Second Fundamental Theorem and Chain Rule - Single Variable Calculus Video Lecture - Engineering Mathematics

welcome back to recitation in this video
segment what I'd like us to do is work
on this following problem find DDX 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 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 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 a 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 DDX 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
in general how do we think about the
chain rule well remember what we do when
we come back here for a second
member 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 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 two X
in front before I finish because
otherwise I get confused so the answer
here is just two x times cosine x
squared now I want to point out really
what we did here this is 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 in 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 add added 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 have
made it
some big polynomial or something more
complicated right and once I do that I
just follow the 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 exact
the same process so I want to point out
I want to point out that this is a this
is a bigger bigger situation than I had
before or a bigger situation than just
this this little problem so just so you
understand that okay
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