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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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00:01:02,390 --> 00:01:06,460
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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00:01:18,200 --> 00:01:21,000
qualitative approach to the
derivative to a more

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quantitative approach.

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00:01:22,540 --> 00:01:24,440
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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00:01:27,600 --> 00:01:28,780
mood over here.

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00:01:28,780 --> 00:01:30,120
Now, the idea is this.

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00:01:30,120 --> 00:01:33,140
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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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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00:02:26,470 --> 00:02:29,240
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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The 'x1 cubed' term here
cancels with the

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00:02:42,120 --> 00:02:43,850
'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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00:02:46,980 --> 00:02:50,170
'delta x'' plus '3x1
'delta x' squared''

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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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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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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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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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00:03:53,780 --> 00:03:55,890
Each of the remaining terms
have a factor of

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00:03:55,890 --> 00:03:57,140
'delta x' in them.

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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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00:04:07,630 --> 00:04:10,190
In particular, since 'x1'
could have been any real

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00:04:10,190 --> 00:04:12,870
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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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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00:04:40,250 --> 00:04:45,100
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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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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00:05:11,830 --> 00:05:16,050
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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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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recognizing that I'm going to
let 'delta x' eventually

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approach 0, I factor out a
'delta x squared' and say,

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lookit, the remaining terms
have the form 'delta x

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squared' times some
finite number.

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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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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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00:06:16,740 --> 00:06:19,620
When we approach the limit, we
are never allowed to equal it.

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'Delta x' never equals 0.

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Consequently, I can cancel a
'delta x' from the denominator

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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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single factor of 'delta
x' in this term.

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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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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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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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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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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?

163
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'.

167
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.

169
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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00:08:24,300 --> 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.

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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

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00:08:49,790 --> 00:08:52,320
the domain of 'g' for
this to make sense.

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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'.

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00:08:59,070 --> 00:09:02,050
In other words, it belongs to
both sets, the domain of 'f'

185
00:09:02,050 --> 00:09:03,340
and the domain of 'g'.

186
00:09:03,340 --> 00:09:06,810
Because if 'x' didn't belong to
at least one of these two,

187
00:09:06,810 --> 00:09:08,540
this expression wouldn't
even make sense.

188
00:09:08,540 --> 00:09:11,510
In other words, 'x' must be a
permissible input to both the

189
00:09:11,510 --> 00:09:13,100
'f' and 'g' machines.

190
00:09:13,100 --> 00:09:16,960
Now, what my claim is, is that
in this particular case, 'h

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00:09:16,960 --> 00:09:17,760
prime' exists.

192
00:09:17,760 --> 00:09:20,730
In other words, 'h' is also
differentiable, and it's

193
00:09:20,730 --> 00:09:24,190
obtained by adding the sum
of these two derivatives.

194
00:09:24,190 --> 00:09:26,730
In other words, this is the
result that's commonly known

195
00:09:26,730 --> 00:09:29,730
as the derivative of a sum is
the sum of the derivatives,

196
00:09:29,730 --> 00:09:31,530
which, of course, sounds like
it should be right.

197
00:09:31,530 --> 00:09:34,110
But I'll comment on that
in a moment also.

198
00:09:34,110 --> 00:09:34,340
what.

199
00:09:34,340 --> 00:09:38,460
I want to show is simply this,
that 'h prime of x1' is simply

200
00:09:38,460 --> 00:09:42,930
'f prime of x1' plus 'g
prime of x1', OK?

201
00:09:42,930 --> 00:09:46,440
And my main purpose is to
highlight the basic

202
00:09:46,440 --> 00:09:46,990
definition.

203
00:09:46,990 --> 00:09:49,070
That's what I want to give
you the drill on.

204
00:09:49,070 --> 00:09:53,670
Namely, how do we compute
'h prime of x1'?

205
00:09:53,670 --> 00:09:57,720
We must look at ''h of 'x1 plus
delta x'' minus 'h of

206
00:09:57,720 --> 00:10:01,770
x1'' over 'delta x', then take
the limit as 'delta x'

207
00:10:01,770 --> 00:10:02,530
approaches 0.

208
00:10:02,530 --> 00:10:05,580
That's the basic definition
that will be used over and

209
00:10:05,580 --> 00:10:06,850
over and over.

210
00:10:06,850 --> 00:10:10,790
At any rate, sparing you all
of the details here, let's

211
00:10:10,790 --> 00:10:11,980
simply observe what?

212
00:10:11,980 --> 00:10:15,940
That 'h of 'x1 plus delta x'',
since 'h' is the sum of 'f'

213
00:10:15,940 --> 00:10:19,410
and 'g', is 'f of 'x1 plus
delta x'' plus 'g of

214
00:10:19,410 --> 00:10:21,200
'x1 plus delta x''.

215
00:10:21,200 --> 00:10:25,910
Consequently, 'h of 'x1 plus
delta x'' minus 'h of x1' is

216
00:10:25,910 --> 00:10:29,970
'f of 'x1 plus delta x'' plus
'g of 'x1 plus delta x''--

217
00:10:29,970 --> 00:10:30,970
that's this part--

218
00:10:30,970 --> 00:10:35,350
minus ''f of x1' plus
'g of x1''.

