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PROFESSOR: Hi.

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Welcome back to recitation.

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In lecture you've been computing
derivatives of functions

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from the limit
definition of derivative.

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So today we're going to
do another example of that

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and do some graphing, as well.

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So I've got a problem
written here on the board.

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So we're defining
a function f of x

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to be 1 over the quantity
1 plus x squared.

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So what I'd like
you to do is graph

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the function of
the curve y equals

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f of x and to compute
the derivative f

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prime of x from the definition.

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So why don't you take
a couple of minutes

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to do that yourself,
then come back,

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and we'll work it out together.

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All right.

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Welcome back.

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So to start off, let's try
graphing this function f of x.

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So one thing you
can always do when

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you start out
graphing a function,

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is to just plot a few points.

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And that'll give you
a very rough sense

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of where the function is, at
least around those points.

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So, for example,
when x is equal to 0

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we have f of 0 is 1 over 1.

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So that's just 1.

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So we've got this
point here, (0, 1).

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And when x is equal
to 1, well, x squared

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is 1, so the denominator is 2,
so the function value is 1/2.

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So, all right I'm not going
to draw this to scale.

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I'm going to put
x equals 1 here.

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And the function value is 1/2,
so this is the point (1, 1/2).

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And, OK.

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We could do one more.

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When x is equal to 2, this
function-- 2 squared is 4,

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so that's 5-- so it's 1/5.

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So I don't know.

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1/5 is smaller than 1/2, right?

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So that's maybe down
here-- so this is something

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like the point (2, 1/5).

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All right.

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But this is a very rough
idea we're getting,

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so we can use some more
sophisticated analysis

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to get a better idea
of what this graph is

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going to look like.

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So the first thing we
could notice, for example,

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is that this is
an even function.

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Right?

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If I change the sign of x,
if I replace x by minus x,

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well, x squared
and minus x squared

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are both equal to x squared.

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So if you replace x by
minus x, the function value

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doesn't change.

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So this is an even function that
has symmetry across the axis

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here.

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So, you know, for
example, I could just

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mirror image these points.

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So these points also have to
be on the graph, the points

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minus 1, 1/2 and minus 2, 1/5.

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And any other part of
the curve that I draw

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will be perfectly mirror imaged.

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Another thing to observe
is that x squared is always

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greater than or equal to 0.

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So 1 plus x squared
is always positive.

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So 1 over 1 plus x squared
is also always positive.

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Also, 1 plus x squared,
it reaches its minimum

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when x is equal to 0 and
then as x gets large,

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either in the positive direction
or in the negative direction,

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this gets larger and larger.

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This, just the 1
plus x squared part.

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So the denominator is
getting larger and larger,

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while the numerator
stays constant.

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The whole fraction gets
smaller and smaller.

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So as x gets bigger, either
bigger positive or bigger

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negative, the function value
will diminish off to 0,

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and it has its maximum
value here at 0.

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Because that's when 1 plus
x squared has its minimum.

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So the function sort of
has its maximum here at 0,

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and then it flattens out.

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And as x gets larger
and larger and larger,

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this goes to infinity, so the
whole fraction goes down to 0.

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But it never reaches it, right?

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Because we said it's
always positive.

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And similarly on the other side.

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OK.

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So that's the graph
of the curve y

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equals 1 over 1 plus x squared.

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Roughly speaking.

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OK.

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So now let's talk about
computing the derivative.

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So right now, to
compute a derivative,

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all you have is the limit
definition of the derivative.

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So when I ask you to
compute the derivative what

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you've got to do is write down
what that definition says.

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That's the limit of a
difference quotient.

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So we have, by definition,
that f prime of x

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is equal to-- well,
it's the limit as delta

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x goes to 0 of some
difference quotient.

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So on the bottom of
the difference quotient

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we just have delta
x, and on the top

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we have f of x plus
delta x minus f of x.

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In our case, we have a
nice formula for f of x.

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So this is equal to the limit
as delta x goes to 0 of 1

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over the quantity 1 plus
x plus delta x quantity

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squared-- oh, I guess I didn't
need that parenthesis there--

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minus 1 over 1 plus x
squared, and the whole thing

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is over delta x.

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So, what would be
really nice, of course,

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is if this were a limit
where we could just

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plug in the value delta x
equals 0 and evaluate it.

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But the way the definition
of a derivative works,

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that never works, right?

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You're always left
with the numerator.

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As delta x goes to
0, that top is always

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going to be f of x minus f
of x and it's going to be 0.

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And the bottom is
always going to be

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delta x going to 0, which is 0.

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So you always, when you have
a differentiable function,

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you always have a
derivative that's

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going to be a limit
of a 0 over 0 form.

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So you need to do some sort
of manipulation in order

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to, in order to get into a
form you can evaluate it.

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What we'd really
like is to manipulate

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this numerator somehow and pull
out, say, a factor of delta x.

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And then that could
cancel with the delta

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x we have in the denominator.

