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--
is 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 alrright, 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 is 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
dealt 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 or 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.