219
00:10:35,350 --> 00:10:38,740
Now, the idea is I want to
divide this by 'delta x'.

220
00:10:38,740 --> 00:10:42,420
I also have some knowledge of
the differentiability of 'f'

221
00:10:42,420 --> 00:10:46,120
and 'g' separately, so I would
like to combine or regroup

222
00:10:46,120 --> 00:10:49,110
these terms to highlight
that fact for me.

223
00:10:49,110 --> 00:10:52,960
So what I do is I group these
two terms together, these two

224
00:10:52,960 --> 00:10:56,140
terms together, and then divide
through by 'delta x'.

225
00:10:56,140 --> 00:11:00,010
In other words, ''h of 'x1 plus
delta x'' minus 'h of x''

226
00:11:00,010 --> 00:11:02,040
over 'delta x' is what?

227
00:11:02,040 --> 00:11:06,030
''f of 'x1 plus delta x'' minus
'f of x1'' over 'delta

228
00:11:06,030 --> 00:11:10,470
x' plus ''g of 'x1 plus
delta x'' minus 'g of

229
00:11:10,470 --> 00:11:12,700
x1'' over 'delta x'.

230
00:11:12,700 --> 00:11:16,550
Now, for my last step, I simply
must take the limit as

231
00:11:16,550 --> 00:11:18,320
'delta x' goes to 0.

232
00:11:18,320 --> 00:11:20,910
Well, just look at this
bracketed expression.

233
00:11:20,910 --> 00:11:23,690
The limit of a sum is the
sum of the limits.

234
00:11:23,690 --> 00:11:25,930
So to take the limit of this
expression, I take the limit

235
00:11:25,930 --> 00:11:28,170
of these two terms separately
and add them.

236
00:11:28,170 --> 00:11:31,790
By definition, the limit of the
bracketed expressions as

237
00:11:31,790 --> 00:11:34,540
'delta x' goes to 0--
by definition--

238
00:11:34,540 --> 00:11:36,750
is 'f prime of x1'.

239
00:11:36,750 --> 00:11:41,190
And as 'delta x' goes to 0,
this term here becomes 'g

240
00:11:41,190 --> 00:11:42,810
prime of x1'.

241
00:11:42,810 --> 00:11:45,400
And by the way, for the last
time, let me give you this

242
00:11:45,400 --> 00:11:46,870
word of caution again.

243
00:11:46,870 --> 00:11:50,520
Do not make the mistake of
saying, gee, when 'delta x'

244
00:11:50,520 --> 00:11:53,180
goes to 0, my numerator is 0.

245
00:11:53,180 --> 00:11:55,190
Therefore, the fraction
must be 0.

246
00:11:55,190 --> 00:11:55,590
No!

247
00:11:55,590 --> 00:11:59,150
When 'delta x' goes to 0, sure,
if you replace 'delta x'

248
00:11:59,150 --> 00:12:03,530
by 0, the numerator is 0, but
so is the denominator.

249
00:12:03,530 --> 00:12:06,410
In other words, this is our old
story that the derivative

250
00:12:06,410 --> 00:12:10,980
becomes a study of 0/0 so if
you let 'delta x' equals 0.

251
00:12:10,980 --> 00:12:12,400
In other words, the whole
point is what?

252
00:12:12,400 --> 00:12:17,040
That as 'delta x' approaches
0, this by definition is 'f

253
00:12:17,040 --> 00:12:18,330
prime of x1'.

254
00:12:18,330 --> 00:12:21,070
This by definition is
'g prime of x1'.

255
00:12:21,070 --> 00:12:26,430
And therefore, we have proven
as a theorem 'h prime of x1'

256
00:12:26,430 --> 00:12:30,380
is 'f prime of x1' plus 'g
prime of x1', that the

257
00:12:30,380 --> 00:12:33,990
derivative of a sum is the
sum of the derivatives.

258
00:12:33,990 --> 00:12:36,960
Now, the main trouble so far is
that every result that I've

259
00:12:36,960 --> 00:12:40,215
proven rigorously for you, you
have probably guessed in

260
00:12:40,215 --> 00:12:41,720
advance that it was
going to happen.

261
00:12:41,720 --> 00:12:45,040
You say who needs this rigorous
stuff when we could

262
00:12:45,040 --> 00:12:47,200
have got the same result
by intuition.

263
00:12:47,200 --> 00:12:49,790
And this is where the rift
between the way the pure

264
00:12:49,790 --> 00:12:53,120
mathematician often teaches
calculus and the way, say, the

265
00:12:53,120 --> 00:12:56,510
applied man teaches calculus
often comes in.

266
00:12:56,510 --> 00:12:59,730
The point is that the places
that you can get into trouble

267
00:12:59,730 --> 00:13:02,110
often aren't stressed
soon enough.

