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PROFESSOR: Hi.
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Today we start the part of our
course that in the old days in
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the traditional treatment
began the
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engineering form of calculus.
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In other words, what we have
done is we have defined the
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derivative, the instantaneous
change, as being a limit of an
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average rate of change.
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We spent a great deal of
time proving certain
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theorems about limits.
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Now what we're going to do is to
analytically take the limit
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definition of a derivative,
which we have already done
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qualitatively, apply our limit
theorems to of this, and
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develop certain formulas for
computing derivatives much
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more quickly from a
computational point of view
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than having to resort to the
original definition.
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00:01:15,840 --> 00:01:18,200
In other words, what we're going
to do is to go from a
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qualitative approach to the
derivative to a more
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00:01:21,000 --> 00:01:22,540
quantitative approach.
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And for this reason, I just
call this lecture the
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'Derivatives of Some Simple
Functions' to create sort of a
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mood over here.
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Now, the idea is this.
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Let's go back to our
basic definition.
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The derivative of 'f of x' when
'x' is equal to 'x1', 'f
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00:01:36,800 --> 00:01:40,150
prime of x1', on is by
definition the limit as 'delta
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00:01:40,150 --> 00:01:44,120
x' approaches 0, ''f of 'x1
plus delta x'' minus 'f of
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00:01:44,120 --> 00:01:45,790
x1'' over 'delta x'.
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00:01:45,790 --> 00:01:48,960
In other words, the average
rate of change of 'f of x'
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00:01:48,960 --> 00:01:53,260
with respect to 'x' as we move
from 'x1' to some new position
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00:01:53,260 --> 00:01:56,560
'x1 + delta x', taking
the limit as 'delta
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00:01:56,560 --> 00:01:57,920
x' approaches 0.
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00:01:57,920 --> 00:02:02,710
Now, just to illustrate this,
let's take, for example, 'f of
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00:02:02,710 --> 00:02:05,140
x' equals 'x cubed'.
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00:02:05,140 --> 00:02:08,270
In this case, if we now compute
'f of 'x1 plus delta
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00:02:08,270 --> 00:02:11,240
x'' minus 'f of x1', recall
that 'f' is the
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00:02:11,240 --> 00:02:12,890
function that does what?
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00:02:12,890 --> 00:02:16,710
Given any input, the output is
the cube of the input, so 'f
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00:02:16,710 --> 00:02:20,990
of 'x1 plus delta x' will be the
cube of ''x1 plus delta x'
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00:02:20,990 --> 00:02:22,230
minus 'f of x1''.
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00:02:22,230 --> 00:02:24,200
That's minus 'x1 cubed'.
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00:02:24,200 --> 00:02:26,470
We now use the binomial
theorem.
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In other words, notice again
how mathematics works.
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You introduce a new definition,
but then all of
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the old prerequisite mathematics
comes back to be
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used as a tool over here.
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00:02:37,440 --> 00:02:39,240
We use the binomial theorem.
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00:02:39,240 --> 00:02:42,120
The 'x1 cubed' term here
cancels with the
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'x1 cubed' term here.
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00:02:43,850 --> 00:02:46,980
And what we're left with is
what? ''3x1 squared' times
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'delta x'' plus '3x1
'delta x' squared''
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00:02:50,170 --> 00:02:52,310
plus 'delta x' cubed'.
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00:02:52,310 --> 00:02:54,760
And to emphasize the next step
that we're going to make,
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let's factor out a 'delta x'.
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00:02:56,960 --> 00:03:00,100
Now, going back to our basic
definition, we must take this
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expression and divide
it by 'delta x'.
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The key idea here is that in the
last step, we're going to
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take the limit as 'delta
x' approaches 0.
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The limit as 'delta x'
approaches 0 means, in
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particular, that 'delta x' is
not allowed to equal 0.
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And as long as 'delta x' is not
0, the 'delta x' in the
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00:03:17,820 --> 00:03:21,350
denominator cancels the 'delta
x' in the numerator to leave
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00:03:21,350 --> 00:03:26,220
'3x1 squared' plus '3x1 'delta
x'' plus 'delta x squared'.
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00:03:26,220 --> 00:03:30,440
And now finally to find what 'f
prime of x1' is, we simply
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take this limit as 'delta
x' approaches 0.
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00:03:33,560 --> 00:03:35,710
And notice what you're doing
over here when you let 'delta
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x' approach 0.
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We're going to be using
the limit theorems.
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Namely, the limit of a sum
is the sum of the limits.
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The limit of a product is the
product of the limits.
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We can then take each of these
limits term by term.
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Notice that the first term, not
depending on 'delta x',
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00:03:51,280 --> 00:03:53,780
it's limit is just
'3x1 squared'.
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Each of the remaining terms
have a factor of
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'delta x' in them.
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00:03:57,140 --> 00:04:00,460
As 'delta x' goes to 0 then,
each of the remaining terms go
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00:04:00,460 --> 00:04:04,390
to 0, and we wind up with, by
our basic definition, that 'f
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00:04:04,390 --> 00:04:07,630
prime of x1' is '3x1 squared'.
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In particular, since 'x1'
could have been any real
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number here, what we
have shown is what?
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00:04:12,870 --> 00:04:20,399
That if 'f of x' is equal to 'x
cubed', then 'f prime of x'
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00:04:20,399 --> 00:04:22,800
is '3x squared'.
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00:04:22,800 --> 00:04:26,410
And the important point here
is simply to observe that
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00:04:26,410 --> 00:04:31,050
we've got this result by
applying the basic definition
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00:04:31,050 --> 00:04:34,160
to a specifically
given function.
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00:04:34,160 --> 00:04:37,720
To generalize this result, let
'f of x' be 'x' to the 'n',
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where 'n' is any positive
whole number.
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And I want to emphasize that
because you're going to see as
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this proof goes on that I really
use the fact then 'n'
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is a positive whole number,
and I'll emphasize to you
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00:04:51,640 --> 00:04:52,650
where I do this.
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00:04:52,650 --> 00:04:55,370
I'm going to mimic the same
thing I did in the special
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00:04:55,370 --> 00:04:57,950
case in our first example
when we simply chose
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00:04:57,950 --> 00:04:59,430
'n' to equal 3.
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00:04:59,430 --> 00:05:02,100
Namely, I compute 'f
of 'x1 plus delta
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00:05:02,100 --> 00:05:04,060
x'' minus 'f of x1'.
