Chain Rule | MIT 18.01SC Single Variable Calculus, Fall 2010

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.