WEBVTT
00:00:00.040 --> 00:00:01.940
NARRATOR: The following content
is provided under a
00:00:01.940 --> 00:00:03.690
Creative Commons license.
00:00:03.690 --> 00:00:06.630
Your support will help MIT
OpenCourseWare continue to
00:00:06.630 --> 00:00:09.990
offer high-quality educational
resources for free.
00:00:09.990 --> 00:00:12.830
To make a donation or to view
additional materials from
00:00:12.830 --> 00:00:16.760
hundreds of MIT courses, visit
MIT OpenCourseWare at
00:00:16.760 --> 00:00:18.010
ocw.mit.edu.
00:00:31.730 --> 00:00:32.720
PROFESSOR: Hi.
00:00:32.720 --> 00:00:36.800
Today we start the part of our
course that in the old days in
00:00:36.800 --> 00:00:38.970
the traditional treatment
began the
00:00:38.970 --> 00:00:40.700
engineering form of calculus.
00:00:40.700 --> 00:00:43.690
In other words, what we have
done is we have defined the
00:00:43.690 --> 00:00:48.020
derivative, the instantaneous
change, as being a limit of an
00:00:48.020 --> 00:00:49.280
average rate of change.
00:00:49.280 --> 00:00:52.480
We spent a great deal of
time proving certain
00:00:52.480 --> 00:00:54.200
theorems about limits.
00:00:54.200 --> 00:00:58.550
Now what we're going to do is to
analytically take the limit
00:00:58.550 --> 00:01:02.390
definition of a derivative,
which we have already done
00:01:02.390 --> 00:01:06.460
qualitatively, apply our limit
theorems to of this, and
00:01:06.460 --> 00:01:10.640
develop certain formulas for
computing derivatives much
00:01:10.640 --> 00:01:13.190
more quickly from a
computational point of view
00:01:13.190 --> 00:01:15.840
than having to resort to the
original definition.
00:01:15.840 --> 00:01:18.200
In other words, what we're going
to do is to go from a
00:01:18.200 --> 00:01:21.000
qualitative approach to the
derivative to a more
00:01:21.000 --> 00:01:22.540
quantitative approach.
00:01:22.540 --> 00:01:24.440
And for this reason, I just
call this lecture the
00:01:24.440 --> 00:01:27.600
'Derivatives of Some Simple
Functions' to create sort of a
00:01:27.600 --> 00:01:28.780
mood over here.
00:01:28.780 --> 00:01:30.120
Now, the idea is this.
00:01:30.120 --> 00:01:33.140
Let's go back to our
basic definition.
00:01:33.140 --> 00:01:36.800
The derivative of 'f of x' when
'x' is equal to 'x1', 'f
00:01:36.800 --> 00:01:40.150
prime of x1', on is by
definition the limit as 'delta
00:01:40.150 --> 00:01:44.120
x' approaches 0, ''f of 'x1
plus delta x'' minus 'f of
00:01:44.120 --> 00:01:45.790
x1'' over 'delta x'.
00:01:45.790 --> 00:01:48.960
In other words, the average
rate of change of 'f of x'
00:01:48.960 --> 00:01:53.260
with respect to 'x' as we move
from 'x1' to some new position
00:01:53.260 --> 00:01:56.560
'x1 + delta x', taking
the limit as 'delta
00:01:56.560 --> 00:01:57.920
x' approaches 0.
00:01:57.920 --> 00:02:02.710
Now, just to illustrate this,
let's take, for example, 'f of
00:02:02.710 --> 00:02:05.140
x' equals 'x cubed'.
00:02:05.140 --> 00:02:08.270
In this case, if we now compute
'f of 'x1 plus delta
00:02:08.270 --> 00:02:11.240
x'' minus 'f of x1', recall
that 'f' is the
00:02:11.240 --> 00:02:12.890
function that does what?
00:02:12.890 --> 00:02:16.710
Given any input, the output is
the cube of the input, so 'f
00:02:16.710 --> 00:02:20.990
of 'x1 plus delta x' will be the
cube of ''x1 plus delta x'
00:02:20.990 --> 00:02:22.230
minus 'f of x1''.
00:02:22.230 --> 00:02:24.200
That's minus 'x1 cubed'.
00:02:24.200 --> 00:02:26.470
We now use the binomial
theorem.
00:02:26.470 --> 00:02:29.240
In other words, notice again
how mathematics works.
00:02:29.240 --> 00:02:32.070
You introduce a new definition,
but then all of
00:02:32.070 --> 00:02:35.930
the old prerequisite mathematics
comes back to be
00:02:35.930 --> 00:02:37.440
used as a tool over here.
00:02:37.440 --> 00:02:39.240
We use the binomial theorem.
00:02:39.240 --> 00:02:42.120
The 'x1 cubed' term here
cancels with the
00:02:42.120 --> 00:02:43.850
'x1 cubed' term here.
00:02:43.850 --> 00:02:46.980
And what we're left with is
what? ''3x1 squared' times
00:02:46.980 --> 00:02:50.170
'delta x'' plus '3x1
'delta x' squared''
00:02:50.170 --> 00:02:52.310
plus 'delta x' cubed'.
00:02:52.310 --> 00:02:54.760
And to emphasize the next step
that we're going to make,
00:02:54.760 --> 00:02:56.960
let's factor out a 'delta x'.
00:02:56.960 --> 00:03:00.100
Now, going back to our basic
definition, we must take this
00:03:00.100 --> 00:03:03.830
expression and divide
it by 'delta x'.
00:03:03.830 --> 00:03:06.300
The key idea here is that in the
last step, we're going to
00:03:06.300 --> 00:03:09.050
take the limit as 'delta
x' approaches 0.
00:03:09.050 --> 00:03:11.340
The limit as 'delta x'
approaches 0 means, in
00:03:11.340 --> 00:03:14.590
particular, that 'delta x' is
not allowed to equal 0.
00:03:14.590 --> 00:03:17.820
And as long as 'delta x' is not
0, the 'delta x' in the
00:03:17.820 --> 00:03:21.350
denominator cancels the 'delta
x' in the numerator to leave
00:03:21.350 --> 00:03:26.220
'3x1 squared' plus '3x1 'delta
x'' plus 'delta x squared'.
00:03:26.220 --> 00:03:30.440
And now finally to find what 'f
prime of x1' is, we simply
00:03:30.440 --> 00:03:33.560
take this limit as 'delta
x' approaches 0.
00:03:33.560 --> 00:03:35.710
And notice what you're doing
over here when you let 'delta
00:03:35.710 --> 00:03:36.610
x' approach 0.
