PROFESSOR: Hi.
Welcome back to recitation.
In lecture you've been computing
derivatives of functions
from the limit
definition of derivative.
So today we're going to
do another example of that
and do some graphing, as well.
So I've got a problem
written here on the board.
So we're defining
a function f of x
to be 1 over the quantity
1 plus x squared.
So what I'd like
you to do is graph
the function of
the curve y equals
f of x and to compute
the derivative f
prime of x from the definition.
So why don't you take
a couple of minutes
to do that yourself,
then come back,
and we'll work it out together.
All right.
Welcome back.
So to start off, let's try
graphing this function f of x.
So one thing you
can always do when
you start out
graphing a function,
is to just plot a few points.
And that'll give you
a very rough sense
of where the function is, at
least around those points.
So, for example,
when x is equal to 0
we have f of 0 is 1 over 1.
So that's just 1.
So we've got this
point here, (0, 1).
And when x is equal
to 1, well, x squared
is 1, so the denominator is 2,
so the function value is 1/2.
So, all right I'm not going
to draw this to scale.
I'm going to put
x equals 1 here.
And the function value is 1/2,
so this is the point (1, 1/2).
And, OK.
We could do one more.
When x is equal to 2, this
function-- 2 squared is 4,
so that's 5-- so it's 1/5.
So I don't know.
1/5 is smaller than 1/2, right?
So that's maybe down
here-- so this is something
like the point (2, 1/5).
All right.
But this is a very rough
idea we're getting,
so we can use some more
sophisticated analysis
to get a better idea
of what this graph is
going to look like.
So the first thing we
could notice, for example,
is that this is
an even function.
Right?
If I change the sign of x,
if I replace x by minus x,
well, x squared
and minus x squared
are both equal to x squared.
So if you replace x by
minus x, the function value
doesn't change.
So this is an even function that
has symmetry across the axis
here.
So, you know, for
example, I could just
mirror image these points.
So these points also have to
be on the graph, the points
minus 1, 1/2 and minus 2, 1/5.
And any other part of
the curve that I draw
will be perfectly mirror imaged.
Another thing to observe
is that x squared is always
greater than or equal to 0.
So 1 plus x squared
is always positive.
So 1 over 1 plus x squared
is also always positive.
Also, 1 plus x squared,
it reaches its minimum
when x is equal to 0 and
then as x gets large,
either in the positive direction
or in the negative direction,
this gets larger and larger.
This, just the 1
plus x squared part.
So the denominator is
getting larger and larger,
while the numerator
stays constant.
The whole fraction gets
smaller and smaller.
So as x gets bigger, either
bigger positive or bigger
negative, the function value
will diminish off to 0,
and it has its maximum
value here at 0.
Because that's when 1 plus
x squared has its minimum.
So the function sort of
has its maximum here at 0,
and then it flattens out.
And as x gets larger
and larger and larger,
this goes to infinity, so the
whole fraction goes down to 0.
But it never reaches it, right?
Because we said it's
always positive.
And similarly on the other side.
OK.
So that's the graph
of the curve y
equals 1 over 1 plus x squared.
Roughly speaking.
OK.
So now let's talk about
computing the derivative.
So right now, to
compute a derivative,
all you have is the limit
definition of the derivative.
So when I ask you to
compute the derivative what
you've got to do is write down
what that definition says.
That's the limit of a
difference quotient.
So we have, by definition,
that f prime of x
is equal to-- well,
it's the limit as delta
x goes to 0 of some
difference quotient.
So on the bottom of
the difference quotient
we just have delta
x, and on the top
we have f of x plus
delta x minus f of x.
In our case, we have a
nice formula for f of x.
So this is equal to the limit
as delta x goes to 0 of 1
over the quantity 1 plus
x plus delta x quantity
squared-- oh, I guess I didn't
need that parenthesis there--
minus 1 over 1 plus x
squared, and the whole thing
is over delta x.
So, what would be
really nice, of course,
is if this were a limit
where we could just
plug in the value delta x
equals 0 and evaluate it.
But the way the definition
of a derivative works,
that never works, right?
You're always left
with the numerator.
As delta x goes to
0, that top is always
going to be f of x minus f
of x and it's going to be 0.
And the bottom is
always going to be
delta x going to 0, which is 0.
So you always, when you have
a differentiable function,
you always have a
derivative that's
going to be a limit
of a 0 over 0 form.
So you need to do some sort
of manipulation in order
to, in order to get into a
form you can evaluate it.