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00:05:04,060 --> 00:05:07,460
Since the output is the n-th
power of the input, that's
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00:05:07,460 --> 00:05:11,830
just 'x1 plus delta x' to the
'n' minus 'x1' to the 'n'.
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Now, as long as 'n' is a
positive whole number, we can
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expand it by the binomial
theorem.
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The first term will be 'x
sub 1' to the n-th.
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That will cancel this 'x
sub 1' to the n-th.
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The next term will be 'n 'x sub
1'' to the 'n - 1' power
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00:05:28,600 --> 00:05:30,260
times 'delta x'.
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00:05:30,260 --> 00:05:31,820
And now I do something
that's a little
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00:05:31,820 --> 00:05:33,270
bit cheating, I guess.
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00:05:33,270 --> 00:05:37,020
What I do next is I observe that
every remaining term in
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00:05:37,020 --> 00:05:41,180
the binomial theorem expansion
here has a factor of at least
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00:05:41,180 --> 00:05:42,730
'delta x squared' in it.
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00:05:42,730 --> 00:05:45,650
So rather than compute all of
these terms individually,
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00:05:45,650 --> 00:05:48,120
recognizing that I'm going to
let 'delta x' eventually
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00:05:48,120 --> 00:05:51,130
approach 0, I factor out a
'delta x squared' and say,
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00:05:51,130 --> 00:05:54,150
lookit, the remaining terms
have the form 'delta x
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00:05:54,150 --> 00:05:58,100
squared' times some
finite number.
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00:05:58,100 --> 00:06:01,150
I don't know what it is, but
it's some finite number.
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00:06:01,150 --> 00:06:02,970
I'll put that-- well, let's just
take a look and see what
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00:06:02,970 --> 00:06:03,900
that means.
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00:06:03,900 --> 00:06:07,320
In particular now, if I take
this expression and divide
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00:06:07,320 --> 00:06:10,690
through by 'delta x', we're
assuming again that 'delta x'
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00:06:10,690 --> 00:06:13,180
is not 0 because we're going to
take the limit as 'delta x'
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00:06:13,180 --> 00:06:14,250
approaches 0.
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00:06:14,250 --> 00:06:16,740
And remember our big harangue
about that.
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When we approach the limit, we
are never allowed to equal it.
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00:06:19,620 --> 00:06:21,240
'Delta x' never equals 0.
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00:06:21,240 --> 00:06:24,320
Consequently, I can cancel a
'delta x' from the denominator
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00:06:24,320 --> 00:06:26,610
with a 'delta x' from
the numerator.
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00:06:26,610 --> 00:06:29,900
Canceling a 'delta x' from the
numerator knocks a 'delta x'
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00:06:29,900 --> 00:06:32,810
altogether out of this term
and leaves me with just a
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00:06:32,810 --> 00:06:35,270
single factor of 'delta
x' in this term.
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00:06:35,270 --> 00:06:40,240
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00:06:40,240 --> 00:06:45,770
What I now do is, referring to
my basic definition, I now
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00:06:45,770 --> 00:06:49,240
take the limit as 'delta
x' approaches 0.
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00:06:49,240 --> 00:06:53,210
As 'delta x' approaches 0, this
term, not depending on
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00:06:53,210 --> 00:06:56,810
'delta x', remains 'nx'
to the 'n - 1'.
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00:06:56,810 --> 00:06:58,800
And here's the key step.
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00:06:58,800 --> 00:07:01,760
The limit of a product is the
product of the limits.
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00:07:01,760 --> 00:07:03,440
This approaches 0.
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00:07:03,440 --> 00:07:06,480
This is some number.
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00:07:06,480 --> 00:07:09,970
And 0 times some number is 0.
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00:07:09,970 --> 00:07:15,920
And therefore, all that's left
here is 'nx1' to the 'n - 1'.
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00:07:15,920 --> 00:07:19,310
Again, since this was any 'x',
what we have is what?
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00:07:19,310 --> 00:07:22,280
If 'f of x' is 'x to the
n', where 'n' is a
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00:07:22,280 --> 00:07:23,390
positive whole number.
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00:07:23,390 --> 00:07:25,060
And where do I use that fact?
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00:07:25,060 --> 00:07:27,630
I use that fact to use
the binomial theorem.
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00:07:27,630 --> 00:07:30,650
Then the derivative is
'nx' to the 'n - 1'.
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00:07:30,650 --> 00:07:33,310
In other words, if you want to
memorize a shortcut now that
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00:07:33,310 --> 00:07:36,940
we know the answer, observe that
it seems to differentiate
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00:07:36,940 --> 00:07:37,940
'x' to the 'n'.
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00:07:37,940 --> 00:07:42,610
All you do is bring the exponent
down and replace it
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00:07:42,610 --> 00:07:45,250
by one less.
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00:07:45,250 --> 00:07:48,030
But for heaven's sakes, don't
look at that as being a proof.
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00:07:48,030 --> 00:07:50,790
Rather, we gave the proof,
then observed what the
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00:07:50,790 --> 00:07:52,750
shortcut recipe was.
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00:07:52,750 --> 00:07:56,600
Don't say isn't it easier just
to bring the exponent down and
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00:07:56,600 --> 00:07:59,330
replace it by one less instead
of going through this whole
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00:07:59,330 --> 00:08:01,260
long harangue over here?
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00:08:01,260 --> 00:08:02,830
No, this is how we prove
that result.
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00:08:02,830 --> 00:08:05,270
In other words, what we have
now shown is that to
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00:08:05,270 --> 00:08:08,660
differentiate 'x' to the 'n' for
any positive integer 'n',
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00:08:08,660 --> 00:08:11,390
the derivative is just
'nx' to the 'n - 1'.
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00:08:11,390 --> 00:08:14,040
So far, we do not know if
that result is true
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00:08:14,040 --> 00:08:15,200
for any other numbers.
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00:08:15,200 --> 00:08:18,150
We'll talk about that a
little bit more later.
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00:08:18,150 --> 00:08:20,360
Well, so much for
that example.
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00:08:20,360 --> 00:08:21,740
Let's look at another one.
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00:08:21,740 --> 00:08:24,300
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00:08:24,300 --> 00:08:27,460
Let's suppose that we have two
functions 'f' and 'g', which
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00:08:27,460 --> 00:08:30,370
are differentiable, say,
at some number 'x1'.
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00:08:30,370 --> 00:08:33,070
That means, among other things,
that 'f prime of x1'
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00:08:33,070 --> 00:08:35,320
and 'g prime of x1' exist.