00:03:36.610 --> 00:03:38.760
We're going to be using
the limit theorems.
00:03:38.760 --> 00:03:42.250
Namely, the limit of a sum
is the sum of the limits.
00:03:42.250 --> 00:03:44.410
The limit of a product is the
product of the limits.
00:03:44.410 --> 00:03:47.810
We can then take each of these
limits term by term.
00:03:47.810 --> 00:03:51.280
Notice that the first term, not
depending on 'delta x',
00:03:51.280 --> 00:03:53.780
it's limit is just
'3x1 squared'.
00:03:53.780 --> 00:03:55.890
Each of the remaining terms
have a factor of
00:03:55.890 --> 00:03:57.140
'delta x' in them.
00:03:57.140 --> 00:04:00.460
As 'delta x' goes to 0 then,
each of the remaining terms go
00:04:00.460 --> 00:04:04.390
to 0, and we wind up with, by
our basic definition, that 'f
00:04:04.390 --> 00:04:07.630
prime of x1' is '3x1 squared'.
00:04:07.630 --> 00:04:10.190
In particular, since 'x1'
could have been any real
00:04:10.190 --> 00:04:12.870
number here, what we
have shown is what?
00:04:12.870 --> 00:04:20.399
That if 'f of x' is equal to 'x
cubed', then 'f prime of x'
00:04:20.399 --> 00:04:22.800
is '3x squared'.
00:04:22.800 --> 00:04:26.410
And the important point here
is simply to observe that
00:04:26.410 --> 00:04:31.050
we've got this result by
applying the basic definition
00:04:31.050 --> 00:04:34.160
to a specifically
given function.
00:04:34.160 --> 00:04:37.720
To generalize this result, let
'f of x' be 'x' to the 'n',
00:04:37.720 --> 00:04:40.250
where 'n' is any positive
whole number.
00:04:40.250 --> 00:04:45.100
And I want to emphasize that
because you're going to see as
00:04:45.100 --> 00:04:48.800
this proof goes on that I really
use the fact then 'n'
00:04:48.800 --> 00:04:51.640
is a positive whole number,
and I'll emphasize to you
00:04:51.640 --> 00:04:52.650
where I do this.
00:04:52.650 --> 00:04:55.370
I'm going to mimic the same
thing I did in the special
00:04:55.370 --> 00:04:57.950
case in our first example
when we simply chose
00:04:57.950 --> 00:04:59.430
'n' to equal 3.
00:04:59.430 --> 00:05:02.100
Namely, I compute 'f
of 'x1 plus delta
00:05:02.100 --> 00:05:04.060
x'' minus 'f of x1'.
00:05:04.060 --> 00:05:07.460
Since the output is the n-th
power of the input, that's
00:05:07.460 --> 00:05:11.830
just 'x1 plus delta x' to the
'n' minus 'x1' to the 'n'.
00:05:11.830 --> 00:05:16.050
Now, as long as 'n' is a
positive whole number, we can
00:05:16.050 --> 00:05:18.620
expand it by the binomial
theorem.
00:05:18.620 --> 00:05:21.470
The first term will be 'x
sub 1' to the n-th.
00:05:21.470 --> 00:05:24.190
That will cancel this 'x
sub 1' to the n-th.
00:05:24.190 --> 00:05:28.600
The next term will be 'n 'x sub
1'' to the 'n - 1' power
00:05:28.600 --> 00:05:30.260
times 'delta x'.
00:05:30.260 --> 00:05:31.820
And now I do something
that's a little
00:05:31.820 --> 00:05:33.270
bit cheating, I guess.
00:05:33.270 --> 00:05:37.020
What I do next is I observe that
every remaining term in
00:05:37.020 --> 00:05:41.180
the binomial theorem expansion
here has a factor of at least
00:05:41.180 --> 00:05:42.730
'delta x squared' in it.
00:05:42.730 --> 00:05:45.650
So rather than compute all of
these terms individually,
00:05:45.650 --> 00:05:48.120
recognizing that I'm going to
let 'delta x' eventually
00:05:48.120 --> 00:05:51.130
approach 0, I factor out a
'delta x squared' and say,
00:05:51.130 --> 00:05:54.150
lookit, the remaining terms
have the form 'delta x
00:05:54.150 --> 00:05:58.100
squared' times some
finite number.
00:05:58.100 --> 00:06:01.150
I don't know what it is, but
it's some finite number.
00:06:01.150 --> 00:06:02.970
I'll put that-- well, let's just
take a look and see what
00:06:02.970 --> 00:06:03.900
that means.
00:06:03.900 --> 00:06:07.320
In particular now, if I take
this expression and divide
00:06:07.320 --> 00:06:10.690
through by 'delta x', we're
assuming again that 'delta x'
00:06:10.690 --> 00:06:13.180
is not 0 because we're going to
take the limit as 'delta x'
00:06:13.180 --> 00:06:14.250
approaches 0.
00:06:14.250 --> 00:06:16.740
And remember our big harangue
about that.
00:06:16.740 --> 00:06:19.620
When we approach the limit, we
are never allowed to equal it.
00:06:19.620 --> 00:06:21.240
'Delta x' never equals 0.
00:06:21.240 --> 00:06:24.320
Consequently, I can cancel a
'delta x' from the denominator
00:06:24.320 --> 00:06:26.610
with a 'delta x' from
the numerator.
00:06:26.610 --> 00:06:29.900
Canceling a 'delta x' from the
numerator knocks a 'delta x'
00:06:29.900 --> 00:06:32.810
altogether out of this term
and leaves me with just a
00:06:32.810 --> 00:06:35.270
single factor of 'delta
x' in this term.
00:06:40.240 --> 00:06:45.770
What I now do is, referring to
my basic definition, I now
00:06:45.770 --> 00:06:49.240
take the limit as 'delta
x' approaches 0.
00:06:49.240 --> 00:06:53.210
As 'delta x' approaches 0, this
term, not depending on
00:06:53.210 --> 00:06:56.810
'delta x', remains 'nx'
to the 'n - 1'.
00:06:56.810 --> 00:06:58.800
And here's the key step.
00:06:58.800 --> 00:07:01.760
The limit of a product is the
product of the limits.
00:07:01.760 --> 00:07:03.440
This approaches 0.
00:07:03.440 --> 00:07:06.480
This is some number.
00:07:06.480 --> 00:07:09.970
And 0 times some number is 0.
00:07:09.970 --> 00:07:15.920
And therefore, all that's left
here is 'nx1' to the 'n - 1'.
00:07:15.920 --> 00:07:19.310
Again, since this was any 'x',
what we have is what?