268
00:13:02,110 --> 00:13:05,370
So before we go any further, let
me show you that you must

269
00:13:05,370 --> 00:13:08,780
be careful, that up until now,
we have been very fortunate

270
00:13:08,780 --> 00:13:11,110
when we say things like the
limit of a product is the

271
00:13:11,110 --> 00:13:13,520
product of the limits, fortunate
in the sense that

272
00:13:13,520 --> 00:13:16,750
they worked out the way the
wording seemed to indicate.

273
00:13:16,750 --> 00:13:21,070
Let me give you an example of
what I mean by saying beware

274
00:13:21,070 --> 00:13:22,610
of self-evident.

275
00:13:22,610 --> 00:13:26,900
I claim, for example, that the
derivative of a product is not

276
00:13:26,900 --> 00:13:29,160
the product of the derivatives
necessarily.

277
00:13:29,160 --> 00:13:32,530
And the best way to prove that
to you is not to say take my

278
00:13:32,530 --> 00:13:33,350
word for it.

279
00:13:33,350 --> 00:13:36,600
This is one of the beauties of
computational mathematics.

280
00:13:36,600 --> 00:13:40,520
You can always show by means of
an example what's going on.

281
00:13:40,520 --> 00:13:44,320
For example, let 'f of
x' be 'x + 1', let 'g

282
00:13:44,320 --> 00:13:46,460
of x' be 'x - 1'.

283
00:13:46,460 --> 00:13:49,380
Now, the derivative of
'x + 1' is simply 1.

284
00:13:49,380 --> 00:13:51,750
The derivative of 'x
- 1' is also 1.

285
00:13:51,750 --> 00:13:54,230
Namely, we differentiate
this as a sum.

286
00:13:54,230 --> 00:13:55,340
The derivative of 'x' is 1.

287
00:13:55,340 --> 00:13:57,870
The derivative of
a constant is 0.

288
00:13:57,870 --> 00:14:02,000
In other words, 'f prime of x'
is 1, 'g prime of x' is 1.

289
00:14:02,000 --> 00:14:05,710
On the other hand, if we first
multiply 'f' and 'g', 'x + 1'

290
00:14:05,710 --> 00:14:09,370
times 'x - 1' is 'x squared
- 1', and the

291
00:14:09,370 --> 00:14:10,740
derivative of that is what?

292
00:14:10,740 --> 00:14:13,420
Well, to differentiate 'x
squared', we just saw.

293
00:14:13,420 --> 00:14:14,850
Bring the exponent down.

294
00:14:14,850 --> 00:14:16,180
Replace it by one less.

295
00:14:16,180 --> 00:14:17,100
That's '2x'.

296
00:14:17,100 --> 00:14:19,170
The derivative of
minus 1 is 0.

297
00:14:19,170 --> 00:14:21,870
In other words, if you multiply
first and then take

298
00:14:21,870 --> 00:14:23,670
the derivative, you get '2x'.

299
00:14:23,670 --> 00:14:26,330
On the other hand, if you
differentiate first and then

300
00:14:26,330 --> 00:14:28,390
take the product, you get 2.

301
00:14:28,390 --> 00:14:31,710
Now, it is not true that '2x'
is a synonym for 1 for all

302
00:14:31,710 --> 00:14:32,790
values of 'x'.

303
00:14:32,790 --> 00:14:38,400
In other words, if you
differentiate first and then

304
00:14:38,400 --> 00:14:41,620
take the product, you get a
different answer than if you

305
00:14:41,620 --> 00:14:44,030
multiply first and then
differentiate.

306
00:14:44,030 --> 00:14:46,480
In other words, self-evident
or not, the thing

307
00:14:46,480 --> 00:14:48,180
happens to be false.

308
00:14:48,180 --> 00:14:51,870
In fact, let's see how the
theory helps us in this case.

309
00:14:51,870 --> 00:14:54,700
See, what I've shown you
here is that this

310
00:14:54,700 --> 00:14:56,710
result isn't true.

311
00:14:56,710 --> 00:14:58,510
Oh, I should say that this
result is true, that the

312
00:14:58,510 --> 00:14:59,880
equality isn't true.

313
00:14:59,880 --> 00:15:02,760
What I haven't shown you
is what is true.

314
00:15:02,760 --> 00:15:05,700
And again, to emphasize
our basic definition,

315
00:15:05,700 --> 00:15:06,870
let's do that now.

316
00:15:06,870 --> 00:15:09,970
Let's show how we can use our
basic definition to find the

317
00:15:09,970 --> 00:15:11,060
correct result.

318
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'.

319
00:15:15,050 --> 00:15:18,480
And we're going to use the
same structure as before.

320
00:15:18,480 --> 00:15:22,720
We are going to compute 'h of
'x1 plus delta x'', subtract

321
00:15:22,720 --> 00:15:26,840
'h of x1' divide by 'delta x',
and take the limit as 'delta

322
00:15:26,840 --> 00:15:27,800
x' approaches 0.

323
00:15:27,800 --> 00:15:29,560
We do that every time.

