed.
It's 3x squared.
There is the answer, but it's
not final because I've got a y
here that doesn't belong.
I've got to get it back to.
X So I have all together 2 times
3 is making 6, and that
y, I have to go back and see
what was y in terms of x.
It was x cubed.
So I have x cubed there, and
here's an x squared,
altogether x to the fifth.
Now, is that the right answer?
In this example, we can
certainly check it because we
know what x cubed squared is.
So x cubed is x times x times
x, and I'm squaring that.
I'm multiplying by itself.
There's another x
times x times x.
Altogether I have x to
the sixth power.
Notice I don't add those.
When I'm squaring x cubed,
I multiply the 2 by
the 3 and get 6.
So z is x to the sixth, and of
course, the derivative of x to
the sixth is 6 times x to the
fifth, one power lower.
OK, I want to do two
more examples.
Let me do one more right away
while we're on a roll.
I'll bring down that board and
take this function, just so
you can spot the
inside function
and the outside function.
So my function z is going to be
1 over the square root of 1
minus x squared.
Such things come up pretty often
so we have to know its
derivative.
We could graph it.
That's a perfectly reasonable
function, and
it's a perfect chain.
The first point is to identify
what's the inside function and
what's the outside.
So inside I'm seeing this
1 minus x squared.
That's the quantity that it'll
be much simpler if I just give
that a single name y.
And then what's the
outside function?
What am I doing to this y?
I'm taking its square root,
so I have y to the 1/2.
But that square root is
in the denominator.
I'm dividing, so it's
y to the minus 1/2.
So z is y to the minus 1/2.
OK, those are functions I'm
totally happy with.
The derivative is what?
dz dy, I won't repeat
the chain rule.
You've got that clearly
in mind.
It's right above.
Let's just put in
the answer here.
dz dy, the derivative, that's
y to some power, so I get
minus 1/2 times y
to what power?
I always go one power lower.
Here the power is minus 1/2.
If I go down by one, I'll
have minus 3/2.
And then I have to have
dy dx, which is easy.
dy dx, y is 1 minus x squared.
The derivative of that
is just minus 2x.
And now I have to assemble
these, put them together, and
get rid of the y.
So the minus 2 cancels
the minus 1/2.
That's nice.
I have an x still here, and
I have y to the minus 3/2.
What's that?
I know what y is, 1 minus x
squared, and so it's that to
the minus 3/2.
I could write it that way.
x times 1 minus x squared--
that's the y--
to the power minus 3/2.
Maybe you like it that way.
I'm totally OK with that.
Or maybe you want
to see it as--
this minus exponent
down here as 3/2.
Either way, both good.
OK, so that's one
more practice.
and I've got one more in mind.
But let me return to this board,
the starting board,
just to justify where did this
chain rule come from.
OK, where do derivatives
come from?
Derivative always start with
small finite steps, with delta
rather than d.
So I start here, I make a change
in x, and I want to
know the change in z.
These are small, but not
zero, not darn small.
OK, all right, those are true
quantities, and for those, I'm
perfectly entitled to divide and
multiply by the change in
y because there will
be a change in y.
When I change x, that produces
a change in g of x.
You remember this was the y.
So this factor--
well, first of all, that's
simply a true
statement for fractions.
But it's the right way.
It's the way we want it.
Because now when I show it, and
in words, it says when I
change x a little, that produces
a change in y, and
the change in y produces
a change in z.
And it's the ratio that we're
after, the ratio between the
original change and
the final change.
So I just put the
inside change up
and divide and multiply.
OK, what am I going to do?
What I always do, whatever body
does with derivatives at
an instant, at a point.
Let delta x go to 0.
Now as delta x goes to 0, delta
y will go to 0, delta z
will go to 0, and we get
a lot of zeroes over 0.
That's what calculus is
prepared to live with.
Because it keeps this ratio.
It doesn't separately think
about 0 and then later 0.
It's looking at the ratio
as things happen.
And that ratio does
approach that.
That was the definition
of the derivative.
This ratio approaches that,
and we get the answer.
This ratio approaches the
derivative we're after.
That in a nutshell
is the thinking
behind the chain rule.
OK, I could discuss it further,
but that's the
essence of it.
OK, now I'm ready to do one more
example that isn't just
so made up.
It's an important one.
And it's one I haven't
tackled before.
My function is going to be e to
the minus x squared over 2.
That's my function.
Shall I call it z?
That's my function of x.
So I want you to identify the
inside function and the
outside function in that change,
take the derivative,
and then let's look at the
graph for this one.
