Uploaded by MIT on 07.01.2011

Transcript:

PROFESSOR: Welcome back to recitation.

In this video segment we're going to look

at the chain rule.

And specifically we're going to answer a question that I've

placed on the board.

I want you to find two values for theta.

So that the d d theta of the function cosine squared theta

to the fourth equals 0.

At this point I should also point out a few things.

One, theta is a variable in this case.

And if you haven't seen it before, frequently we use it

when we're dealing with trigonometric functions.

Theta often represents the variable that measures angle.

So if you haven't seen theta before, this is

what it looks like.

That's how you say it.

And I'm actually doing the same things we've done before.

I'm taking the derivative with respect to the variable theta,

of a function of theta.

The other thing I want to point out, which I think we

know already, but just to be sure, is it that this squared

here, cosine squared theta to the fourth, means I'm taking

cosine of theta to the fourth and I'm actually squaring it.

So with that knowledge I would like us to use the chain rule

to find two values for theta where this

derivative is equal to 0.

So I'll give you a moment to take a stab at finding the

derivative, setting it equal to 0 and finding some values

for theta that make this equation, the

derivative equal 0.

And we'll come back and then I'll work it out as well.

OK, so the first thing we obviously need to do is be

able to take the derivative of this function on

the left-hand side.

And I should say it's a little more complicated than an

example you saw in the lecture, because in the

lecture you were given an example

with just two functions.

So I want to write it a little differently just to show very

obviously what the three functions are we're composing.

So this function I can rewrite it as cosine theta to the

fourth-- does that look like a 4, there we go--

and then I square that whole thing.

That's really what the function is.

OK?

So we can see what is the outermost function here?

The outermost function is actually the quantity of

something squared.

So the outermost function is x squared.

What's the next function in, in this composition?

The next function in is the cosine function.

And then the last function in is the function taking

something, raising it to the fourth.

So it's very important you understand sort of the

composition, which is the outermost function, which is

the innermost function?

In order to do this chain rule.

Now as you saw in the recitation, you were given the

example, you had y as a function of t-- sorry, not in

recitation--

in the lecture you were given, you were given y as a function

of t or a function of x and then you had to put one other

variable in the middle.

Here we're going to have a

composition of three functions.

So we need to have two other things sort of in the middle.

So let's write this out.

First, the outermost function, we'll call the whole thing y.

So y is equal to x squared is the outermost function.

So then this whole thing is x now.

So then we'll write x is equal to cosine of w.

And then w is equal to theta to the fourth.

OK.

Again this is what you saw before in the lecture.

So you write the outermost function.

And then that function is a function of

cosine of another one.

Cosine of w and w is a function of

theta to the fourth.

So if I put all these back together, I have theta to the

fourth in here, and then I square that, and I come back

to the function I wanted.

Now we know from the lecture what we need to do to find,

essentially, dy d theta.

That's what we're looking for.

So dy d theta, you'll see, you remember from the lecture,

should be dy dx times dx dw times dw d theta.

So it's slightly more complicated than we saw before

because there's one more term.

OK.

So now let's work out what these are.

Well dy dx is fairly straightforward.

dy dx is just 2x.

And then dx dw, well what's the derivative

of the cosine function?

The derivative of cosine is negative sine.

So this is times negative sine of w times--

and what's dw d theta?

So w is the function theta to the fourth.

So dw d theta is 4 theta to the third.

Now when you look at this you should remember, x we sort of

inserted into the problem to make the problem easier, and w

we inserted into the problem to make the problem easier for

us to follow.

So we don't want all these x's and w's.

We want everything in terms of, in terms of theta.

But what is w?

Well w is theta to the fourth.

So I can replace that by theta to the fourth.

And what is x?

x is cosine w.

But w is theta to the fourth.

So x is actually cosine of theta to the fourth.

OK?

So I get 2 cosine of theta to the fourth

times negative sine--

w again is theta to the fourth--

of theta to the fourth times 4 theta to the third.

Let's make this nicer.

I'll bring the coefficient and the theta to the third in

front and this minus sign in front.

I get negative 8 theta to the third cosine of theta to

fourth sine of theta to the fourth.

OK, and then the problem asks to find where the derivative

is equal to 0.

Find two values.

Now why did I ask you to find two values?

Because one is very easy.

If this thing is set equal to 0, one value should stand out

right away for theta that makes this product 0.

And that is theta equals 0.

So theta equals 0 is our easiest answer.

So if you didn't do that, then you wanted the more

challenging stuff, you could have done the

other things also.

So what about cosine theta to the fourth?

If we want a product of three things to be 0, then at least

one of them has to be 0.

So I could have set cosine theta to the

fourth equal to 0.

And for what values of--

an angle I should just say-- for what values of theta to

the fourth is this going to be zero?

Well this means that theta to the fourth is equal to either

maybe pi over 2, or you could add another pi.

OK?

So we could say, well let's just say that's one example.

Theta to the fourth equals pi over 2 would work right?

Because it cosine of pi over 2 is equal to 0.

OK?

And so then we could have said theta is equal to pi

over 2 to the 1/4.

That's another example there.

We could have also done, we could have added pi to this

and gotten another answer.

But I only asked for two.

So I guess that I'm allowed to stop there.