324
00:15:29,560 --> 00:15:32,210
All that changes is the property
of the particular

325
00:15:32,210 --> 00:15:33,760
function that we're
dealing with.

326
00:15:33,760 --> 00:15:37,860
In any event, if I do this, 'h
of 'x1 plus delta x'' is

327
00:15:37,860 --> 00:15:42,760
obtained by replacing 'x' in
here by 'x1 plus delta x'.

328
00:15:42,760 --> 00:15:46,000
And 'h of x1' is obtained by
replacing 'x' by 'x1' over

329
00:15:46,000 --> 00:15:50,140
here, so I wind up with 'f of
'x1 plus delta x'' times 'g of

330
00:15:50,140 --> 00:15:55,430
'x1 plus delta x'' minus 'f
of x1' times 'g of x1'.

331
00:15:55,430 --> 00:15:59,230
And now the problem is I would
like somehow to be able to get

332
00:15:59,230 --> 00:16:03,880
terms involving 'f of 'x1 plus
delta x'' minus 'f of x1', 'g

333
00:16:03,880 --> 00:16:07,810
of 'x1 plus delta x'' minus 'g
of x1', and I use the same

334
00:16:07,810 --> 00:16:11,550
trick here that I used when I
proved that the limit of a

335
00:16:11,550 --> 00:16:13,480
product is the product
of the limits.

336
00:16:13,480 --> 00:16:18,050
Namely, I am going to add in
and subtract the same term,

337
00:16:18,050 --> 00:16:19,770
and this will help me factor.

338
00:16:19,770 --> 00:16:22,020
In other words, what I'm going
to do here is this.

339
00:16:22,020 --> 00:16:23,590
I write down this term.

340
00:16:23,590 --> 00:16:27,370
Now I say to myself I would like
an 'f of x1' term to go

341
00:16:27,370 --> 00:16:29,610
with the 'f of 'x1
plus delta x''.

342
00:16:29,610 --> 00:16:32,280
So what I do is I write
minus 'f of x1'.

343
00:16:32,280 --> 00:16:35,400
Then I observe that 'g of 'x1
plus delta x'' is a factor

344
00:16:35,400 --> 00:16:38,010
here, so I put that
in over here.

345
00:16:38,010 --> 00:16:39,830
See, 'g of 'x1 plus delta x''.

346
00:16:39,830 --> 00:16:42,720
See, in other words, I can now
factor out a 'g of 'x1 plus

347
00:16:42,720 --> 00:16:46,130
delta x'' from these two terms
and have the formula that I

348
00:16:46,130 --> 00:16:47,940
want left over.

349
00:16:47,940 --> 00:16:50,250
But, of course, this term
doesn't belong here.

350
00:16:50,250 --> 00:16:53,700
I just added it myself, or I
should say subtracted it, so

351
00:16:53,700 --> 00:16:56,790
now I put it in with the
opposite sign, namely, plus 'f

352
00:16:56,790 --> 00:16:59,260
of x1' 'g of 'x1
plus delta x''.

353
00:16:59,260 --> 00:17:01,840
And now put this term back
in here: minus 'f of

354
00:17:01,840 --> 00:17:03,700
x1' times 'g of x1'.

355
00:17:03,700 --> 00:17:06,420
In other words, all I've done
is rewritten this, but by

356
00:17:06,420 --> 00:17:10,420
adding and subtracting the same
term, which does what?

357
00:17:10,420 --> 00:17:14,700
Dividing through by 'delta x',
I factor out a 'g of 'x1 plus

358
00:17:14,700 --> 00:17:17,440
delta x' from these two terms.

359
00:17:17,440 --> 00:17:21,359
That leaves me with ''f of 'x1
plus delta x'' minus 'f of x1'

360
00:17:21,359 --> 00:17:25,040
over 'delta x'' because I'm
dividing through by 'delta x'.

361
00:17:25,040 --> 00:17:28,910
Now, what I do is I factor out
'f of x1' from here and divide

362
00:17:28,910 --> 00:17:32,500
what's left through by 'delta
x'-- that gives me 'f of x1'--

363
00:17:32,500 --> 00:17:36,360
times the quantity ''g of 'x1
plus delta x'' minus 'g of

364
00:17:36,360 --> 00:17:38,720
x1'' over 'delta x'.

365
00:17:38,720 --> 00:17:40,170
I get this thing over here.

366
00:17:40,170 --> 00:17:42,130
Now, my final step is what?

367
00:17:42,130 --> 00:17:45,180
I take the limit as 'delta
x' approaches 0.

368
00:17:45,180 --> 00:17:47,350
Well, let's do the
easy part first.

369
00:17:47,350 --> 00:17:52,750
As 'delta x' approaches 0, this
becomes 'f prime of x1'.

370
00:17:52,750 --> 00:17:57,830
And as 'delta x' approaches 0,
this becomes 'g prime of x1'.

371
00:17:57,830 --> 00:18:00,260
This, of course, remains
'f of x1'.