180
00:08:35,320 --> 00:08:38,679
Now define a new function
'h' to be the sum of
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00:08:38,679 --> 00:08:40,450
the given two functions.
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00:08:40,450 --> 00:08:42,600
By the way, we have to be
very, very careful here.
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00:08:42,600 --> 00:08:45,910
In general, 'f' and 'g' could
have different domains.
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00:08:45,910 --> 00:08:49,790
Notice that 'x' has to belong to
both the domain of 'f' and
185
00:08:49,790 --> 00:08:52,320
the domain of 'g' for
this to make sense.
186
00:08:52,320 --> 00:08:55,650
Consequently, I write down over
here that 'x' belongs to
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00:08:55,650 --> 00:08:59,070
the domain of 'f' intersect
domain 'g'.
188
00:08:59,070 --> 00:09:02,050
In other words, it belongs to
both sets, the domain of 'f'
189
00:09:02,050 --> 00:09:03,340
and the domain of 'g'.
190
00:09:03,340 --> 00:09:06,810
Because if 'x' didn't belong to
at least one of these two,
191
00:09:06,810 --> 00:09:08,540
this expression wouldn't
even make sense.
192
00:09:08,540 --> 00:09:11,510
In other words, 'x' must be a
permissible input to both the
193
00:09:11,510 --> 00:09:13,100
'f' and 'g' machines.
194
00:09:13,100 --> 00:09:16,960
Now, what my claim is, is that
in this particular case, 'h
195
00:09:16,960 --> 00:09:17,760
prime' exists.
196
00:09:17,760 --> 00:09:20,730
In other words, 'h' is also
differentiable, and it's
197
00:09:20,730 --> 00:09:24,190
obtained by adding the sum
of these two derivatives.
198
00:09:24,190 --> 00:09:26,730
In other words, this is the
result that's commonly known
199
00:09:26,730 --> 00:09:29,730
as the derivative of a sum is
the sum of the derivatives,
200
00:09:29,730 --> 00:09:31,530
which, of course, sounds like
it should be right.
201
00:09:31,530 --> 00:09:34,110
But I'll comment on that
in a moment also.
202
00:09:34,110 --> 00:09:34,340
what.
203
00:09:34,340 --> 00:09:38,460
I want to show is simply this,
that 'h prime of x1' is simply
204
00:09:38,460 --> 00:09:42,930
'f prime of x1' plus 'g
prime of x1', OK?
205
00:09:42,930 --> 00:09:46,440
And my main purpose is to
highlight the basic
206
00:09:46,440 --> 00:09:46,990
definition.
207
00:09:46,990 --> 00:09:49,070
That's what I want to give
you the drill on.
208
00:09:49,070 --> 00:09:53,670
Namely, how do we compute
'h prime of x1'?
209
00:09:53,670 --> 00:09:57,720
We must look at ''h of 'x1 plus
delta x'' minus 'h of
210
00:09:57,720 --> 00:10:01,770
x1'' over 'delta x', then take
the limit as 'delta x'
211
00:10:01,770 --> 00:10:02,530
approaches 0.
212
00:10:02,530 --> 00:10:05,580
That's the basic definition
that will be used over and
213
00:10:05,580 --> 00:10:06,850
over and over.
214
00:10:06,850 --> 00:10:10,790
At any rate, sparing you all
of the details here, let's
215
00:10:10,790 --> 00:10:11,980
simply observe what?
216
00:10:11,980 --> 00:10:15,940
That 'h of 'x1 plus delta x'',
since 'h' is the sum of 'f'
217
00:10:15,940 --> 00:10:19,410
and 'g', is 'f of 'x1 plus
delta x'' plus 'g of
218
00:10:19,410 --> 00:10:21,200
'x1 plus delta x''.
219
00:10:21,200 --> 00:10:25,910
Consequently, 'h of 'x1 plus
delta x'' minus 'h of x1' is
220
00:10:25,910 --> 00:10:29,970
'f of 'x1 plus delta x'' plus
'g of 'x1 plus delta x''--
221
00:10:29,970 --> 00:10:30,970
that's this part--
222
00:10:30,970 --> 00:10:35,350
minus ''f of x1' plus
'g of x1''.
223
00:10:35,350 --> 00:10:38,740
Now, the idea is I want to
divide this by 'delta x'.
224
00:10:38,740 --> 00:10:42,420
I also have some knowledge of
the differentiability of 'f'
225
00:10:42,420 --> 00:10:46,120
and 'g' separately, so I would
like to combine or regroup
226
00:10:46,120 --> 00:10:49,110
these terms to highlight
that fact for me.
227
00:10:49,110 --> 00:10:52,960
So what I do is I group these
two terms together, these two
228
00:10:52,960 --> 00:10:56,140
terms together, and then divide
through by 'delta x'.
229
00:10:56,140 --> 00:11:00,010
In other words, ''h of 'x1 plus
delta x'' minus 'h of x''
230
00:11:00,010 --> 00:11:02,040
over 'delta x' is what?
231
00:11:02,040 --> 00:11:06,030
''f of 'x1 plus delta x'' minus
'f of x1'' over 'delta
232
00:11:06,030 --> 00:11:10,470
x' plus ''g of 'x1 plus
delta x'' minus 'g of
233
00:11:10,470 --> 00:11:12,700
x1'' over 'delta x'.
234
00:11:12,700 --> 00:11:16,550
Now, for my last step, I simply
must take the limit as
235
00:11:16,550 --> 00:11:18,320
'delta x' goes to 0.
236
00:11:18,320 --> 00:11:20,910
Well, just look at this
bracketed expression.
237
00:11:20,910 --> 00:11:23,690
The limit of a sum is the
sum of the limits.
238
00:11:23,690 --> 00:11:25,930
So to take the limit of this
expression, I take the limit
239
00:11:25,930 --> 00:11:28,170
of these two terms separately
and add them.
240
00:11:28,170 --> 00:11:31,790
By definition, the limit of the
bracketed expressions as
241
00:11:31,790 --> 00:11:34,540
'delta x' goes to 0--
by definition--
242
00:11:34,540 --> 00:11:36,750
is 'f prime of x1'.
243
00:11:36,750 --> 00:11:41,190
And as 'delta x' goes to 0,
this term here becomes 'g
244
00:11:41,190 --> 00:11:42,810
prime of x1'.
245
00:11:42,810 --> 00:11:45,400
And by the way, for the last
time, let me give you this
246
00:11:45,400 --> 00:11:46,870
word of caution again.