What we'd really
like is to manipulate
this numerator somehow and pull
out, say, a factor of delta x.
And then that could
cancel with the delta
x we have in the denominator.
Something, some trick like that.
Some algebraic or
other manipulation
to make this into a form where
we can plug in and evaluate.
So all right, so
right here there's
sort of only one manipulation
that's natural to do,
which is we can add these
two fractions together.
So let's do that, and we
can rewrite this limit.
The limit is delta x goes to 0.
All right, I'm going to
pull this 1 over delta x
out front just to
make everything
look a little bit nicer.
It's 1 over delta x times-- OK.
I want to, you know,
subtract these two fractions.
I want to put them over
a common denominator,
so the denominator
is just going to be
the product of the denominator.
So that's 1 plus x plus
delta x quantity squared,
times 1 plus x squared.
OK.
And so this fraction is 1 plus
x squared over that denominator.
And the second one is 1
plus x plus delta x quantity
squared over that
common denominator.
OK.
So we still haven't
got where we want
to be yet because we still have
this 1 over delta x hanging
out.
So OK, so we have to,
you know, keep going.
And so here, I guess
there's a-- this
is sort of a problem
that forces us
a little bit in one direction.
You, know, there's not much we
can do with the denominator,
but here in the numerator we
can expand this out and start
combining stuff.
So let's do that.
So this is equal to-- all right,
well, the limit hangs out--
the limit as delta
x goes to 0 of 1
over delta x times--
OK, so 1 plus x squared
minus-- all right, so if
you expand out x plus delta
x quantity squared, using
your favorite, either FOIL
or the binomial theorem
or just whatever you like,
however you like to multiply two
binomials-- so we get a minus 1
minus x squared minus 2x times
delta x minus delta x squared.
That's the top.
OK, and we haven't
changed the bottom.
It's still 1 plus x plus delta x
squared times 1 plus x squared.
OK.
Well, so what?
OK, so now some nice stuff
is starting to happen,
which is this 1 and this
minus 1 are going to cancel,
and this x squared and
this minus x squared
are going to cancel.
And then after we
cancel those terms
we see that in the
numerator here,
everything is going to have
a factor of delta x, right?
These four are going to
cancel, and we'll just
be left with these two
terms, both of which
are divisible by delta x.
So that's where
this cancellation
we've been looking for
is going to come from.
So let's keep going.
So we cancel those, they
subtract, give us 0.
This limit is equal to the
limit delta x goes to 0-- OK.
And then we can divide this
delta x from the denominator
in, and what we're
left with upstairs
is minus 2x minus delta
x, the whole thing
over the same
denominator, still.
1 plus x plus delta x squared
times 1 plus x squared.
All right.
Great.
So we've done this manipulation.
We finally found a delta x that
we could cancel with that delta
x we started with
in the denominator.
And now this limit is no longer
this 0 over 0 form, right?
When delta x goes to 0,
the top goes to minus 2x.
And the bottom-- well let's see,
this delta x just goes to 0,
so it's 1 plus x squared
times 1 plus x squared.
So that's not 0 over 0.
We can just plug in to evaluate.
So this is, just works out
to a minus 2x over-- OK,
1 plus x squared times 1 plus
x squared is 1 plus x squared,
quantity squared.
And so this is the derivative
that we were looking for.
This is, just to remind you what
that was, that's d over dx of 1
over the quantity
1 plus x squared.
Now, if you wanted, you
could check this a little bit
by looking at the graph and
looking at this function
and just making sure
that it makes sense.
So for example, this
function, this derivative
has the property that
it's 0 when x is 0.
And that's the only time it's 0.
And if we go back
and look at the graph
that we drew over here, we
see that's also a property
that this graph has, right?
It has this horizontal
tangent line there,
and then it diminishes
off to the right and it,
on the left side it
increases, then it
has that horizontal tangent
line, and then it decreases.
And so if we go back to this
function we see, yes indeed,
when x is negative, this
whole thing is positive.
And then at 0 it's 0, and
then it's negative thereafter.
And similarly, you could
note that this function here
is an odd function.
If you change the
sign of x, that
changes the sign of this
whole expression, and so OK,
and so that makes
perfect sense back here.
The symmetry of this curve
is such that, you know,
if we look at a tangent
line to the left of 0
and the symmetric tangent
line to the right of 0,
they're mirror
images of each other.
So their slopes are exactly
negatives of each other.
So that's a nice way you
can sort of put the two
different pieces of this
problem together in order
to double check your work.
So that's that.