372
00:18:00,260 --> 00:18:03,470
And I guess you would argue
intuitively that as 'delta x'

373
00:18:03,470 --> 00:18:07,580
approaches 0, 'x1 plus delta
x' approaches 'x1'.

374
00:18:07,580 --> 00:18:10,060
Therefore, 'g of 'x1
plus delta x''

375
00:18:10,060 --> 00:18:11,940
approaches 'g of x1'.

376
00:18:11,940 --> 00:18:14,170
That would give you this
result over here.

377
00:18:14,170 --> 00:18:17,160
I have put a little asterisk
over here because I want to

378
00:18:17,160 --> 00:18:19,080
make a footnote about this.

379
00:18:19,080 --> 00:18:21,560
You know, even though you would
have allowed me to say

380
00:18:21,560 --> 00:18:24,850
that this approaches 'g of x1'
as 'delta x' approaches 0, it

381
00:18:24,850 --> 00:18:29,130
is not in general true that we
can be this sloppy about this.

382
00:18:29,130 --> 00:18:32,560
I want to mention that later,
but meanwhile, I don't want to

383
00:18:32,560 --> 00:18:35,200
obscure the result that
I'm driving at.

384
00:18:35,200 --> 00:18:38,480
Namely, assuming that this is
a proper step, what we have

385
00:18:38,480 --> 00:18:42,250
shown now is that to
differentiate a product, you

386
00:18:42,250 --> 00:18:44,090
differentiate the
first factor and

387
00:18:44,090 --> 00:18:46,190
multiply that by the second.

388
00:18:46,190 --> 00:18:49,850
Then add on to that the first
factor times the derivative of

389
00:18:49,850 --> 00:18:50,760
the second.

390
00:18:50,760 --> 00:18:53,820
Now, that may not be
self-evident, but what we have

391
00:18:53,820 --> 00:18:58,080
shown is that self-evident or
not, this result follows

392
00:18:58,080 --> 00:19:01,200
inescapably from our basic
definitions and

393
00:19:01,200 --> 00:19:02,730
our previous theorems.

394
00:19:02,730 --> 00:19:05,940
And by the way, to finish off
the example that we started

395
00:19:05,940 --> 00:19:12,890
where 'f of x' was 'x + 1' and
'g of x' was 'x - 1', notice

396
00:19:12,890 --> 00:19:16,740
that if we use this recipe,
'f prime of x1' is 1.

397
00:19:16,740 --> 00:19:19,540
'g of x1' is 'x1 - 1'.

398
00:19:19,540 --> 00:19:22,300
'f of x1' is 'x1 + 1'.

399
00:19:22,300 --> 00:19:25,380
'g prime of x1' is 1.

400
00:19:25,380 --> 00:19:28,040
And if we now combine all of
these terms, we get twice

401
00:19:28,040 --> 00:19:32,180
'x1', and that's exactly what
the derivative of the product

402
00:19:32,180 --> 00:19:32,710
should have been.

403
00:19:32,710 --> 00:19:34,720
In other words, just to
summarize this thing off,

404
00:19:34,720 --> 00:19:38,760
notice that when we were back
over here, this was the result

405
00:19:38,760 --> 00:19:40,350
that we were supposed to get.

406
00:19:40,350 --> 00:19:42,520
In other words, to differentiate
a product, our

407
00:19:42,520 --> 00:19:45,100
basic formula tells us that
it's the first times the

408
00:19:45,100 --> 00:19:47,280
derivative of the second plus
the second times the

409
00:19:47,280 --> 00:19:49,150
derivative of the first.

410
00:19:49,150 --> 00:19:51,770
And let me just go back to this
parenthetical footnote

411
00:19:51,770 --> 00:19:55,130
that I wanted to mention
for you.

412
00:19:55,130 --> 00:19:59,940
It is not always that the
limit of 'g of t' as 't'

413
00:19:59,940 --> 00:20:02,800
approaches 'a' is 'g of a'.

414
00:20:02,800 --> 00:20:06,530
Remember, our whole study of
limits said you cannot replace

415
00:20:06,530 --> 00:20:09,920
't' by 'a' or 'a' by
't' automatically.

416
00:20:09,920 --> 00:20:12,570
In fact what, type of examples
did we see that this got us

417
00:20:12,570 --> 00:20:13,360
into trouble?

418
00:20:13,360 --> 00:20:17,510
For example, let 'g of t'
be ''t squared - 1'

419
00:20:17,510 --> 00:20:21,190
over 't - 1'', OK?

420
00:20:21,190 --> 00:20:23,940
Then the limit of 'g of t' as
't' approaches 1, well, as

421
00:20:23,940 --> 00:20:27,250
long as 't' is not 1,
't - 1' is not 0.

422
00:20:27,250 --> 00:20:29,690
We can cancel the 't - 1' from
the numerator and the

423
00:20:29,690 --> 00:20:30,670
denominator.

424
00:20:30,670 --> 00:20:33,210
That leaves us with 't + 1'.

425
00:20:33,210 --> 00:20:37,010
As 't' approaches 1, 't
+ 1' approaches 2.