00:07:19.310 --> 00:07:22.280
If 'f of x' is 'x to the
n', where 'n' is a
00:07:22.280 --> 00:07:23.390
positive whole number.
00:07:23.390 --> 00:07:25.060
And where do I use that fact?
00:07:25.060 --> 00:07:27.630
I use that fact to use
the binomial theorem.
00:07:27.630 --> 00:07:30.650
Then the derivative is
'nx' to the 'n - 1'.
00:07:30.650 --> 00:07:33.310
In other words, if you want to
memorize a shortcut now that
00:07:33.310 --> 00:07:36.940
we know the answer, observe that
it seems to differentiate
00:07:36.940 --> 00:07:37.940
'x' to the 'n'.
00:07:37.940 --> 00:07:42.610
All you do is bring the exponent
down and replace it
00:07:42.610 --> 00:07:45.250
by one less.
00:07:45.250 --> 00:07:48.030
But for heaven's sakes, don't
look at that as being a proof.
00:07:48.030 --> 00:07:50.790
Rather, we gave the proof,
then observed what the
00:07:50.790 --> 00:07:52.750
shortcut recipe was.
00:07:52.750 --> 00:07:56.600
Don't say isn't it easier just
to bring the exponent down and
00:07:56.600 --> 00:07:59.330
replace it by one less instead
of going through this whole
00:07:59.330 --> 00:08:01.260
long harangue over here?
00:08:01.260 --> 00:08:02.830
No, this is how we prove
that result.
00:08:02.830 --> 00:08:05.270
In other words, what we have
now shown is that to
00:08:05.270 --> 00:08:08.660
differentiate 'x' to the 'n' for
any positive integer 'n',
00:08:08.660 --> 00:08:11.390
the derivative is just
'nx' to the 'n - 1'.
00:08:11.390 --> 00:08:14.040
So far, we do not know if
that result is true
00:08:14.040 --> 00:08:15.200
for any other numbers.
00:08:15.200 --> 00:08:18.150
We'll talk about that a
little bit more later.
00:08:18.150 --> 00:08:20.360
Well, so much for
that example.
00:08:20.360 --> 00:08:21.740
Let's look at another one.
00:08:24.300 --> 00:08:27.460
Let's suppose that we have two
functions 'f' and 'g', which
00:08:27.460 --> 00:08:30.370
are differentiable, say,
at some number 'x1'.
00:08:30.370 --> 00:08:33.070
That means, among other things,
that 'f prime of x1'
00:08:33.070 --> 00:08:35.320
and 'g prime of x1' exist.
00:08:35.320 --> 00:08:38.679
Now define a new function
'h' to be the sum of
00:08:38.679 --> 00:08:40.450
the given two functions.
00:08:40.450 --> 00:08:42.600
By the way, we have to be
very, very careful here.
00:08:42.600 --> 00:08:45.910
In general, 'f' and 'g' could
have different domains.
00:08:45.910 --> 00:08:49.790
Notice that 'x' has to belong to
both the domain of 'f' and
00:08:49.790 --> 00:08:52.320
the domain of 'g' for
this to make sense.
00:08:52.320 --> 00:08:55.650
Consequently, I write down over
here that 'x' belongs to
00:08:55.650 --> 00:08:59.070
the domain of 'f' intersect
domain 'g'.
00:08:59.070 --> 00:09:02.050
In other words, it belongs to
both sets, the domain of 'f'
00:09:02.050 --> 00:09:03.340
and the domain of 'g'.
00:09:03.340 --> 00:09:06.810
Because if 'x' didn't belong to
at least one of these two,
00:09:06.810 --> 00:09:08.540
this expression wouldn't
even make sense.
00:09:08.540 --> 00:09:11.510
In other words, 'x' must be a
permissible input to both the
00:09:11.510 --> 00:09:13.100
'f' and 'g' machines.
00:09:13.100 --> 00:09:16.960
Now, what my claim is, is that
in this particular case, 'h
00:09:16.960 --> 00:09:17.760
prime' exists.
00:09:17.760 --> 00:09:20.730
In other words, 'h' is also
differentiable, and it's
00:09:20.730 --> 00:09:24.190
obtained by adding the sum
of these two derivatives.
00:09:24.190 --> 00:09:26.730
In other words, this is the
result that's commonly known
00:09:26.730 --> 00:09:29.730
as the derivative of a sum is
the sum of the derivatives,
00:09:29.730 --> 00:09:31.530
which, of course, sounds like
it should be right.
00:09:31.530 --> 00:09:34.110
But I'll comment on that
in a moment also.
00:09:34.110 --> 00:09:34.340
what.
00:09:34.340 --> 00:09:38.460
I want to show is simply this,
that 'h prime of x1' is simply
00:09:38.460 --> 00:09:42.930
'f prime of x1' plus 'g
prime of x1', OK?
00:09:42.930 --> 00:09:46.440
And my main purpose is to
highlight the basic
00:09:46.440 --> 00:09:46.990
definition.
00:09:46.990 --> 00:09:49.070
That's what I want to give
you the drill on.
00:09:49.070 --> 00:09:53.670
Namely, how do we compute
'h prime of x1'?
00:09:53.670 --> 00:09:57.720
We must look at ''h of 'x1 plus
delta x'' minus 'h of
00:09:57.720 --> 00:10:01.770
x1'' over 'delta x', then take
the limit as 'delta x'
00:10:01.770 --> 00:10:02.530
approaches 0.
00:10:02.530 --> 00:10:05.580
That's the basic definition
that will be used over and
00:10:05.580 --> 00:10:06.850
over and over.
00:10:06.850 --> 00:10:10.790
At any rate, sparing you all
of the details here, let's
00:10:10.790 --> 00:10:11.980
simply observe what?
00:10:11.980 --> 00:10:15.940
That 'h of 'x1 plus delta x'',
since 'h' is the sum of 'f'
00:10:15.940 --> 00:10:19.410
and 'g', is 'f of 'x1 plus
delta x'' plus 'g of
00:10:19.410 --> 00:10:21.200
'x1 plus delta x''.
00:10:21.200 --> 00:10:25.910
Consequently, 'h of 'x1 plus
delta x'' minus 'h of x1' is
00:10:25.910 --> 00:10:29.970
'f of 'x1 plus delta x'' plus
'g of 'x1 plus delta x''--
00:10:29.970 --> 00:10:30.970
that's this part--
00:10:30.970 --> 00:10:35.350
minus ''f of x1' plus
'g of x1''.
00:10:35.350 --> 00:10:38.740
Now, the idea is I want to
divide this by 'delta x'.
