PROFESSOR: Welcome
back to recitation.
In this video segment,
what I'd like us to do
is work on this
following problem.
Find d/dx of the integral from
0 to x squared, cosine t dt.
I'm going to give you a
moment to think about it
and then I'll come back
and show you how I do it.
OK, welcome back.
Hopefully you were able to
make some headway on this.
Let's look at the problem and
see how we would break it down.
Well we know from the
fundamental theorem of calculus
that you saw in the
lecture, that if up here
instead of x
squared we had an x,
then the problem would
be easy to solve.
We'd just use the fundamental
theorem of calculus, the answer
would be cosine x.
But of course, we don't have
an x, we have an x squared.
That's why I gave
you this problem.
And we need to figure out how
to solve this problem when
there's a different function
up here besides just x.
What we're going to
use is we're going
to combine the fundamental
theorem of calculus
and the chain rule.
So, let's start off with
how we would do this
if it were the integral
from 0 to x, as I mentioned.
So we'll define
capital F of x to be
equal to the integral
from 0 to x cosine t dt.
And then we know that f prime
of x is equal to cosine x.
Now the problem is, we don't
just have this, as I mentioned.
What we actually have--
let me write this down--
we have F of x squared.
Right?
That's what this--
sorry let me highlight
what I mean-- this boxed
thing is F of x squared.
So we took F of x and we
evaluated it at x squared.
That's what we get in the box.
And so if we want to find
d/dx of F of x squared,
it really is just
the chain rule.
We really want to think of this
as a composition of functions.
The first, the outside
function is capital F
and the inside
function is x squared.
So just in general, how do we
think about the chain rule?
Well remember what
we do-- let me
come back here for a
second-- remember what
we do is we take the derivative
of the outside function,
we evaluate it at
the inside function
and then we take the derivative
of the inside function
and we multiply
those two together.
So all I have to
do is figure out,
what is the following thing?
We know d/dx the
quantity F of x squared
should be equal to
F prime evaluated
at x squared times 2x.
That's just what we
said earlier, right?
It's the derivative
of F evaluated
at x squared times the
derivative of x squared.
So now I just have to
figure out what this is.
Well let's go back
to the other side
and see what we wrote
that F prime was.
If we come over here, we see
F prime at x is just cosine x.
So F prime at x squared is going
to be this function evaluated
at x squared.
That's just cosine of x squared.
So we see over here, we just
get cosine x squared times 2x.
And because I'm a
mathematician, I
want to write the 2x in
front before I finish.
Because otherwise
I get confused.
So the answer here is just
2x times cosine x squared.
Now I want to point out
really what we did here.
This is the answer to
this particular problem,
but we can now solve
problems in general,
when I put any function up
here, any function of x up here.
Right?
Ultimately, all I did was I
used the fundamental theorem
of calculus and the chain rule.
So any function I
put up here, I can
do exactly the same process.
I would define F of x to
be this type of thing,
the way we would define it
for the fundamental theorem
of calculus.
I would know what
F prime of x was.
And then I would
have to evaluate F
at a, at this function up
here, whatever I put up there.
So in this case
it was x squared.
I could have made
it natural log x.
I could've made it some big
polynomial or something more
complicated.
Right?
And once I do that, I just
follow this same process.
Now, instead of
the x squared here
I would have that
other function.
So I'd evaluate capital
F at whatever function
that is times the
derivative of that function.
It's exactly the same process.
So I want to point out that this
is a bigger situation than I
had before, or a bigger
situation than just
this little problem.
So, just so you understand that.
OK?
So again, I just want
to say one more time.
Now you know how
to solve problems
where you have any other
function of x up here
and you want to
take the derivative
of this kind of
expression of an integral
with another function
of x up there.
All right, I think
I'll stop there.