426
00:20:37,010 --> 00:20:39,610
In other words, the limit of 'g
of t' as 't' approaches 1

427
00:20:39,610 --> 00:20:43,350
is 2, but 'g of 1' doesn't
even exist.

428
00:20:43,350 --> 00:20:45,800
'g of 1' is 0/0.

429
00:20:45,800 --> 00:20:49,260
In other words, it's not true
here that as 't' approaches 1,

430
00:20:49,260 --> 00:20:52,420
'g of t' approaches 'g of 1'.

431
00:20:52,420 --> 00:20:55,440
In other words, you can say
that when 'x1' approaches

432
00:20:55,440 --> 00:20:58,820
'x2', 'f of x1' approaches
is 'f of x2'.

433
00:20:58,820 --> 00:21:00,610
It doesn't have to be true.

434
00:21:00,610 --> 00:21:04,590
Why then could I use it in
my particular example?

435
00:21:04,590 --> 00:21:08,860
And by the way, the situation in
which the limit of 'g of t'

436
00:21:08,860 --> 00:21:12,250
as 't' approaches 'a' is 'g of
a' comes under the heading of

437
00:21:12,250 --> 00:21:16,160
a subject called continuity, and
that is a later lecture in

438
00:21:16,160 --> 00:21:17,240
this particular block.

439
00:21:17,240 --> 00:21:19,240
We'll talk about it in
more detail then.

440
00:21:19,240 --> 00:21:22,510
But for the time being, let me
show you why we could use this

441
00:21:22,510 --> 00:21:24,050
in our present case.

442
00:21:24,050 --> 00:21:25,860
Again, we use a trick.

443
00:21:25,860 --> 00:21:29,780
We want to show that as 'x1 plus
delta x' gets close to

444
00:21:29,780 --> 00:21:34,300
'x1', 'g of 'x1 plus delta x''
gets close to 'g of x1'.

445
00:21:34,300 --> 00:21:36,700
And to do that, that's the
same as showing that this

446
00:21:36,700 --> 00:21:40,860
difference gets close to 0 as
'delta x' gets close to 0.

447
00:21:40,860 --> 00:21:43,580
What we do is utilize
the fact, and

448
00:21:43,580 --> 00:21:44,800
this is very important.

449
00:21:44,800 --> 00:21:48,600
We utilize the fact that
'g' is differentiable.

450
00:21:48,600 --> 00:21:50,820
And what we do is-- and it
doesn't look like a very

451
00:21:50,820 --> 00:21:52,450
significant step, but it is.

452
00:21:52,450 --> 00:21:55,725
What we do is we take this
expression and both multiply

453
00:21:55,725 --> 00:21:58,370
it and divide it by 'delta x'.

454
00:21:58,370 --> 00:22:01,330
I don't know if you can see the
method to my madness here.

455
00:22:01,330 --> 00:22:04,780
The whole thing I'm setting up
here is that when I do this,

456
00:22:04,780 --> 00:22:07,970
the bracketed expression now
looks like the thing that

457
00:22:07,970 --> 00:22:11,380
yields 'g prime of x1' when
you take the limit.

458
00:22:11,380 --> 00:22:13,320
And now that's exactly
what I'm going to do.

459
00:22:13,320 --> 00:22:16,090
I'm going to take the limit of
this expression as 'delta x'

460
00:22:16,090 --> 00:22:17,790
approaches 0.

461
00:22:17,790 --> 00:22:19,975
The limit of a product is the
product of the limits.

462
00:22:23,640 --> 00:22:26,240
As 'delta x' is allowed to
approach 0, this bracketed

463
00:22:26,240 --> 00:22:29,930
expression by definition becomes
'g prime of x1'.

464
00:22:29,930 --> 00:22:32,510
And obviously, this is just
the limit of 'delta x' as

465
00:22:32,510 --> 00:22:35,860
'delta x' approaches
0, which is 0.

466
00:22:35,860 --> 00:22:38,050
The fact that 'g' is
differentiable means that this

467
00:22:38,050 --> 00:22:42,110
is a finite number, and any
finite number times 0 is 0.

468
00:22:42,110 --> 00:22:45,230
In other words, the fact that
'g' was differentiable tells

469
00:22:45,230 --> 00:22:48,890
us that this limit is 0, and
that's the same as saying that

470
00:22:48,890 --> 00:22:54,680
the limit 'g of 'x1 plus delta
x'' as 'delta x' approaches 0

471
00:22:54,680 --> 00:22:57,170
is, in fact, 'g of x1'.

472
00:22:57,170 --> 00:23:00,700
As self-evident as it seems,
this is not a true result if

473
00:23:00,700 --> 00:23:03,600
the function 'g' is not
differentiable, or it may not

474
00:23:03,600 --> 00:23:07,200
be a true result if 'g'
is not differentiable.

475
00:23:07,200 --> 00:23:09,320
Well, so much for that.

476
00:23:09,320 --> 00:23:12,620
There is one more recipe that's
very important called

477
00:23:12,620 --> 00:23:14,310
the quotient rule.