00:10:38.740 --> 00:10:42.420
I also have some knowledge of
the differentiability of 'f'
00:10:42.420 --> 00:10:46.120
and 'g' separately, so I would
like to combine or regroup
00:10:46.120 --> 00:10:49.110
these terms to highlight
that fact for me.
00:10:49.110 --> 00:10:52.960
So what I do is I group these
two terms together, these two
00:10:52.960 --> 00:10:56.140
terms together, and then divide
through by 'delta x'.
00:10:56.140 --> 00:11:00.010
In other words, ''h of 'x1 plus
delta x'' minus 'h of x''
00:11:00.010 --> 00:11:02.040
over 'delta x' is what?
00:11:02.040 --> 00:11:06.030
''f of 'x1 plus delta x'' minus
'f of x1'' over 'delta
00:11:06.030 --> 00:11:10.470
x' plus ''g of 'x1 plus
delta x'' minus 'g of
00:11:10.470 --> 00:11:12.700
x1'' over 'delta x'.
00:11:12.700 --> 00:11:16.550
Now, for my last step, I simply
must take the limit as
00:11:16.550 --> 00:11:18.320
'delta x' goes to 0.
00:11:18.320 --> 00:11:20.910
Well, just look at this
bracketed expression.
00:11:20.910 --> 00:11:23.690
The limit of a sum is the
sum of the limits.
00:11:23.690 --> 00:11:25.930
So to take the limit of this
expression, I take the limit
00:11:25.930 --> 00:11:28.170
of these two terms separately
and add them.
00:11:28.170 --> 00:11:31.790
By definition, the limit of the
bracketed expressions as
00:11:31.790 --> 00:11:34.540
'delta x' goes to 0--
by definition--
00:11:34.540 --> 00:11:36.750
is 'f prime of x1'.
00:11:36.750 --> 00:11:41.190
And as 'delta x' goes to 0,
this term here becomes 'g
00:11:41.190 --> 00:11:42.810
prime of x1'.
00:11:42.810 --> 00:11:45.400
And by the way, for the last
time, let me give you this
00:11:45.400 --> 00:11:46.870
word of caution again.
00:11:46.870 --> 00:11:50.520
Do not make the mistake of
saying, gee, when 'delta x'
00:11:50.520 --> 00:11:53.180
goes to 0, my numerator is 0.
00:11:53.180 --> 00:11:55.190
Therefore, the fraction
must be 0.
00:11:55.190 --> 00:11:55.590
No!
00:11:55.590 --> 00:11:59.150
When 'delta x' goes to 0, sure,
if you replace 'delta x'
00:11:59.150 --> 00:12:03.530
by 0, the numerator is 0, but
so is the denominator.
00:12:03.530 --> 00:12:06.410
In other words, this is our old
story that the derivative
00:12:06.410 --> 00:12:10.980
becomes a study of 0/0 so if
you let 'delta x' equals 0.
00:12:10.980 --> 00:12:12.400
In other words, the whole
point is what?
00:12:12.400 --> 00:12:17.040
That as 'delta x' approaches
0, this by definition is 'f
00:12:17.040 --> 00:12:18.330
prime of x1'.
00:12:18.330 --> 00:12:21.070
This by definition is
'g prime of x1'.
00:12:21.070 --> 00:12:26.430
And therefore, we have proven
as a theorem 'h prime of x1'
00:12:26.430 --> 00:12:30.380
is 'f prime of x1' plus 'g
prime of x1', that the
00:12:30.380 --> 00:12:33.990
derivative of a sum is the
sum of the derivatives.
00:12:33.990 --> 00:12:36.960
Now, the main trouble so far is
that every result that I've
00:12:36.960 --> 00:12:40.215
proven rigorously for you, you
have probably guessed in
00:12:40.215 --> 00:12:41.720
advance that it was
going to happen.
00:12:41.720 --> 00:12:45.040
You say who needs this rigorous
stuff when we could
00:12:45.040 --> 00:12:47.200
have got the same result
by intuition.
00:12:47.200 --> 00:12:49.790
And this is where the rift
between the way the pure
00:12:49.790 --> 00:12:53.120
mathematician often teaches
calculus and the way, say, the
00:12:53.120 --> 00:12:56.510
applied man teaches calculus
often comes in.
00:12:56.510 --> 00:12:59.730
The point is that the places
that you can get into trouble
00:12:59.730 --> 00:13:02.110
often aren't stressed
soon enough.
00:13:02.110 --> 00:13:05.370
So before we go any further, let
me show you that you must
00:13:05.370 --> 00:13:08.780
be careful, that up until now,
we have been very fortunate
00:13:08.780 --> 00:13:11.110
when we say things like the
limit of a product is the
00:13:11.110 --> 00:13:13.520
product of the limits, fortunate
in the sense that
00:13:13.520 --> 00:13:16.750
they worked out the way the
wording seemed to indicate.
00:13:16.750 --> 00:13:21.070
Let me give you an example of
what I mean by saying beware
00:13:21.070 --> 00:13:22.610
of self-evident.
00:13:22.610 --> 00:13:26.900
I claim, for example, that the
derivative of a product is not
00:13:26.900 --> 00:13:29.160
the product of the derivatives
necessarily.
00:13:29.160 --> 00:13:32.530
And the best way to prove that
to you is not to say take my
00:13:32.530 --> 00:13:33.350
word for it.
00:13:33.350 --> 00:13:36.600
This is one of the beauties of
computational mathematics.
00:13:36.600 --> 00:13:40.520
You can always show by means of
an example what's going on.
00:13:40.520 --> 00:13:44.320
For example, let 'f of
x' be 'x + 1', let 'g
00:13:44.320 --> 00:13:46.460
of x' be 'x - 1'.
00:13:46.460 --> 00:13:49.380
Now, the derivative of
'x + 1' is simply 1.
00:13:49.380 --> 00:13:51.750
The derivative of 'x
- 1' is also 1.
00:13:51.750 --> 00:13:54.230
Namely, we differentiate
this as a sum.
00:13:54.230 --> 00:13:55.340
The derivative of 'x' is 1.
00:13:55.340 --> 00:13:57.870
The derivative of
a constant is 0.
00:13:57.870 --> 00:14:02.000
In other words, 'f prime of x'
is 1, 'g prime of x' is 1.
00:14:02.000 --> 00:14:05.710
On the other hand, if we first
multiply 'f' and 'g', 'x + 1'
00:14:05.710 --> 00:14:09.370
times 'x - 1' is 'x squared
- 1', and the
00:14:09.370 --> 00:14:10.740
derivative of that is what?