247
00:11:46,870 --> 00:11:50,520
Do not make the mistake of
saying, gee, when 'delta x'
248
00:11:50,520 --> 00:11:53,180
goes to 0, my numerator is 0.
249
00:11:53,180 --> 00:11:55,190
Therefore, the fraction
must be 0.
250
00:11:55,190 --> 00:11:55,590
No!
251
00:11:55,590 --> 00:11:59,150
When 'delta x' goes to 0, sure,
if you replace 'delta x'
252
00:11:59,150 --> 00:12:03,530
by 0, the numerator is 0, but
so is the denominator.
253
00:12:03,530 --> 00:12:06,410
In other words, this is our old
story that the derivative
254
00:12:06,410 --> 00:12:10,980
becomes a study of 0/0 so if
you let 'delta x' equals 0.
255
00:12:10,980 --> 00:12:12,400
In other words, the whole
point is what?
256
00:12:12,400 --> 00:12:17,040
That as 'delta x' approaches
0, this by definition is 'f
257
00:12:17,040 --> 00:12:18,330
prime of x1'.
258
00:12:18,330 --> 00:12:21,070
This by definition is
'g prime of x1'.
259
00:12:21,070 --> 00:12:26,430
And therefore, we have proven
as a theorem 'h prime of x1'
260
00:12:26,430 --> 00:12:30,380
is 'f prime of x1' plus 'g
prime of x1', that the
261
00:12:30,380 --> 00:12:33,990
derivative of a sum is the
sum of the derivatives.
262
00:12:33,990 --> 00:12:36,960
Now, the main trouble so far is
that every result that I've
263
00:12:36,960 --> 00:12:40,215
proven rigorously for you, you
have probably guessed in
264
00:12:40,215 --> 00:12:41,720
advance that it was
going to happen.
265
00:12:41,720 --> 00:12:45,040
You say who needs this rigorous
stuff when we could
266
00:12:45,040 --> 00:12:47,200
have got the same result
by intuition.
267
00:12:47,200 --> 00:12:49,790
And this is where the rift
between the way the pure
268
00:12:49,790 --> 00:12:53,120
mathematician often teaches
calculus and the way, say, the
269
00:12:53,120 --> 00:12:56,510
applied man teaches calculus
often comes in.
270
00:12:56,510 --> 00:12:59,730
The point is that the places
that you can get into trouble
271
00:12:59,730 --> 00:13:02,110
often aren't stressed
soon enough.
272
00:13:02,110 --> 00:13:05,370
So before we go any further, let
me show you that you must
273
00:13:05,370 --> 00:13:08,780
be careful, that up until now,
we have been very fortunate
274
00:13:08,780 --> 00:13:11,110
when we say things like the
limit of a product is the
275
00:13:11,110 --> 00:13:13,520
product of the limits, fortunate
in the sense that
276
00:13:13,520 --> 00:13:16,750
they worked out the way the
wording seemed to indicate.
277
00:13:16,750 --> 00:13:21,070
Let me give you an example of
what I mean by saying beware
278
00:13:21,070 --> 00:13:22,610
of self-evident.
279
00:13:22,610 --> 00:13:26,900
I claim, for example, that the
derivative of a product is not
280
00:13:26,900 --> 00:13:29,160
the product of the derivatives
necessarily.
281
00:13:29,160 --> 00:13:32,530
And the best way to prove that
to you is not to say take my
282
00:13:32,530 --> 00:13:33,350
word for it.
283
00:13:33,350 --> 00:13:36,600
This is one of the beauties of
computational mathematics.
284
00:13:36,600 --> 00:13:40,520
You can always show by means of
an example what's going on.
285
00:13:40,520 --> 00:13:44,320
For example, let 'f of
x' be 'x + 1', let 'g
286
00:13:44,320 --> 00:13:46,460
of x' be 'x - 1'.
287
00:13:46,460 --> 00:13:49,380
Now, the derivative of
'x + 1' is simply 1.
288
00:13:49,380 --> 00:13:51,750
The derivative of 'x
- 1' is also 1.
289
00:13:51,750 --> 00:13:54,230
Namely, we differentiate
this as a sum.
290
00:13:54,230 --> 00:13:55,340
The derivative of 'x' is 1.
291
00:13:55,340 --> 00:13:57,870
The derivative of
a constant is 0.
292
00:13:57,870 --> 00:14:02,000
In other words, 'f prime of x'
is 1, 'g prime of x' is 1.
293
00:14:02,000 --> 00:14:05,710
On the other hand, if we first
multiply 'f' and 'g', 'x + 1'
294
00:14:05,710 --> 00:14:09,370
times 'x - 1' is 'x squared
- 1', and the
295
00:14:09,370 --> 00:14:10,740
derivative of that is what?
296
00:14:10,740 --> 00:14:13,420
Well, to differentiate 'x
squared', we just saw.
297
00:14:13,420 --> 00:14:14,850
Bring the exponent down.
298
00:14:14,850 --> 00:14:16,180
Replace it by one less.
299
00:14:16,180 --> 00:14:17,100
That's '2x'.
300
00:14:17,100 --> 00:14:19,170
The derivative of
minus 1 is 0.
301
00:14:19,170 --> 00:14:21,870
In other words, if you multiply
first and then take
302
00:14:21,870 --> 00:14:23,670
the derivative, you get '2x'.
303
00:14:23,670 --> 00:14:26,330
On the other hand, if you
differentiate first and then
304
00:14:26,330 --> 00:14:28,390
take the product, you get 2.
305
00:14:28,390 --> 00:14:31,710
Now, it is not true that '2x'
is a synonym for 1 for all
306
00:14:31,710 --> 00:14:32,790
values of 'x'.
307
00:14:32,790 --> 00:14:38,400
In other words, if you
differentiate first and then
308
00:14:38,400 --> 00:14:41,620
take the product, you get a
different answer than if you
309
00:14:41,620 --> 00:14:44,030
multiply first and then
differentiate.
310
00:14:44,030 --> 00:14:46,480
In other words, self-evident
or not, the thing
311
00:14:46,480 --> 00:14:48,180
happens to be false.
312
00:14:48,180 --> 00:14:51,870
In fact, let's see how the
theory helps us in this case.
313
00:14:51,870 --> 00:14:54,700
See, what I've shown you
here is that this
314
00:14:54,700 --> 00:14:56,710
result isn't true.
315
00:14:56,710 --> 00:14:58,510
Oh, I should say that this
result is true, that the
316
00:14:58,510 --> 00:14:59,880
equality isn't true.
317
00:14:59,880 --> 00:15:02,760
What I haven't shown you
is what is true.