478
00:23:14,310 --> 00:23:17,770
The one we just did was called
the product rule, how do you

479
00:23:17,770 --> 00:23:19,770
differentiate the product
of two functions.

480
00:23:19,770 --> 00:23:22,830
There is an analogous type of
recipe for differentiating the

481
00:23:22,830 --> 00:23:24,410
quotient of two functions.

482
00:23:24,410 --> 00:23:27,100
And since the proof is very
much the same as what I've

483
00:23:27,100 --> 00:23:29,860
already done, I leave
the details to you.

484
00:23:29,860 --> 00:23:31,910
They're in the textbook.

485
00:23:31,910 --> 00:23:34,060
You can go through that in
more detail if you wish.

486
00:23:34,060 --> 00:23:35,480
But the result is this.

487
00:23:35,480 --> 00:23:38,430
If 'f' and 'g' are
differentiable functions, the

488
00:23:38,430 --> 00:23:40,910
quotient is also
differentiable.

489
00:23:40,910 --> 00:23:43,150
And the way you get the result--
and again, notice how

490
00:23:43,150 --> 00:23:44,990
nonintuitive this is.

491
00:23:44,990 --> 00:23:46,900
Get this in terms of logic
if you're trying

492
00:23:46,900 --> 00:23:48,760
to memorize it logically.

493
00:23:48,760 --> 00:23:50,980
You take the denominator
times the

494
00:23:50,980 --> 00:23:52,730
derivative of the numerator.

495
00:23:52,730 --> 00:23:56,840
And you subtract off the
numerator times the derivative

496
00:23:56,840 --> 00:23:58,300
of the denominator.

497
00:23:58,300 --> 00:24:01,020
And then you divide the whole
thing by the square of the

498
00:24:01,020 --> 00:24:02,590
denominator.

499
00:24:02,590 --> 00:24:05,550
See, again, notice how
overwhelming this course

500
00:24:05,550 --> 00:24:08,820
becomes if you try to memorize
every result.

501
00:24:08,820 --> 00:24:09,980
Rather, do what?

502
00:24:09,980 --> 00:24:13,300
Memorize the basic definition
and derive these results.

503
00:24:13,300 --> 00:24:16,470
And believe me, once you use
these results long enough,

504
00:24:16,470 --> 00:24:17,810
you'll memorize them
automatically

505
00:24:17,810 --> 00:24:19,440
just by repeated use.

506
00:24:19,440 --> 00:24:22,390
Well, at any rate, let me give
you an example of using this.

507
00:24:22,390 --> 00:24:25,490
Remember before we could only
differentiate 'x' to the n if

508
00:24:25,490 --> 00:24:27,050
the exponent was positive?

509
00:24:27,050 --> 00:24:30,280
Suppose 'f of x' is 'x' to the
'minus n' where 'n' is now a

510
00:24:30,280 --> 00:24:31,490
positive integer.

511
00:24:31,490 --> 00:24:33,250
That means 'minus
n' is negative.

512
00:24:33,250 --> 00:24:35,660
How would we differentiate
this?

513
00:24:35,660 --> 00:24:38,160
And the idea here is we say
lookit, if this had been a

514
00:24:38,160 --> 00:24:40,190
positive 'n', we could
handle this.

515
00:24:40,190 --> 00:24:43,140
But 'x' to the 'minus n'
by definition is 1

516
00:24:43,140 --> 00:24:44,550
over 'x' to the 'n'.

517
00:24:44,550 --> 00:24:47,510
We know how to differentiate 'x'
to the 'n' when 'n' is a

518
00:24:47,510 --> 00:24:48,370
positive number.

519
00:24:48,370 --> 00:24:49,820
What do I have now?

520
00:24:49,820 --> 00:24:51,100
I have a quotient.

521
00:24:51,100 --> 00:24:52,750
So I use the quotient rule.

522
00:24:52,750 --> 00:24:54,230
'f prime of x' is what?

523
00:24:54,230 --> 00:24:57,130
It's my denominator, which is
'x' to the 'n', times the

524
00:24:57,130 --> 00:24:58,740
derivative of my numerator.

525
00:24:58,740 --> 00:25:01,800
My numerator is a constant so
it's the derivative of 0,

526
00:25:01,800 --> 00:25:04,300
minus the numerator--
that's minus 1--

527
00:25:04,300 --> 00:25:07,070
times the derivative
of the denominator.

528
00:25:07,070 --> 00:25:09,920
But for a positive integer
'n', we know that the

529
00:25:09,920 --> 00:25:14,130
derivative of 'x' to the 'n'
is 'nx' to the 'n - 1', all

530
00:25:14,130 --> 00:25:15,870
over the square of
the denominator.

531
00:25:15,870 --> 00:25:19,410
But the square of 'x' to the
'n' is 'x' to the '2n'.

532
00:25:19,410 --> 00:25:21,720
And if I now simplify
this, I have what?

533
00:25:21,720 --> 00:25:22,890
This is 0.

534
00:25:22,890 --> 00:25:24,580
This is 'minus n'.