00:14:10.740 --> 00:14:13.420
Well, to differentiate 'x
squared', we just saw.
00:14:13.420 --> 00:14:14.850
Bring the exponent down.
00:14:14.850 --> 00:14:16.180
Replace it by one less.
00:14:16.180 --> 00:14:17.100
That's '2x'.
00:14:17.100 --> 00:14:19.170
The derivative of
minus 1 is 0.
00:14:19.170 --> 00:14:21.870
In other words, if you multiply
first and then take
00:14:21.870 --> 00:14:23.670
the derivative, you get '2x'.
00:14:23.670 --> 00:14:26.330
On the other hand, if you
differentiate first and then
00:14:26.330 --> 00:14:28.390
take the product, you get 2.
00:14:28.390 --> 00:14:31.710
Now, it is not true that '2x'
is a synonym for 1 for all
00:14:31.710 --> 00:14:32.790
values of 'x'.
00:14:32.790 --> 00:14:38.400
In other words, if you
differentiate first and then
00:14:38.400 --> 00:14:41.620
take the product, you get a
different answer than if you
00:14:41.620 --> 00:14:44.030
multiply first and then
differentiate.
00:14:44.030 --> 00:14:46.480
In other words, self-evident
or not, the thing
00:14:46.480 --> 00:14:48.180
happens to be false.
00:14:48.180 --> 00:14:51.870
In fact, let's see how the
theory helps us in this case.
00:14:51.870 --> 00:14:54.700
See, what I've shown you
here is that this
00:14:54.700 --> 00:14:56.710
result isn't true.
00:14:56.710 --> 00:14:58.510
Oh, I should say that this
result is true, that the
00:14:58.510 --> 00:14:59.880
equality isn't true.
00:14:59.880 --> 00:15:02.760
What I haven't shown you
is what is true.
00:15:02.760 --> 00:15:05.700
And again, to emphasize
our basic definition,
00:15:05.700 --> 00:15:06.870
let's do that now.
00:15:06.870 --> 00:15:09.970
Let's show how we can use our
basic definition to find the
00:15:09.970 --> 00:15:11.060
correct result.
00:15:11.060 --> 00:15:15.050
What we do is we let 'h of x'
equal 'f of x' times 'g of x'.
00:15:15.050 --> 00:15:18.480
And we're going to use the
same structure as before.
00:15:18.480 --> 00:15:22.720
We are going to compute 'h of
'x1 plus delta x'', subtract
00:15:22.720 --> 00:15:26.840
'h of x1' divide by 'delta x',
and take the limit as 'delta
00:15:26.840 --> 00:15:27.800
x' approaches 0.
00:15:27.800 --> 00:15:29.560
We do that every time.
00:15:29.560 --> 00:15:32.210
All that changes is the property
of the particular
00:15:32.210 --> 00:15:33.760
function that we're
dealing with.
00:15:33.760 --> 00:15:37.860
In any event, if I do this, 'h
of 'x1 plus delta x'' is
00:15:37.860 --> 00:15:42.760
obtained by replacing 'x' in
here by 'x1 plus delta x'.
00:15:42.760 --> 00:15:46.000
And 'h of x1' is obtained by
replacing 'x' by 'x1' over
00:15:46.000 --> 00:15:50.140
here, so I wind up with 'f of
'x1 plus delta x'' times 'g of
00:15:50.140 --> 00:15:55.430
'x1 plus delta x'' minus 'f
of x1' times 'g of x1'.
00:15:55.430 --> 00:15:59.230
And now the problem is I would
like somehow to be able to get
00:15:59.230 --> 00:16:03.880
terms involving 'f of 'x1 plus
delta x'' minus 'f of x1', 'g
00:16:03.880 --> 00:16:07.810
of 'x1 plus delta x'' minus 'g
of x1', and I use the same
00:16:07.810 --> 00:16:11.550
trick here that I used when I
proved that the limit of a
00:16:11.550 --> 00:16:13.480
product is the product
of the limits.
00:16:13.480 --> 00:16:18.050
Namely, I am going to add in
and subtract the same term,
00:16:18.050 --> 00:16:19.770
and this will help me factor.
00:16:19.770 --> 00:16:22.020
In other words, what I'm going
to do here is this.
00:16:22.020 --> 00:16:23.590
I write down this term.
00:16:23.590 --> 00:16:27.370
Now I say to myself I would like
an 'f of x1' term to go
00:16:27.370 --> 00:16:29.610
with the 'f of 'x1
plus delta x''.
00:16:29.610 --> 00:16:32.280
So what I do is I write
minus 'f of x1'.
00:16:32.280 --> 00:16:35.400
Then I observe that 'g of 'x1
plus delta x'' is a factor
00:16:35.400 --> 00:16:38.010
here, so I put that
in over here.
00:16:38.010 --> 00:16:39.830
See, 'g of 'x1 plus delta x''.
00:16:39.830 --> 00:16:42.720
See, in other words, I can now
factor out a 'g of 'x1 plus
00:16:42.720 --> 00:16:46.130
delta x'' from these two terms
and have the formula that I
00:16:46.130 --> 00:16:47.940
want left over.
00:16:47.940 --> 00:16:50.250
But, of course, this term
doesn't belong here.
00:16:50.250 --> 00:16:53.700
I just added it myself, or I
should say subtracted it, so
00:16:53.700 --> 00:16:56.790
now I put it in with the
opposite sign, namely, plus 'f
00:16:56.790 --> 00:16:59.260
of x1' 'g of 'x1
plus delta x''.
00:16:59.260 --> 00:17:01.840
And now put this term back
in here: minus 'f of
00:17:01.840 --> 00:17:03.700
x1' times 'g of x1'.
00:17:03.700 --> 00:17:06.420
In other words, all I've done
is rewritten this, but by
00:17:06.420 --> 00:17:10.420
adding and subtracting the same
term, which does what?
00:17:10.420 --> 00:17:14.700
Dividing through by 'delta x',
I factor out a 'g of 'x1 plus
00:17:14.700 --> 00:17:17.440
delta x' from these two terms.
00:17:17.440 --> 00:17:21.359
That leaves me with ''f of 'x1
plus delta x'' minus 'f of x1'
00:17:21.359 --> 00:17:25.040
over 'delta x'' because I'm
dividing through by 'delta x'.
00:17:25.040 --> 00:17:28.910
Now, what I do is I factor out
'f of x1' from here and divide
00:17:28.910 --> 00:17:32.500
what's left through by 'delta
x'-- that gives me 'f of x1'--
00:17:32.500 --> 00:17:36.360
times the quantity ''g of 'x1
plus delta x'' minus 'g of
00:17:36.360 --> 00:17:38.720
x1'' over 'delta x'.