318
00:15:02,760 --> 00:15:05,700
And again, to emphasize
our basic definition,
319
00:15:05,700 --> 00:15:06,870
let's do that now.
320
00:15:06,870 --> 00:15:09,970
Let's show how we can use our
basic definition to find the
321
00:15:09,970 --> 00:15:11,060
correct result.
322
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'.
323
00:15:15,050 --> 00:15:18,480
And we're going to use the
same structure as before.
324
00:15:18,480 --> 00:15:22,720
We are going to compute 'h of
'x1 plus delta x'', subtract
325
00:15:22,720 --> 00:15:26,840
'h of x1' divide by 'delta x',
and take the limit as 'delta
326
00:15:26,840 --> 00:15:27,800
x' approaches 0.
327
00:15:27,800 --> 00:15:29,560
We do that every time.
328
00:15:29,560 --> 00:15:32,210
All that changes is the property
of the particular
329
00:15:32,210 --> 00:15:33,760
function that we're
dealing with.
330
00:15:33,760 --> 00:15:37,860
In any event, if I do this, 'h
of 'x1 plus delta x'' is
331
00:15:37,860 --> 00:15:42,760
obtained by replacing 'x' in
here by 'x1 plus delta x'.
332
00:15:42,760 --> 00:15:46,000
And 'h of x1' is obtained by
replacing 'x' by 'x1' over
333
00:15:46,000 --> 00:15:50,140
here, so I wind up with 'f of
'x1 plus delta x'' times 'g of
334
00:15:50,140 --> 00:15:55,430
'x1 plus delta x'' minus 'f
of x1' times 'g of x1'.
335
00:15:55,430 --> 00:15:59,230
And now the problem is I would
like somehow to be able to get
336
00:15:59,230 --> 00:16:03,880
terms involving 'f of 'x1 plus
delta x'' minus 'f of x1', 'g
337
00:16:03,880 --> 00:16:07,810
of 'x1 plus delta x'' minus 'g
of x1', and I use the same
338
00:16:07,810 --> 00:16:11,550
trick here that I used when I
proved that the limit of a
339
00:16:11,550 --> 00:16:13,480
product is the product
of the limits.
340
00:16:13,480 --> 00:16:18,050
Namely, I am going to add in
and subtract the same term,
341
00:16:18,050 --> 00:16:19,770
and this will help me factor.
342
00:16:19,770 --> 00:16:22,020
In other words, what I'm going
to do here is this.
343
00:16:22,020 --> 00:16:23,590
I write down this term.
344
00:16:23,590 --> 00:16:27,370
Now I say to myself I would like
an 'f of x1' term to go
345
00:16:27,370 --> 00:16:29,610
with the 'f of 'x1
plus delta x''.
346
00:16:29,610 --> 00:16:32,280
So what I do is I write
minus 'f of x1'.
347
00:16:32,280 --> 00:16:35,400
Then I observe that 'g of 'x1
plus delta x'' is a factor
348
00:16:35,400 --> 00:16:38,010
here, so I put that
in over here.
349
00:16:38,010 --> 00:16:39,830
See, 'g of 'x1 plus delta x''.
350
00:16:39,830 --> 00:16:42,720
See, in other words, I can now
factor out a 'g of 'x1 plus
351
00:16:42,720 --> 00:16:46,130
delta x'' from these two terms
and have the formula that I
352
00:16:46,130 --> 00:16:47,940
want left over.
353
00:16:47,940 --> 00:16:50,250
But, of course, this term
doesn't belong here.
354
00:16:50,250 --> 00:16:53,700
I just added it myself, or I
should say subtracted it, so
355
00:16:53,700 --> 00:16:56,790
now I put it in with the
opposite sign, namely, plus 'f
356
00:16:56,790 --> 00:16:59,260
of x1' 'g of 'x1
plus delta x''.
357
00:16:59,260 --> 00:17:01,840
And now put this term back
in here: minus 'f of
358
00:17:01,840 --> 00:17:03,700
x1' times 'g of x1'.
359
00:17:03,700 --> 00:17:06,420
In other words, all I've done
is rewritten this, but by
360
00:17:06,420 --> 00:17:10,420
adding and subtracting the same
term, which does what?
361
00:17:10,420 --> 00:17:14,700
Dividing through by 'delta x',
I factor out a 'g of 'x1 plus
362
00:17:14,700 --> 00:17:17,440
delta x' from these two terms.
363
00:17:17,440 --> 00:17:21,359
That leaves me with ''f of 'x1
plus delta x'' minus 'f of x1'
364
00:17:21,359 --> 00:17:25,040
over 'delta x'' because I'm
dividing through by 'delta x'.
365
00:17:25,040 --> 00:17:28,910
Now, what I do is I factor out
'f of x1' from here and divide
366
00:17:28,910 --> 00:17:32,500
what's left through by 'delta
x'-- that gives me 'f of x1'--
367
00:17:32,500 --> 00:17:36,360
times the quantity ''g of 'x1
plus delta x'' minus 'g of
368
00:17:36,360 --> 00:17:38,720
x1'' over 'delta x'.
369
00:17:38,720 --> 00:17:40,170
I get this thing over here.
370
00:17:40,170 --> 00:17:42,130
Now, my final step is what?
371
00:17:42,130 --> 00:17:45,180
I take the limit as 'delta
x' approaches 0.
372
00:17:45,180 --> 00:17:47,350
Well, let's do the
easy part first.
373
00:17:47,350 --> 00:17:52,750
As 'delta x' approaches 0, this
becomes 'f prime of x1'.
374
00:17:52,750 --> 00:17:57,830
And as 'delta x' approaches 0,
this becomes 'g prime of x1'.
375
00:17:57,830 --> 00:18:00,260
This, of course, remains
'f of x1'.
376
00:18:00,260 --> 00:18:03,470
And I guess you would argue
intuitively that as 'delta x'
377
00:18:03,470 --> 00:18:07,580
approaches 0, 'x1 plus delta
x' approaches 'x1'.
378
00:18:07,580 --> 00:18:10,060
Therefore, 'g of 'x1
plus delta x''
379
00:18:10,060 --> 00:18:11,940
approaches 'g of x1'.
380
00:18:11,940 --> 00:18:14,170
That would give you this
result over here.
381
00:18:14,170 --> 00:18:17,160
I have put a little asterisk
over here because I want to
382
00:18:17,160 --> 00:18:19,080
make a footnote about this.