The graph of this one is a
familiar important graph.
But it's quite an interesting
function.
OK, so what are you
going to take?
This often happens.
We have e to the something,
e to some function.
So that's our inside
function up there.
Our function y, inside function,
is going to be minus
x squared over 2, that quantity
that's sitting up there.
And then z, the outside
function, is
just e to the y, right?
So two very, very simple
functions have gone into this
chain and produced this
e to the minus x
squared over 2 function.
OK, I'm going to ask you for
the derivative, and you're
going to do it.
No problem.
So dz dx, let's use
the chain rule.
Again, it's sitting
right above.
dz dy, so I'm going to take the
slope, the derivative of
the outside function dz dy,
which is e to the y.
And that has that remarkable
property, which is why we care
about it, why we named it,
why we created it.
The derivative of
that is itself.
And the derivative of minus
x squared over 2 is--
that's a picnic, right?--
is a minus.
x squared, we'll
bring down a 2.
Cancel that 2, it'll
be minus x.
That's the derivative of
minus x squared over 2.
Notice the result is negative.
This function is at
least out where--
if x is positive, the whole
slope is negative, and the
graph is going downwards.
And now what's--
everybody knows this
final step.
I can't leave the answer
like that because
it's got a y in it.
I have to put in what
y is, and it is--
can I write the minus x first?
Because it's easier to write it
in front of this e to the
y, which is e to the minus
x squared over 2.
So that's the derivative
we wanted.
Now I want to think about
that function a bit.
OK, notice that we started
with an e to the minus
something, and we ended with
an e to the minus something
with other factors.
This is typical for
exponentials.
Exponentials, the derivative
stays with that exponent.
We could even take the
derivative of that, and we
would again have some
expression.
Well, let's do it in a minute,
the derivative of that.
OK, I'd like to graph these
functions, the original
function z and the slope
of the z function.
OK, so let's see.
x can have any sign.
x can go for this--
now, I'm graphing this.
OK, so what do I expect?
I can certainly figure out
the point x equals 0.
At x equals 0, I have e
to the 0, which is 1.
So at x equals 0, it's 1.
OK, now at x equals to 1, it
has dropped to something.
And also at x equals minus
1, notice the symmetry.
This function is going to be--
this graph is going to be
symmetric around the y-axis
because I've got x squared.
The right official name
for that is we
have an even function.
It's even when it's same
for x and for minus x.
OK, so what's happening
at x equal 1?
That's e to the minus 1/2.
Whew!
I should have looked
ahead to figure out
what that number is.
Whatever.
It's smaller than 1, certainly,
because it's e to
the minus something.
So let me put it there, and
it'll be here, too.
And now rather than a particular
value, what's your
impression of the whole graph?
The whole graph is--It's
symmetric, so it's going to
start like this, and it's
going to start sinking.
And then it's going to sink.
Let me try to get through
that point.
Look here.
As x gets large, say x is even
just 3 or 4 or 1000, I'm
squaring it, so I'm getting
9 or 16 or 1000000.
And then divide by 2.
No problem.
And then e to the minus is--
I mean, so e to the thousandth
would be off
that board by miles.
e to the minus 1000 is a very
small number and getting
smaller fast. So this is going
to get-- but never touches 0,
so it's going to--
well, let's see.
I want to make it symmetric, and
then I want to somehow I
made it touch because this
darn finite chalk.
I couldn't leave
a little space.
But to your eye it touches.
If we had even fine print, you
couldn't see that distance.
So this is that curve, which was
meant to be symmetric, is
the famous bell-shaped curve.
It's the most important curve
for gamblers, for
mathematicians who work
in probability.
That bell-shaped curve will come
up, and you'll see in a
later lecture a connection
between how calculus enters in
probability, and it enters
for this function.
OK, now what's this slope?
What's the slope of
that function?
Again, symmetric, or maybe
anti-symmetric, because I have
this factor x.
So what's the slope?
The slope starts at 0.
So here's x again.
I'm graphing now the slope,
so this was z.
Now I'm going to graph
the slope of this.
OK, the slope starts out at 0,
as we see from this picture.
Now we can see, as I go forward
here, the slope is
always negative.
The slope is going down.
Here it starts out--
yeah, so the slope is 0 there.
The slope is becoming more
and more negative.
Let's see.
The slope is becoming more and
more negative, maybe up to
some point.
Actually, I believe it's
that point where
the slope is becoming--
then it becomes less negative.
It's always negative.
I think that the slope goes down
to that point x equals 1,
and that's where the slope
is as steep as it gets.