00:17:38.720 --> 00:17:40.170
I get this thing over here.
00:17:40.170 --> 00:17:42.130
Now, my final step is what?
00:17:42.130 --> 00:17:45.180
I take the limit as 'delta
x' approaches 0.
00:17:45.180 --> 00:17:47.350
Well, let's do the
easy part first.
00:17:47.350 --> 00:17:52.750
As 'delta x' approaches 0, this
becomes 'f prime of x1'.
00:17:52.750 --> 00:17:57.830
And as 'delta x' approaches 0,
this becomes 'g prime of x1'.
00:17:57.830 --> 00:18:00.260
This, of course, remains
'f of x1'.
00:18:00.260 --> 00:18:03.470
And I guess you would argue
intuitively that as 'delta x'
00:18:03.470 --> 00:18:07.580
approaches 0, 'x1 plus delta
x' approaches 'x1'.
00:18:07.580 --> 00:18:10.060
Therefore, 'g of 'x1
plus delta x''
00:18:10.060 --> 00:18:11.940
approaches 'g of x1'.
00:18:11.940 --> 00:18:14.170
That would give you this
result over here.
00:18:14.170 --> 00:18:17.160
I have put a little asterisk
over here because I want to
00:18:17.160 --> 00:18:19.080
make a footnote about this.
00:18:19.080 --> 00:18:21.560
You know, even though you would
have allowed me to say
00:18:21.560 --> 00:18:24.850
that this approaches 'g of x1'
as 'delta x' approaches 0, it
00:18:24.850 --> 00:18:29.130
is not in general true that we
can be this sloppy about this.
00:18:29.130 --> 00:18:32.560
I want to mention that later,
but meanwhile, I don't want to
00:18:32.560 --> 00:18:35.200
obscure the result that
I'm driving at.
00:18:35.200 --> 00:18:38.480
Namely, assuming that this is
a proper step, what we have
00:18:38.480 --> 00:18:42.250
shown now is that to
differentiate a product, you
00:18:42.250 --> 00:18:44.090
differentiate the
first factor and
00:18:44.090 --> 00:18:46.190
multiply that by the second.
00:18:46.190 --> 00:18:49.850
Then add on to that the first
factor times the derivative of
00:18:49.850 --> 00:18:50.760
the second.
00:18:50.760 --> 00:18:53.820
Now, that may not be
self-evident, but what we have
00:18:53.820 --> 00:18:58.080
shown is that self-evident or
not, this result follows
00:18:58.080 --> 00:19:01.200
inescapably from our basic
definitions and
00:19:01.200 --> 00:19:02.730
our previous theorems.
00:19:02.730 --> 00:19:05.940
And by the way, to finish off
the example that we started
00:19:05.940 --> 00:19:12.890
where 'f of x' was 'x + 1' and
'g of x' was 'x - 1', notice
00:19:12.890 --> 00:19:16.740
that if we use this recipe,
'f prime of x1' is 1.
00:19:16.740 --> 00:19:19.540
'g of x1' is 'x1 - 1'.
00:19:19.540 --> 00:19:22.300
'f of x1' is 'x1 + 1'.
00:19:22.300 --> 00:19:25.380
'g prime of x1' is 1.
00:19:25.380 --> 00:19:28.040
And if we now combine all of
these terms, we get twice
00:19:28.040 --> 00:19:32.180
'x1', and that's exactly what
the derivative of the product
00:19:32.180 --> 00:19:32.710
should have been.
00:19:32.710 --> 00:19:34.720
In other words, just to
summarize this thing off,
00:19:34.720 --> 00:19:38.760
notice that when we were back
over here, this was the result
00:19:38.760 --> 00:19:40.350
that we were supposed to get.
00:19:40.350 --> 00:19:42.520
In other words, to differentiate
a product, our
00:19:42.520 --> 00:19:45.100
basic formula tells us that
it's the first times the
00:19:45.100 --> 00:19:47.280
derivative of the second plus
the second times the
00:19:47.280 --> 00:19:49.150
derivative of the first.
00:19:49.150 --> 00:19:51.770
And let me just go back to this
parenthetical footnote
00:19:51.770 --> 00:19:55.130
that I wanted to mention
for you.
00:19:55.130 --> 00:19:59.940
It is not always that the
limit of 'g of t' as 't'
00:19:59.940 --> 00:20:02.800
approaches 'a' is 'g of a'.
00:20:02.800 --> 00:20:06.530
Remember, our whole study of
limits said you cannot replace
00:20:06.530 --> 00:20:09.920
't' by 'a' or 'a' by
't' automatically.
00:20:09.920 --> 00:20:12.570
In fact what, type of examples
did we see that this got us
00:20:12.570 --> 00:20:13.360
into trouble?
00:20:13.360 --> 00:20:17.510
For example, let 'g of t'
be ''t squared - 1'
00:20:17.510 --> 00:20:21.190
over 't - 1'', OK?
00:20:21.190 --> 00:20:23.940
Then the limit of 'g of t' as
't' approaches 1, well, as
00:20:23.940 --> 00:20:27.250
long as 't' is not 1,
't - 1' is not 0.
00:20:27.250 --> 00:20:29.690
We can cancel the 't - 1' from
the numerator and the
00:20:29.690 --> 00:20:30.670
denominator.
00:20:30.670 --> 00:20:33.210
That leaves us with 't + 1'.
00:20:33.210 --> 00:20:37.010
As 't' approaches 1, 't
+ 1' approaches 2.
00:20:37.010 --> 00:20:39.610
In other words, the limit of 'g
of t' as 't' approaches 1
00:20:39.610 --> 00:20:43.350
is 2, but 'g of 1' doesn't
even exist.
00:20:43.350 --> 00:20:45.800
'g of 1' is 0/0.
00:20:45.800 --> 00:20:49.260
In other words, it's not true
here that as 't' approaches 1,
00:20:49.260 --> 00:20:52.420
'g of t' approaches 'g of 1'.
00:20:52.420 --> 00:20:55.440
In other words, you can say
that when 'x1' approaches
00:20:55.440 --> 00:20:58.820
'x2', 'f of x1' approaches
is 'f of x2'.
00:20:58.820 --> 00:21:00.610
It doesn't have to be true.
00:21:00.610 --> 00:21:04.590
Why then could I use it in
my particular example?