383
00:18:19,080 --> 00:18:21,560
You know, even though you would
have allowed me to say
384
00:18:21,560 --> 00:18:24,850
that this approaches 'g of x1'
as 'delta x' approaches 0, it
385
00:18:24,850 --> 00:18:29,130
is not in general true that we
can be this sloppy about this.
386
00:18:29,130 --> 00:18:32,560
I want to mention that later,
but meanwhile, I don't want to
387
00:18:32,560 --> 00:18:35,200
obscure the result that
I'm driving at.
388
00:18:35,200 --> 00:18:38,480
Namely, assuming that this is
a proper step, what we have
389
00:18:38,480 --> 00:18:42,250
shown now is that to
differentiate a product, you
390
00:18:42,250 --> 00:18:44,090
differentiate the
first factor and
391
00:18:44,090 --> 00:18:46,190
multiply that by the second.
392
00:18:46,190 --> 00:18:49,850
Then add on to that the first
factor times the derivative of
393
00:18:49,850 --> 00:18:50,760
the second.
394
00:18:50,760 --> 00:18:53,820
Now, that may not be
self-evident, but what we have
395
00:18:53,820 --> 00:18:58,080
shown is that self-evident or
not, this result follows
396
00:18:58,080 --> 00:19:01,200
inescapably from our basic
definitions and
397
00:19:01,200 --> 00:19:02,730
our previous theorems.
398
00:19:02,730 --> 00:19:05,940
And by the way, to finish off
the example that we started
399
00:19:05,940 --> 00:19:12,890
where 'f of x' was 'x + 1' and
'g of x' was 'x - 1', notice
400
00:19:12,890 --> 00:19:16,740
that if we use this recipe,
'f prime of x1' is 1.
401
00:19:16,740 --> 00:19:19,540
'g of x1' is 'x1 - 1'.
402
00:19:19,540 --> 00:19:22,300
'f of x1' is 'x1 + 1'.
403
00:19:22,300 --> 00:19:25,380
'g prime of x1' is 1.
404
00:19:25,380 --> 00:19:28,040
And if we now combine all of
these terms, we get twice
405
00:19:28,040 --> 00:19:32,180
'x1', and that's exactly what
the derivative of the product
406
00:19:32,180 --> 00:19:32,710
should have been.
407
00:19:32,710 --> 00:19:34,720
In other words, just to
summarize this thing off,
408
00:19:34,720 --> 00:19:38,760
notice that when we were back
over here, this was the result
409
00:19:38,760 --> 00:19:40,350
that we were supposed to get.
410
00:19:40,350 --> 00:19:42,520
In other words, to differentiate
a product, our
411
00:19:42,520 --> 00:19:45,100
basic formula tells us that
it's the first times the
412
00:19:45,100 --> 00:19:47,280
derivative of the second plus
the second times the
413
00:19:47,280 --> 00:19:49,150
derivative of the first.
414
00:19:49,150 --> 00:19:51,770
And let me just go back to this
parenthetical footnote
415
00:19:51,770 --> 00:19:55,130
that I wanted to mention
for you.
416
00:19:55,130 --> 00:19:59,940
It is not always that the
limit of 'g of t' as 't'
417
00:19:59,940 --> 00:20:02,800
approaches 'a' is 'g of a'.
418
00:20:02,800 --> 00:20:06,530
Remember, our whole study of
limits said you cannot replace
419
00:20:06,530 --> 00:20:09,920
't' by 'a' or 'a' by
't' automatically.
420
00:20:09,920 --> 00:20:12,570
In fact what, type of examples
did we see that this got us
421
00:20:12,570 --> 00:20:13,360
into trouble?
422
00:20:13,360 --> 00:20:17,510
For example, let 'g of t'
be ''t squared - 1'
423
00:20:17,510 --> 00:20:21,190
over 't - 1'', OK?
424
00:20:21,190 --> 00:20:23,940
Then the limit of 'g of t' as
't' approaches 1, well, as
425
00:20:23,940 --> 00:20:27,250
long as 't' is not 1,
't - 1' is not 0.
426
00:20:27,250 --> 00:20:29,690
We can cancel the 't - 1' from
the numerator and the
427
00:20:29,690 --> 00:20:30,670
denominator.
428
00:20:30,670 --> 00:20:33,210
That leaves us with 't + 1'.
429
00:20:33,210 --> 00:20:37,010
As 't' approaches 1, 't
+ 1' approaches 2.
430
00:20:37,010 --> 00:20:39,610
In other words, the limit of 'g
of t' as 't' approaches 1
431
00:20:39,610 --> 00:20:43,350
is 2, but 'g of 1' doesn't
even exist.
432
00:20:43,350 --> 00:20:45,800
'g of 1' is 0/0.
433
00:20:45,800 --> 00:20:49,260
In other words, it's not true
here that as 't' approaches 1,
434
00:20:49,260 --> 00:20:52,420
'g of t' approaches 'g of 1'.
435
00:20:52,420 --> 00:20:55,440
In other words, you can say
that when 'x1' approaches
436
00:20:55,440 --> 00:20:58,820
'x2', 'f of x1' approaches
is 'f of x2'.
437
00:20:58,820 --> 00:21:00,610
It doesn't have to be true.
438
00:21:00,610 --> 00:21:04,590
Why then could I use it in
my particular example?
439
00:21:04,590 --> 00:21:08,860
And by the way, the situation in
which the limit of 'g of t'
440
00:21:08,860 --> 00:21:12,250
as 't' approaches 'a' is 'g of
a' comes under the heading of
441
00:21:12,250 --> 00:21:16,160
a subject called continuity, and
that is a later lecture in
442
00:21:16,160 --> 00:21:17,240
this particular block.
443
00:21:17,240 --> 00:21:19,240
We'll talk about it in
more detail then.
444
00:21:19,240 --> 00:21:22,510
But for the time being, let me
show you why we could use this
445
00:21:22,510 --> 00:21:24,050
in our present case.
446
00:21:24,050 --> 00:21:25,860
Again, we use a trick.
447
00:21:25,860 --> 00:21:29,780
We want to show that as 'x1 plus
delta x' gets close to
448
00:21:29,780 --> 00:21:34,300
'x1', 'g of 'x1 plus delta x''
gets close to 'g of x1'.
449
00:21:34,300 --> 00:21:36,700
And to do that, that's the
same as showing that this
450
00:21:36,700 --> 00:21:40,860
difference gets close to 0 as
'delta x' gets close to 0.
451
00:21:40,860 --> 00:21:43,580
What we do is utilize
the fact, and
452
00:21:43,580 --> 00:21:44,800
this is very important.
