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PROFESSOR: Welcome
back to recitation.
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In this video
segment we're going
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to look at the chain rule.
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And specifically we're
going to answer a question
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that I've placed on the board.
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I want you to find
two values for theta
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so that d d theta of
the function cosine
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squared theta to
the fourth equals 0.
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At this point I should also
point out a few things.
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One, theta is a
variable in this case.
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And if you haven't
seen it before,
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frequently we use it
when we're dealing
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with trigonometric functions.
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Theta often represents the
variable that measures angle.
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So if you haven't
seen theta before,
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this is what it looks like.
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That's how you say it.
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And I'm actually doing the
same things we've done before.
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I'm taking the derivative with
respect to the variable theta,
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of a function of theta.
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The other thing I
want to point out,
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which I think we know
already, but just to be sure,
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is it that this squared
here, cosine squared theta
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to the fourth, means I'm taking
cosine of theta to the fourth
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and I'm actually squaring it.
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So with that
knowledge I would like
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us to use the chain rule to
find two values for theta
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where this derivative
is equal to 0.
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So I'll give you a
moment to take a stab
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at finding the derivative,
setting it equal to 0
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and finding some values for
theta that make this equation,
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the derivative equal 0.
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And we'll come back and then
I'll work it out as well.
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OK, so the first thing
we obviously need to do
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is be able to take the
derivative of this function
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on the left-hand side.
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And I should say it's a
little more complicated
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than an example you saw in the
lecture, because in the lecture
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you were given an example
with just two functions.
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So I want to write it a
little differently just
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to show very obviously what
the three functions are
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we're composing.
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So this function
I can rewrite it
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as cosine theta to the fourth
does that look like a 4,
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there we go and then I
square that whole thing.
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That's really what
the function is.
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OK?
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So we can see what is the
outermost function here?
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The outermost function
is actually the quantity
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of something squared.
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So the outermost
function is x squared.
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What's the next function
in, in this composition?
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The next function in
is the cosine function.
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And then the last function in is
the function taking something,
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raising it to the fourth.
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So it's very important
you understand
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sort of the
composition, which is
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the outermost function, which
is the innermost function?
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In order to do this chain rule.
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Now as you saw in
recitation, you
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were given the example dy-- you
had y as as a function of t--
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sorry, not in recitation.
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In the lecture, you were
given y as a function of t
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or a function of
x and then you had
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to put one other
variable in the middle.
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Here we're going to have a
composition of three functions.
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So we need to have two other
things sort of in the middle.
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So let's write this out.
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First, the outermost function,
we'll call the whole thing y.
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So y is equal to x squared
is the outermost function.
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So then this whole
thing is x now.
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So then we'll write x
is equal to cosine of w.
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And then w is equal to
theta to the fourth.
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OK.
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Again this is what you
saw before in the lecture.
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So you write the
outermost function.
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And then that
function is a function
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of cosine of another
one, cosine of w,
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and w is a function of
theta to the fourth.
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So if I put all
these back together,
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I have theta to the fourth in
here, and then I square that,
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and I come back to
the function I wanted.
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Now we know from the lecture
what we need to do to find,
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essentially, dy/d theta.
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That's what we're looking for.
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So dy/d theta, you'll see,
you remember from the lecture,
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should be dy/dx times
dx/dw times dw/d theta.
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So it's slightly
more complicated
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than we saw before because
there's one more term.
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OK.
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So now let's work
out what these are.
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Well dy/dx is fairly
straightforward.
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dy/dx is just 2x.
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And then dx/dw, well
what's the derivative
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of the cosine function?
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The derivative of
cosine is negative sine.
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So this is times negative
sine of w times--
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and what's dw/d theta?
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So w is the function
theta to the fourth.
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So dw/d theta is 4
theta to the third.
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Now when you look at
this you should remember,
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x we sort of inserted
into the problem
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to make the problem easier, and
w we inserted into the problem
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to make the problem
easier for us to follow.
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So we don't want all
these x's and w's.
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We want everything in terms
of, in terms of theta.
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But what is w?
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Well w is theta to the fourth.
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So I can replace that
by theta to the fourth.
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And what is x?
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x is cosine w.
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But w is theta to the fourth.
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So x is actually cosine
of theta to the fourth.
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OK?
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So I get 2 cosine of theta
to the fourth times negative
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sine-- w again is theta
to the fourth-- of theta
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to the fourth times
4 theta to the third.
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Let's make this nicer.
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I'll bring the
coefficient and the theta
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to the third in front and
this minus sign in front.
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I get negative 8 theta to
the third cosine of theta
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to fourth sine of
theta to the fourth.
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OK, and then the problem
asks to find where
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the derivative is equal to 0.
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Find two values.
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Now why did I ask you
to find two values?
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Because one is very easy.
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If this thing is set
equal to 0, one value
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should stand out
right away for theta
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that makes this product 0.
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And that is theta equals 0.
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So theta equals 0 is
our easiest answer.
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So if you didn't
do that, then you
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wanted the more
challenging stuff,
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you could have done
the other things also.
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So what about cosine
theta to the fourth?
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If we want a product of
three things to be 0,
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then at least one
of them has to be 0.
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So I could have set cosine
theta to the fourth equal to 0.
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And for what values
of-- an angle
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I should just say-- for what
values of theta to the fourth
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is this going to be 0?
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Well this means that
theta to the fourth
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is equal to either maybe pi over
2, or you could add another pi.
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OK?
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So we could say, well let's
just say that's one example.
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Theta to the fourth equals
pi over 2 would work right?
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Because cosine of pi
over 2 is equal to 0.
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OK?
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And so then we could
have said theta is
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equal to pi over 2 to the 1/4.
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That's another example there.
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We could have also done, we
could have added pi to this
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and gotten another answer.
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But I only asked for two.
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So I guess that I'm
allowed to stop there.