00:21:04.590 --> 00:21:08.860
And by the way, the situation in
which the limit of 'g of t'
00:21:08.860 --> 00:21:12.250
as 't' approaches 'a' is 'g of
a' comes under the heading of
00:21:12.250 --> 00:21:16.160
a subject called continuity, and
that is a later lecture in
00:21:16.160 --> 00:21:17.240
this particular block.
00:21:17.240 --> 00:21:19.240
We'll talk about it in
more detail then.
00:21:19.240 --> 00:21:22.510
But for the time being, let me
show you why we could use this
00:21:22.510 --> 00:21:24.050
in our present case.
00:21:24.050 --> 00:21:25.860
Again, we use a trick.
00:21:25.860 --> 00:21:29.780
We want to show that as 'x1 plus
delta x' gets close to
00:21:29.780 --> 00:21:34.300
'x1', 'g of 'x1 plus delta x''
gets close to 'g of x1'.
00:21:34.300 --> 00:21:36.700
And to do that, that's the
same as showing that this
00:21:36.700 --> 00:21:40.860
difference gets close to 0 as
'delta x' gets close to 0.
00:21:40.860 --> 00:21:43.580
What we do is utilize
the fact, and
00:21:43.580 --> 00:21:44.800
this is very important.
00:21:44.800 --> 00:21:48.600
We utilize the fact that
'g' is differentiable.
00:21:48.600 --> 00:21:50.820
And what we do is-- and it
doesn't look like a very
00:21:50.820 --> 00:21:52.450
significant step, but it is.
00:21:52.450 --> 00:21:55.725
What we do is we take this
expression and both multiply
00:21:55.725 --> 00:21:58.370
it and divide it by 'delta x'.
00:21:58.370 --> 00:22:01.330
I don't know if you can see the
method to my madness here.
00:22:01.330 --> 00:22:04.780
The whole thing I'm setting up
here is that when I do this,
00:22:04.780 --> 00:22:07.970
the bracketed expression now
looks like the thing that
00:22:07.970 --> 00:22:11.380
yields 'g prime of x1' when
you take the limit.
00:22:11.380 --> 00:22:13.320
And now that's exactly
what I'm going to do.
00:22:13.320 --> 00:22:16.090
I'm going to take the limit of
this expression as 'delta x'
00:22:16.090 --> 00:22:17.790
approaches 0.
00:22:17.790 --> 00:22:19.975
The limit of a product is the
product of the limits.
00:22:23.640 --> 00:22:26.240
As 'delta x' is allowed to
approach 0, this bracketed
00:22:26.240 --> 00:22:29.930
expression by definition becomes
'g prime of x1'.
00:22:29.930 --> 00:22:32.510
And obviously, this is just
the limit of 'delta x' as
00:22:32.510 --> 00:22:35.860
'delta x' approaches
0, which is 0.
00:22:35.860 --> 00:22:38.050
The fact that 'g' is
differentiable means that this
00:22:38.050 --> 00:22:42.110
is a finite number, and any
finite number times 0 is 0.
00:22:42.110 --> 00:22:45.230
In other words, the fact that
'g' was differentiable tells
00:22:45.230 --> 00:22:48.890
us that this limit is 0, and
that's the same as saying that
00:22:48.890 --> 00:22:54.680
the limit 'g of 'x1 plus delta
x'' as 'delta x' approaches 0
00:22:54.680 --> 00:22:57.170
is, in fact, 'g of x1'.
00:22:57.170 --> 00:23:00.700
As self-evident as it seems,
this is not a true result if
00:23:00.700 --> 00:23:03.600
the function 'g' is not
differentiable, or it may not
00:23:03.600 --> 00:23:07.200
be a true result if 'g'
is not differentiable.
00:23:07.200 --> 00:23:09.320
Well, so much for that.
00:23:09.320 --> 00:23:12.620
There is one more recipe that's
very important called
00:23:12.620 --> 00:23:14.310
the quotient rule.
00:23:14.310 --> 00:23:17.770
The one we just did was called
the product rule, how do you
00:23:17.770 --> 00:23:19.770
differentiate the product
of two functions.
00:23:19.770 --> 00:23:22.830
There is an analogous type of
recipe for differentiating the
00:23:22.830 --> 00:23:24.410
quotient of two functions.
00:23:24.410 --> 00:23:27.100
And since the proof is very
much the same as what I've
00:23:27.100 --> 00:23:29.860
already done, I leave
the details to you.
00:23:29.860 --> 00:23:31.910
They're in the textbook.
00:23:31.910 --> 00:23:34.060
You can go through that in
more detail if you wish.
00:23:34.060 --> 00:23:35.480
But the result is this.
00:23:35.480 --> 00:23:38.430
If 'f' and 'g' are
differentiable functions, the
00:23:38.430 --> 00:23:40.910
quotient is also
differentiable.
00:23:40.910 --> 00:23:43.150
And the way you get the result--
and again, notice how
00:23:43.150 --> 00:23:44.990
nonintuitive this is.
00:23:44.990 --> 00:23:46.900
Get this in terms of logic
if you're trying
00:23:46.900 --> 00:23:48.760
to memorize it logically.
00:23:48.760 --> 00:23:50.980
You take the denominator
times the
00:23:50.980 --> 00:23:52.730
derivative of the numerator.
00:23:52.730 --> 00:23:56.840
And you subtract off the
numerator times the derivative
00:23:56.840 --> 00:23:58.300
of the denominator.
00:23:58.300 --> 00:24:01.020
And then you divide the whole
thing by the square of the
00:24:01.020 --> 00:24:02.590
denominator.
00:24:02.590 --> 00:24:05.550
See, again, notice how
overwhelming this course
00:24:05.550 --> 00:24:08.820
becomes if you try to memorize
every result.
00:24:08.820 --> 00:24:09.980
Rather, do what?
00:24:09.980 --> 00:24:13.300
Memorize the basic definition
and derive these results.
00:24:13.300 --> 00:24:16.470
And believe me, once you use
these results long enough,
00:24:16.470 --> 00:24:17.810
you'll memorize them
automatically
00:24:17.810 --> 00:24:19.440
just by repeated use.
00:24:19.440 --> 00:24:22.390
Well, at any rate, let me give
you an example of using this.
00:24:22.390 --> 00:24:25.490
Remember before we could only
differentiate 'x' to the n if
00:24:25.490 --> 00:24:27.050
the exponent was positive?
00:24:27.050 --> 00:24:30.280
Suppose 'f of x' is 'x' to the
'minus n' where 'n' is now a
00:24:30.280 --> 00:24:31.490
positive integer.
00:24:31.490 --> 00:24:33.250
That means 'minus
n' is negative.
00:24:33.250 --> 00:24:35.660
How would we differentiate
this?