453
00:21:44,800 --> 00:21:48,600
We utilize the fact that
'g' is differentiable.
454
00:21:48,600 --> 00:21:50,820
And what we do is-- and it
doesn't look like a very
455
00:21:50,820 --> 00:21:52,450
significant step, but it is.
456
00:21:52,450 --> 00:21:55,725
What we do is we take this
expression and both multiply
457
00:21:55,725 --> 00:21:58,370
it and divide it by 'delta x'.
458
00:21:58,370 --> 00:22:01,330
I don't know if you can see the
method to my madness here.
459
00:22:01,330 --> 00:22:04,780
The whole thing I'm setting up
here is that when I do this,
460
00:22:04,780 --> 00:22:07,970
the bracketed expression now
looks like the thing that
461
00:22:07,970 --> 00:22:11,380
yields 'g prime of x1' when
you take the limit.
462
00:22:11,380 --> 00:22:13,320
And now that's exactly
what I'm going to do.
463
00:22:13,320 --> 00:22:16,090
I'm going to take the limit of
this expression as 'delta x'
464
00:22:16,090 --> 00:22:17,790
approaches 0.
465
00:22:17,790 --> 00:22:19,975
The limit of a product is the
product of the limits.
466
00:22:19,975 --> 00:22:23,640
467
00:22:23,640 --> 00:22:26,240
As 'delta x' is allowed to
approach 0, this bracketed
468
00:22:26,240 --> 00:22:29,930
expression by definition becomes
'g prime of x1'.
469
00:22:29,930 --> 00:22:32,510
And obviously, this is just
the limit of 'delta x' as
470
00:22:32,510 --> 00:22:35,860
'delta x' approaches
0, which is 0.
471
00:22:35,860 --> 00:22:38,050
The fact that 'g' is
differentiable means that this
472
00:22:38,050 --> 00:22:42,110
is a finite number, and any
finite number times 0 is 0.
473
00:22:42,110 --> 00:22:45,230
In other words, the fact that
'g' was differentiable tells
474
00:22:45,230 --> 00:22:48,890
us that this limit is 0, and
that's the same as saying that
475
00:22:48,890 --> 00:22:54,680
the limit 'g of 'x1 plus delta
x'' as 'delta x' approaches 0
476
00:22:54,680 --> 00:22:57,170
is, in fact, 'g of x1'.
477
00:22:57,170 --> 00:23:00,700
As self-evident as it seems,
this is not a true result if
478
00:23:00,700 --> 00:23:03,600
the function 'g' is not
differentiable, or it may not
479
00:23:03,600 --> 00:23:07,200
be a true result if 'g'
is not differentiable.
480
00:23:07,200 --> 00:23:09,320
Well, so much for that.
481
00:23:09,320 --> 00:23:12,620
There is one more recipe that's
very important called
482
00:23:12,620 --> 00:23:14,310
the quotient rule.
483
00:23:14,310 --> 00:23:17,770
The one we just did was called
the product rule, how do you
484
00:23:17,770 --> 00:23:19,770
differentiate the product
of two functions.
485
00:23:19,770 --> 00:23:22,830
There is an analogous type of
recipe for differentiating the
486
00:23:22,830 --> 00:23:24,410
quotient of two functions.
487
00:23:24,410 --> 00:23:27,100
And since the proof is very
much the same as what I've
488
00:23:27,100 --> 00:23:29,860
already done, I leave
the details to you.
489
00:23:29,860 --> 00:23:31,910
They're in the textbook.
490
00:23:31,910 --> 00:23:34,060
You can go through that in
more detail if you wish.
491
00:23:34,060 --> 00:23:35,480
But the result is this.
492
00:23:35,480 --> 00:23:38,430
If 'f' and 'g' are
differentiable functions, the
493
00:23:38,430 --> 00:23:40,910
quotient is also
differentiable.
494
00:23:40,910 --> 00:23:43,150
And the way you get the result--
and again, notice how
495
00:23:43,150 --> 00:23:44,990
nonintuitive this is.
496
00:23:44,990 --> 00:23:46,900
Get this in terms of logic
if you're trying
497
00:23:46,900 --> 00:23:48,760
to memorize it logically.
498
00:23:48,760 --> 00:23:50,980
You take the denominator
times the
499
00:23:50,980 --> 00:23:52,730
derivative of the numerator.
500
00:23:52,730 --> 00:23:56,840
And you subtract off the
numerator times the derivative
501
00:23:56,840 --> 00:23:58,300
of the denominator.
502
00:23:58,300 --> 00:24:01,020
And then you divide the whole
thing by the square of the
503
00:24:01,020 --> 00:24:02,590
denominator.
504
00:24:02,590 --> 00:24:05,550
See, again, notice how
overwhelming this course
505
00:24:05,550 --> 00:24:08,820
becomes if you try to memorize
every result.
506
00:24:08,820 --> 00:24:09,980
Rather, do what?
507
00:24:09,980 --> 00:24:13,300
Memorize the basic definition
and derive these results.
508
00:24:13,300 --> 00:24:16,470
And believe me, once you use
these results long enough,
509
00:24:16,470 --> 00:24:17,810
you'll memorize them
automatically
510
00:24:17,810 --> 00:24:19,440
just by repeated use.
511
00:24:19,440 --> 00:24:22,390
Well, at any rate, let me give
you an example of using this.
512
00:24:22,390 --> 00:24:25,490
Remember before we could only
differentiate 'x' to the n if
513
00:24:25,490 --> 00:24:27,050
the exponent was positive?
514
00:24:27,050 --> 00:24:30,280
Suppose 'f of x' is 'x' to the
'minus n' where 'n' is now a
515
00:24:30,280 --> 00:24:31,490
positive integer.
516
00:24:31,490 --> 00:24:33,250
That means 'minus
n' is negative.
517
00:24:33,250 --> 00:24:35,660
How would we differentiate
this?
518
00:24:35,660 --> 00:24:38,160
And the idea here is we say
lookit, if this had been a
519
00:24:38,160 --> 00:24:40,190
positive 'n', we could
handle this.
520
00:24:40,190 --> 00:24:43,140
But 'x' to the 'minus n'
by definition is 1
521
00:24:43,140 --> 00:24:44,550
over 'x' to the 'n'.
522
00:24:44,550 --> 00:24:47,510
We know how to differentiate 'x'
to the 'n' when 'n' is a
523
00:24:47,510 --> 00:24:48,370
positive number.