535
00:25:24,580 --> 00:25:27,070
This is 'x' to the 'n - 1'.

536
00:25:27,070 --> 00:25:29,590
This comes upstairs
as a 'minus 2n'.

537
00:25:29,590 --> 00:25:35,060
So I have 'minus x' to
the 'n - 1 - 2n'.

538
00:25:35,060 --> 00:25:36,520
That's the same as what?

539
00:25:36,520 --> 00:25:41,380
'Minus nx' to the
'minus 'n - 1'.

540
00:25:41,380 --> 00:25:45,050
And by the way, this is a rather
interesting result now

541
00:25:45,050 --> 00:25:46,020
that we've proven it.

542
00:25:46,020 --> 00:25:49,100
Suppose we said to a person,
let's just bring down the

543
00:25:49,100 --> 00:25:51,970
exponential and replace
it by one less.

544
00:25:51,970 --> 00:25:55,460
Well, if we brought down the
exponent, that's 'minus n'.

545
00:25:55,460 --> 00:25:58,540
And if we replaced 'minus n'
by one less, that would be

546
00:25:58,540 --> 00:26:00,750
'minus 'n - 1'.

547
00:26:00,750 --> 00:26:03,150
And lo and behold, that's
precisely the

548
00:26:03,150 --> 00:26:04,740
result that we got.

549
00:26:04,740 --> 00:26:07,610
In other words, by using the
quotient rule, for example, we

550
00:26:07,610 --> 00:26:10,460
can now show that the rule
that says bring down the

551
00:26:10,460 --> 00:26:15,180
exponent and replace it by one
less applies for all integers,

552
00:26:15,180 --> 00:26:17,990
not just the positive
integers.

553
00:26:17,990 --> 00:26:20,960
In later lectures, we will
show that it applies to

554
00:26:20,960 --> 00:26:24,830
fractional exponents as well
as to any real number

555
00:26:24,830 --> 00:26:26,660
exponents as we go along.

556
00:26:26,660 --> 00:26:28,770
But I just wanted you to
see the structure here.

557
00:26:28,770 --> 00:26:32,370
And in fact, I think in terms
of what this lesson is all

558
00:26:32,370 --> 00:26:35,210
about, we have given
enough examples.

559
00:26:35,210 --> 00:26:39,490
I think we might just as well
summarize here, and let it be,

560
00:26:39,490 --> 00:26:42,340
and let the rest come from the
reading material and the

561
00:26:42,340 --> 00:26:45,090
learning exercises in general.

562
00:26:45,090 --> 00:26:46,920
And the summary is
simply this.

563
00:26:46,920 --> 00:26:51,880
The basic definition of a
derivative is 'f prime of x1'

564
00:26:51,880 --> 00:26:55,670
is the limit as 'delta x'
approaches 0, ''f of 'x1 plus

565
00:26:55,670 --> 00:26:59,540
delta x'' minus 'f of
x1'' over 'delta x'.

566
00:26:59,540 --> 00:27:02,440
That basic definition
never changes.

567
00:27:02,440 --> 00:27:07,890
We always mean this when
we write this.

568
00:27:07,890 --> 00:27:12,290
But what is important is that
this basic definition can be

569
00:27:12,290 --> 00:27:16,570
manipulated to yield, and now
I use quotation marks

570
00:27:16,570 --> 00:27:19,170
"convenient," because I don't
know how convenient something

571
00:27:19,170 --> 00:27:20,890
has to be before it's
really convenient,

572
00:27:20,890 --> 00:27:22,550
but convenient what?

573
00:27:22,550 --> 00:27:24,630
Recipes.

574
00:27:24,630 --> 00:27:25,300
Meaning what?

575
00:27:25,300 --> 00:27:28,410
Convenient ways to find
the derivative.

576
00:27:28,410 --> 00:27:30,300
You see, what's going to happen
in the remainder of our

577
00:27:30,300 --> 00:27:33,030
course as we shall soon see in
the next lecture, in fact,

578
00:27:33,030 --> 00:27:37,140
what we're going to start doing
is using derivatives the

579
00:27:37,140 --> 00:27:39,520
same as they always were
intended to be used.

580
00:27:39,520 --> 00:27:43,500
But now we are going to have
much sharper computational

581
00:27:43,500 --> 00:27:47,090
techniques for finding these
derivatives more rapidly than

582
00:27:47,090 --> 00:27:49,880
having to resort to the basic
limit definition.

583
00:27:49,880 --> 00:27:51,990
Well, enough about
that for now.

584
00:27:51,990 --> 00:27:53,380
Until next time, goodbye.

585
00:27:56,410 --> 00:27:58,940
NARRATOR: Funding for the
publication of this video was

586
00:27:58,940 --> 00:28:03,660
provided by the Gabriella and
Paul Rosenbaum Foundation.

587
00:28:03,660 --> 00:28:07,830
Help OCW continue to provide
free and open access to MIT

588
00:28:07,830 --> 00:28:12,030
courses by making a donation
at ocw.mit.edu/donate.