00:24:35.660 --> 00:24:38.160
And the idea here is we say
lookit, if this had been a
00:24:38.160 --> 00:24:40.190
positive 'n', we could
handle this.
00:24:40.190 --> 00:24:43.140
But 'x' to the 'minus n'
by definition is 1
00:24:43.140 --> 00:24:44.550
over 'x' to the 'n'.
00:24:44.550 --> 00:24:47.510
We know how to differentiate 'x'
to the 'n' when 'n' is a
00:24:47.510 --> 00:24:48.370
positive number.
00:24:48.370 --> 00:24:49.820
What do I have now?
00:24:49.820 --> 00:24:51.100
I have a quotient.
00:24:51.100 --> 00:24:52.750
So I use the quotient rule.
00:24:52.750 --> 00:24:54.230
'f prime of x' is what?
00:24:54.230 --> 00:24:57.130
It's my denominator, which is
'x' to the 'n', times the
00:24:57.130 --> 00:24:58.740
derivative of my numerator.
00:24:58.740 --> 00:25:01.800
My numerator is a constant so
it's the derivative of 0,
00:25:01.800 --> 00:25:04.300
minus the numerator--
that's minus 1--
00:25:04.300 --> 00:25:07.070
times the derivative
of the denominator.
00:25:07.070 --> 00:25:09.920
But for a positive integer
'n', we know that the
00:25:09.920 --> 00:25:14.130
derivative of 'x' to the 'n'
is 'nx' to the 'n - 1', all
00:25:14.130 --> 00:25:15.870
over the square of
the denominator.
00:25:15.870 --> 00:25:19.410
But the square of 'x' to the
'n' is 'x' to the '2n'.
00:25:19.410 --> 00:25:21.720
And if I now simplify
this, I have what?
00:25:21.720 --> 00:25:22.890
This is 0.
00:25:22.890 --> 00:25:24.580
This is 'minus n'.
00:25:24.580 --> 00:25:27.070
This is 'x' to the 'n - 1'.
00:25:27.070 --> 00:25:29.590
This comes upstairs
as a 'minus 2n'.
00:25:29.590 --> 00:25:35.060
So I have 'minus x' to
the 'n - 1 - 2n'.
00:25:35.060 --> 00:25:36.520
That's the same as what?
00:25:36.520 --> 00:25:41.380
'Minus nx' to the
'minus 'n - 1'.
00:25:41.380 --> 00:25:45.050
And by the way, this is a rather
interesting result now
00:25:45.050 --> 00:25:46.020
that we've proven it.
00:25:46.020 --> 00:25:49.100
Suppose we said to a person,
let's just bring down the
00:25:49.100 --> 00:25:51.970
exponential and replace
it by one less.
00:25:51.970 --> 00:25:55.460
Well, if we brought down the
exponent, that's 'minus n'.
00:25:55.460 --> 00:25:58.540
And if we replaced 'minus n'
by one less, that would be
00:25:58.540 --> 00:26:00.750
'minus 'n - 1'.
00:26:00.750 --> 00:26:03.150
And lo and behold, that's
precisely the
00:26:03.150 --> 00:26:04.740
result that we got.
00:26:04.740 --> 00:26:07.610
In other words, by using the
quotient rule, for example, we
00:26:07.610 --> 00:26:10.460
can now show that the rule
that says bring down the
00:26:10.460 --> 00:26:15.180
exponent and replace it by one
less applies for all integers,
00:26:15.180 --> 00:26:17.990
not just the positive
integers.
00:26:17.990 --> 00:26:20.960
In later lectures, we will
show that it applies to
00:26:20.960 --> 00:26:24.830
fractional exponents as well
as to any real number
00:26:24.830 --> 00:26:26.660
exponents as we go along.
00:26:26.660 --> 00:26:28.770
But I just wanted you to
see the structure here.
00:26:28.770 --> 00:26:32.370
And in fact, I think in terms
of what this lesson is all
00:26:32.370 --> 00:26:35.210
about, we have given
enough examples.
00:26:35.210 --> 00:26:39.490
I think we might just as well
summarize here, and let it be,
00:26:39.490 --> 00:26:42.340
and let the rest come from the
reading material and the
00:26:42.340 --> 00:26:45.090
learning exercises in general.
00:26:45.090 --> 00:26:46.920
And the summary is
simply this.
00:26:46.920 --> 00:26:51.880
The basic definition of a
derivative is 'f prime of x1'
00:26:51.880 --> 00:26:55.670
is the limit as 'delta x'
approaches 0, ''f of 'x1 plus
00:26:55.670 --> 00:26:59.540
delta x'' minus 'f of
x1'' over 'delta x'.
00:26:59.540 --> 00:27:02.440
That basic definition
never changes.
00:27:02.440 --> 00:27:07.890
We always mean this when
we write this.
00:27:07.890 --> 00:27:12.290
But what is important is that
this basic definition can be
00:27:12.290 --> 00:27:16.570
manipulated to yield, and now
I use quotation marks
00:27:16.570 --> 00:27:19.170
"convenient," because I don't
know how convenient something
00:27:19.170 --> 00:27:20.890
has to be before it's
really convenient,
00:27:20.890 --> 00:27:22.550
but convenient what?
00:27:22.550 --> 00:27:24.630
Recipes.
00:27:24.630 --> 00:27:25.300
Meaning what?
00:27:25.300 --> 00:27:28.410
Convenient ways to find
the derivative.
00:27:28.410 --> 00:27:30.300
You see, what's going to happen
in the remainder of our
00:27:30.300 --> 00:27:33.030
course as we shall soon see in
the next lecture, in fact,
00:27:33.030 --> 00:27:37.140
what we're going to start doing
is using derivatives the
00:27:37.140 --> 00:27:39.520
same as they always were
intended to be used.
00:27:39.520 --> 00:27:43.500
But now we are going to have
much sharper computational
00:27:43.500 --> 00:27:47.090
techniques for finding these
derivatives more rapidly than
00:27:47.090 --> 00:27:49.880
having to resort to the basic
limit definition.
00:27:49.880 --> 00:27:51.990
Well, enough about
that for now.
00:27:51.990 --> 00:27:53.380
Until next time, goodbye.
00:27:56.410 --> 00:27:58.940
NARRATOR: Funding for the
publication of this video was
00:27:58.940 --> 00:28:03.660
provided by the Gabriella and
Paul Rosenbaum Foundation.
00:28:03.660 --> 00:28:07.830
Help OCW continue to provide
free and open access to MIT
00:28:07.830 --> 00:28:12.030
courses by making a donation
at ocw.mit.edu/donate.