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Something, some trick like that.

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Some algebraic or
other manipulation

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to make this into a form where
we can plug in and evaluate.

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So all right, so
right here there's

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sort of only one manipulation
that's natural to do,

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which is we can add these
two fractions together.

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So let's do that, and we
can rewrite this limit.

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The limit is delta x goes to 0.

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All right, I'm going to
pull this 1 over delta x

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out front just to
make everything

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look a little bit nicer.

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It's 1 over delta x times-- OK.

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I want to, you know,
subtract these two fractions.

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I want to put them over
a common denominator,

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so the denominator
is just going to be

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the product of the denominator.

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So that's 1 plus x plus
delta x quantity squared,

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times 1 plus x squared.

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OK.

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And so this fraction is 1 plus
x squared over that denominator.

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And the second one is 1
plus x plus delta x quantity

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squared over that
common denominator.

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OK.

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So we still haven't
got where we want

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to be yet because we still have
this 1 over delta x hanging

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out.

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So OK, so we have to,
you know, keep going.

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And so here, I guess
there's a-- this

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is sort of a problem
that forces us

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a little bit in one direction.

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You, know, there's not much we
can do with the denominator,

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but here in the numerator we
can expand this out and start

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combining stuff.

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So let's do that.

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So this is equal to-- all right,
well, the limit hangs out--

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the limit as delta
x goes to 0 of 1

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over delta x times--
OK, so 1 plus x squared

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minus-- all right, so if
you expand out x plus delta

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x quantity squared, using
your favorite, either FOIL

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or the binomial theorem
or just whatever you like,

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however you like to multiply two
binomials-- so we get a minus 1

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minus x squared minus 2x times
delta x minus delta x squared.

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That's the top.

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OK, and we haven't
changed the bottom.

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It's still 1 plus x plus delta x
squared times 1 plus x squared.

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OK.

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Well, so what?

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OK, so now some nice stuff
is starting to happen,

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which is this 1 and this
minus 1 are going to cancel,

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and this x squared and
this minus x squared

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are going to cancel.

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And then after we
cancel those terms

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we see that in the
numerator here,

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everything is going to have
a factor of delta x, right?

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These four are going to
cancel, and we'll just

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be left with these two
terms, both of which

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are divisible by delta x.

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So that's where
this cancellation

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we've been looking for
is going to come from.

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So let's keep going.

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So we cancel those, they
subtract, give us 0.

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This limit is equal to the
limit delta x goes to 0-- OK.

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And then we can divide this
delta x from the denominator

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in, and what we're
left with upstairs

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is minus 2x minus delta
x, the whole thing

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over the same
denominator, still.

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1 plus x plus delta x squared
times 1 plus x squared.

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All right.

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Great.

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So we've done this manipulation.

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We finally found a delta x that
we could cancel with that delta

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x we started with
in the denominator.

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And now this limit is no longer
this 0 over 0 form, right?

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When delta x goes to 0,
the top goes to minus 2x.

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And the bottom-- well let's see,
this delta x just goes to 0,

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so it's 1 plus x squared
times 1 plus x squared.

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So that's not 0 over 0.

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We can just plug in to evaluate.

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So this is, just works out
to a minus 2x over-- OK,

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1 plus x squared times 1 plus
x squared is 1 plus x squared,

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quantity squared.

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And so this is the derivative
that we were looking for.

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This is, just to remind you what
that was, that's d over dx of 1

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over the quantity
1 plus x squared.

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Now, if you wanted, you
could check this a little bit

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00:11:01,670 --> 00:11:04,240
by looking at the graph and
looking at this function

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and just making sure
that it makes sense.

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00:11:06,630 --> 00:11:11,260
So for example, this
function, this derivative

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has the property that
it's 0 when x is 0.

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And that's the only time it's 0.

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And if we go back
and look at the graph

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that we drew over here, we
see that's also a property

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that this graph has, right?

224
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It has this horizontal
tangent line there,

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and then it diminishes
off to the right and it,

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on the left side it
increases, then it

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has that horizontal tangent
line, and then it decreases.

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And so if we go back to this
function we see, yes indeed,

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when x is negative, this
whole thing is positive.

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And then at 0 it's 0, and
then it's negative thereafter.

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And similarly, you could
note that this function here

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is an odd function.

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00:11:52,310 --> 00:11:54,710
If you change the
sign of x, that

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changes the sign of this
whole expression, and so OK,

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and so that makes
perfect sense back here.

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00:12:01,940 --> 00:12:05,910
The symmetry of this curve
is such that, you know,

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if we look at a tangent
line to the left of 0

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and the symmetric tangent
line to the right of 0,

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they're mirror
images of each other.

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So their slopes are exactly
negatives of each other.

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So that's a nice way you
can sort of put the two

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different pieces of this
problem together in order

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00:12:23,640 --> 00:12:25,230
to double check your work.

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So that's that.

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00:12:26,981 --> 00:12:27,481