And then the slope comes
up again, but the slope
never gets to 0.
We're always going downhill,
but very slightly.
Oh, well, of course, I expect
to be close to that line
because this e to the minus
x squared over 2
is getting so small.
And then over here, I think
this will be symmetric.
Here the slopes are positive.
Ah!
Look at that!
Here we had an even function,
symmetric across 0.
Here its slope turns
out to be--
and this could not
be an accident.
Its slope turns out to
be an odd function,
anti-symmetric across 0.
Now, it just was.
This is an odd function, because
if I change x, I
change the sign of
that function.
OK, now if you will give me
another moment, I'll ask you
about the second derivative.
Maybe this is the first time
we've done the second
derivative.
What do you think the second
derivative is?
The second derivative is the
derivative of the derivative,
the slope of the slope.
My classical calculus problem
starts with function one,
produces function two,
height to slope.
Now when I take another
derivative, I'm starting with
this function one, and over here
will be a function two.
So this was dz dx, and now here
is going to be the second
derivative.
Second derivative.
And we'll give it a nice
notation, nice symbol.
It's not dz dx, all squared.
That's not what I'm doing.
I'm taking the derivative
of this.
So I'm taking--
well, the derivative
of that, I could--
I'm going to give a whole
return to the second
derivative.
It's a big deal.
I'll just say how I write
it: dz dx squared.
That's the second derivative.
It's the slope of
this function.
And I guess what I want is would
you know how to take the
slope of that function?
Can we just think what
would go into that,
and I'll put it here?
Let me put that function here.
minus x e to the minus
x squared over 2.
Slope of that, derivative
of that.
What do I see there?
I see a product.
I see that times that.
So I'm going to use
the product rule.
But then I also see that in this
factor, in this minus x
squared over 2, I see a chain.
In fact, it's exactly
my original chain.
I know how to deal
with that chain.
So I'm going to use
the product rule
and the chain rule.
And that's the point that once
we have our list of rules,
these are now what we might
call four simple rules.
We know those guys: sum,
difference, product, quotient.
And now we're doing the chain
rule, but we have to be
prepared as here for a product,
and then one of these
factors is a chain.
All right, can we do it?
So the derivative, slope.
Well, slope of slope, because
this was the original slope.
OK, so it's the first factor
times the derivative of the
second factor.
And that's the chain,
but that's the one
we've already done.
So the derivative of that is
what we already computed, and
what was it?
It was that.
So the second factor was minus
x e to the minus x
squared over 2.
So this is--
can I just like remember
this is f dg dx
in the product rule.
And the product, this is--
here is a product
of f times g.
So f times dg dx, and
now I need g times--
this was g, and this is
df dx, or it will be.
What's df dx?
Phooey on this old example.
Gone.
OK, df dx, well, f is minus
x. df dx is just minus 1.
Simple.
All right, put the
pieces together.
We have, as I expected we would,
everything has this
factor e to the minus
x squared over 2.
That's controlling everything,
but the question is what's
it-- so here we have a minus
1; is that right?
And here, we have a
plus x squared.
So I think we have x squared
minus 1 times that.
OK, so we computed a
second derivative.
Ha!
Two things I want to do,
one with this example.
The second derivative
will switch sign.
If I graph the darn thing--
suppose I tried to graph that?
When x is 0, this thing
is negative.
What is that telling me?
So this is the second group.
This is telling me that
the slope is going
downwards at the start.
I see it.
But then at x equal 1, that
second derivative, because of
this x squared minus 1
factor, is up to 0.
It's going to take time with
this second derivative.
That's the slope of the slope.
That's this point here.
Here is the slope.
Now, at that point,
its slope is 0.
And after that point, its
slope is upwards.
We're getting something
like this.
The slope of the slope, and
it'll go evenly upwards, and
then so on.
Ha!
You see that we've got the
derivative code, the slope,
but we've got a little more
thinking to do for the slope
of the slope, the rate of change
of the rate of change.
Then you really have
calculus straight.
And a challenge that I don't
want to try right now would be
what's the chain rule for
the second derivative?
Ha!
I'll leave that as a challenge
for professors who might or
might not be able to do it.
OK, we've introduced the second
derivative here at the
end of a lecture.
The key central idea of the
lecture was the chain rule to
give us that derivative.
Good!
Thank you.
NARRATOR: This has been
a production of MIT
OpenCourseWare and
Gilbert Strang.
Funding for this video was
provided by the Lord
Foundation.
To help OCW continue to provide
free and open access
to MIT courses, please
make a donation at
ocw.mit.edu/donate.