524
00:24:48,370 --> 00:24:49,820
What do I have now?
525
00:24:49,820 --> 00:24:51,100
I have a quotient.
526
00:24:51,100 --> 00:24:52,750
So I use the quotient rule.
527
00:24:52,750 --> 00:24:54,230
'f prime of x' is what?
528
00:24:54,230 --> 00:24:57,130
It's my denominator, which is
'x' to the 'n', times the
529
00:24:57,130 --> 00:24:58,740
derivative of my numerator.
530
00:24:58,740 --> 00:25:01,800
My numerator is a constant so
it's the derivative of 0,
531
00:25:01,800 --> 00:25:04,300
minus the numerator--
that's minus 1--
532
00:25:04,300 --> 00:25:07,070
times the derivative
of the denominator.
533
00:25:07,070 --> 00:25:09,920
But for a positive integer
'n', we know that the
534
00:25:09,920 --> 00:25:14,130
derivative of 'x' to the 'n'
is 'nx' to the 'n - 1', all
535
00:25:14,130 --> 00:25:15,870
over the square of
the denominator.
536
00:25:15,870 --> 00:25:19,410
But the square of 'x' to the
'n' is 'x' to the '2n'.
537
00:25:19,410 --> 00:25:21,720
And if I now simplify
this, I have what?
538
00:25:21,720 --> 00:25:22,890
This is 0.
539
00:25:22,890 --> 00:25:24,580
This is 'minus n'.
540
00:25:24,580 --> 00:25:27,070
This is 'x' to the 'n - 1'.
541
00:25:27,070 --> 00:25:29,590
This comes upstairs
as a 'minus 2n'.
542
00:25:29,590 --> 00:25:35,060
So I have 'minus x' to
the 'n - 1 - 2n'.
543
00:25:35,060 --> 00:25:36,520
That's the same as what?
544
00:25:36,520 --> 00:25:41,380
'Minus nx' to the
'minus 'n - 1'.
545
00:25:41,380 --> 00:25:45,050
And by the way, this is a rather
interesting result now
546
00:25:45,050 --> 00:25:46,020
that we've proven it.
547
00:25:46,020 --> 00:25:49,100
Suppose we said to a person,
let's just bring down the
548
00:25:49,100 --> 00:25:51,970
exponential and replace
it by one less.
549
00:25:51,970 --> 00:25:55,460
Well, if we brought down the
exponent, that's 'minus n'.
550
00:25:55,460 --> 00:25:58,540
And if we replaced 'minus n'
by one less, that would be
551
00:25:58,540 --> 00:26:00,750
'minus 'n - 1'.
552
00:26:00,750 --> 00:26:03,150
And lo and behold, that's
precisely the
553
00:26:03,150 --> 00:26:04,740
result that we got.
554
00:26:04,740 --> 00:26:07,610
In other words, by using the
quotient rule, for example, we
555
00:26:07,610 --> 00:26:10,460
can now show that the rule
that says bring down the
556
00:26:10,460 --> 00:26:15,180
exponent and replace it by one
less applies for all integers,
557
00:26:15,180 --> 00:26:17,990
not just the positive
integers.
558
00:26:17,990 --> 00:26:20,960
In later lectures, we will
show that it applies to
559
00:26:20,960 --> 00:26:24,830
fractional exponents as well
as to any real number
560
00:26:24,830 --> 00:26:26,660
exponents as we go along.
561
00:26:26,660 --> 00:26:28,770
But I just wanted you to
see the structure here.
562
00:26:28,770 --> 00:26:32,370
And in fact, I think in terms
of what this lesson is all
563
00:26:32,370 --> 00:26:35,210
about, we have given
enough examples.
564
00:26:35,210 --> 00:26:39,490
I think we might just as well
summarize here, and let it be,
565
00:26:39,490 --> 00:26:42,340
and let the rest come from the
reading material and the
566
00:26:42,340 --> 00:26:45,090
learning exercises in general.
567
00:26:45,090 --> 00:26:46,920
And the summary is
simply this.
568
00:26:46,920 --> 00:26:51,880
The basic definition of a
derivative is 'f prime of x1'
569
00:26:51,880 --> 00:26:55,670
is the limit as 'delta x'
approaches 0, ''f of 'x1 plus
570
00:26:55,670 --> 00:26:59,540
delta x'' minus 'f of
x1'' over 'delta x'.
571
00:26:59,540 --> 00:27:02,440
That basic definition
never changes.
572
00:27:02,440 --> 00:27:07,890
We always mean this when
we write this.
573
00:27:07,890 --> 00:27:12,290
But what is important is that
this basic definition can be
574
00:27:12,290 --> 00:27:16,570
manipulated to yield, and now
I use quotation marks
575
00:27:16,570 --> 00:27:19,170
"convenient," because I don't
know how convenient something
576
00:27:19,170 --> 00:27:20,890
has to be before it's
really convenient,
577
00:27:20,890 --> 00:27:22,550
but convenient what?
578
00:27:22,550 --> 00:27:24,630
Recipes.
579
00:27:24,630 --> 00:27:25,300
Meaning what?
580
00:27:25,300 --> 00:27:28,410
Convenient ways to find
the derivative.
581
00:27:28,410 --> 00:27:30,300
You see, what's going to happen
in the remainder of our
582
00:27:30,300 --> 00:27:33,030
course as we shall soon see in
the next lecture, in fact,
583
00:27:33,030 --> 00:27:37,140
what we're going to start doing
is using derivatives the
584
00:27:37,140 --> 00:27:39,520
same as they always were
intended to be used.
585
00:27:39,520 --> 00:27:43,500
But now we are going to have
much sharper computational
586
00:27:43,500 --> 00:27:47,090
techniques for finding these
derivatives more rapidly than
587
00:27:47,090 --> 00:27:49,880
having to resort to the basic
limit definition.
588
00:27:49,880 --> 00:27:51,990
Well, enough about
that for now.
589
00:27:51,990 --> 00:27:53,380
Until next time, goodbye.
590
00:27:53,380 --> 00:27:56,410
591
00:27:56,410 --> 00:27:58,940
NARRATOR: Funding for the
publication of this video was
592
00:27:58,940 --> 00:28:03,660
provided by the Gabriella and
Paul Rosenbaum Foundation.
593
00:28:03,660 --> 00:28:07,830
Help OCW continue to provide
free and open access to MIT
594
00:28:07,830 --> 00:28:12,030
courses by making a donation
at ocw.mit.edu/donate.
595
00:28:12,030 --> 00:28:16,767