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PROFESSOR: Welcome
back to recitation.

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In this video segment,
what I'd like us to do

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is work on this
following problem.

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Find d/dx of the integral from
0 to x squared, cosine t dt.

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I'm going to give you a
moment to think about it

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and then I'll come back
and show you how I do it.

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OK, welcome back.

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Hopefully you were able to
make some headway on this.

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Let's look at the problem and
see how we would break it down.

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Well we know from the
fundamental theorem of calculus

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that you saw in the
lecture, that if up here

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instead of x
squared we had an x,

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then the problem would
be easy to solve.

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We'd just use the fundamental
theorem of calculus, the answer

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would be cosine x.

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But of course, we don't have
an x, we have an x squared.

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That's why I gave
you this problem.

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And we need to figure out how
to solve this problem when

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there's a different function
up here besides just x.

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What we're going to
use is we're going

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to combine the fundamental
theorem of calculus

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and the chain rule.

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So, let's start off with
how we would do this

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if it were the integral
from 0 to x, as I mentioned.

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So we'll define
capital F of x to be

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equal to the integral
from 0 to x cosine t dt.

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And then we know that f prime
of x is equal to cosine x.

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Now the problem is, we don't
just have this, as I mentioned.

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What we actually have--
let me write this down--

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we have F of x squared.

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Right?

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That's what this--
sorry let me highlight

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what I mean-- this boxed
thing is F of x squared.

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So we took F of x and we
evaluated it at x squared.

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That's what we get in the box.

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And so if we want to find
d/dx of F of x squared,

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it really is just
the chain rule.

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We really want to think of this
as a composition of functions.

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The first, the outside
function is capital F

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and the inside
function is x squared.

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So just in general, how do we
think about the chain rule?

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Well remember what
we do-- let me

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come back here for a
second-- remember what

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we do is we take the derivative
of the outside function,

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we evaluate it at
the inside function

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and then we take the derivative
of the inside function

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and we multiply
those two together.

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So all I have to
do is figure out,

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what is the following thing?

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We know d/dx the
quantity F of x squared

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should be equal to
F prime evaluated

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at x squared times 2x.

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That's just what we
said earlier, right?

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It's the derivative
of F evaluated

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at x squared times the
derivative of x squared.

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So now I just have to
figure out what this is.

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Well let's go back
to the other side

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and see what we wrote
that F prime was.

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If we come over here, we see
F prime at x is just cosine x.

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So F prime at x squared is going
to be this function evaluated

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at x squared.

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That's just cosine of x squared.

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So we see over here, we just
get cosine x squared times 2x.

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And because I'm a
mathematician, I

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want to write the 2x in
front before I finish.

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Because otherwise
I get confused.

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So the answer here is just
2x times cosine x squared.

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Now I want to point out
really what we did here.

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This is the answer to
this particular problem,

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but we can now solve
problems in general,

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when I put any function up
here, any function of x up here.

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Right?

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Ultimately, all I did was I
used the fundamental theorem

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of calculus and the chain rule.

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So any function I
put up here, I can

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do exactly the same process.

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I would define F of x to
be this type of thing,

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the way we would define it
for the fundamental theorem

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of calculus.

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I would know what
F prime of x was.

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And then I would
have to evaluate F

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at a, at this function up
here, whatever I put up there.

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So in this case
it was x squared.

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I could have made
it natural log x.

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I could've made it some big
polynomial or something more

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complicated.

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Right?

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And once I do that, I just
follow this same process.

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Now, instead of
the x squared here

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I would have that
other function.

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So I'd evaluate capital
F at whatever function

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that is times the
derivative of that function.

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It's exactly the same process.

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So I want to point out that this
is a bigger situation than I

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had before, or a bigger
situation than just

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this little problem.

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So, just so you understand that.

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OK?

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So again, I just want
to say one more time.

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Now you know how
to solve problems

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where you have any other
function of x up here

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and you want to
take the derivative

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of this kind of
expression of an integral

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with another function
of x up there.

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All right, I think
I'll stop there.