0:00:00.840,0:00:04.090
So I've been requested to do[br]the proof of the derivative of

0:00:04.090,0:00:06.300
the square root of x, so I[br]thought I would do a quick

0:00:06.300,0:00:08.300
video on the proof of the[br]derivative of the

0:00:08.300,0:00:10.370
square root of x.

0:00:10.370,0:00:13.680
So we know from the definition[br]of a derivative that the

0:00:13.680,0:00:22.280
derivative of the function[br]square root of x, that is equal

0:00:22.280,0:00:26.520
to-- let me switch colors, just[br]for a variety-- that's equal to

0:00:26.520,0:00:33.080
the limit as delta[br]x approaches 0.

0:00:33.080,0:00:35.595
And you know, some people[br]say h approaches 0,

0:00:35.595,0:00:36.360
or d approaches 0.

0:00:36.360,0:00:37.450
I just use delta x.

0:00:37.450,0:00:39.450
So the change in x over 0.

0:00:39.450,0:00:41.830
And then we say f of x[br]plus delta x, so in this

0:00:41.830,0:00:42.910
case this is f of x.

0:00:42.910,0:00:52.260
So it's the square root of x[br]plus delta x minus f of x,

0:00:52.260,0:00:54.640
which in this case it's[br]square root of x.

0:00:54.640,0:00:57.140
All of that over the change[br]in x, over delta x.

0:01:00.040,0:01:02.580
Right now when I look at that,[br]there's not much simplification

0:01:02.580,0:01:04.945
I can do to make this come out[br]with something meaningful.

0:01:09.940,0:01:12.540
I'm going to multiply the[br]numerator and the denominator

0:01:12.540,0:01:13.790
by the conjugate of the[br]numerator is what

0:01:13.790,0:01:14.200
I mean by that.

0:01:14.200,0:01:15.480
Let me rewrite it.

0:01:15.480,0:01:19.740
Limit is delta x approaching[br]0-- I'm just rewriting

0:01:19.740,0:01:21.280
what I have here.

0:01:21.280,0:01:26.650
So I said the square root[br]of x plus delta x minus

0:01:26.650,0:01:28.610
square root of x.

0:01:28.610,0:01:31.200
All of that over delta x.

0:01:31.200,0:01:34.490
And I'm going to multiply[br]that-- after switching colors--

0:01:34.490,0:01:41.840
times square root of x plus[br]delta x plus the square root of

0:01:41.840,0:01:48.260
x, over the square root of x[br]plus delta x plus the

0:01:48.260,0:01:49.250
square root of x.

0:01:49.250,0:01:53.420
This is just 1, so I could of[br]course multiply that times-- if

0:01:53.420,0:01:57.110
we assume that x and delta x[br]aren't both 0, this is a

0:01:57.110,0:01:59.090
defined number and[br]this will be 1.

0:01:59.090,0:02:00.010
And we can do that.

0:02:00.010,0:02:02.130
This is 1/1, we're just[br]multiplying it times this

0:02:02.130,0:02:10.900
equation, and we get limit[br]as delta x approaches 0.

0:02:10.900,0:02:13.510
This is a minus b[br]times a plus b.

0:02:13.510,0:02:15.360
Let me do little aside here.

0:02:15.360,0:02:20.880
Let me say a plus b times[br]a minus b is equal to a

0:02:20.880,0:02:23.150
squared minus b squared.

0:02:23.150,0:02:26.600
So this is a plus b[br]times a minus b.

0:02:26.600,0:02:29.410
So it's going to be[br]equal to a squared.

0:02:29.410,0:02:32.010
So what's this quantity squared[br]or this quantity squared,

0:02:32.010,0:02:33.180
either one, these are my a's.

0:02:33.180,0:02:35.450
Well it's just going[br]to be x plus delta x.

0:02:35.450,0:02:39.430
So we get x plus delta x.

0:02:39.430,0:02:41.050
And then what's b squared?

0:02:41.050,0:02:46.380
So minus square root of[br]x is b in this analogy.

0:02:46.380,0:02:50.640
So square root of x[br]squared is just x.

0:02:50.640,0:02:56.760
And all of that over delta[br]x times square root of x

0:02:56.760,0:03:04.210
plus delta x plus the[br]square root of x.

0:03:04.210,0:03:05.900
Let's see what[br]simplification we can do.

0:03:05.900,0:03:08.580
Well we have an x and[br]then a minus x, so those

0:03:08.580,0:03:11.480
cancel out. x minus x.

0:03:11.480,0:03:13.460
And then we're left in the[br]numerator and the denominator,

0:03:13.460,0:03:15.690
all we have is a delta x here[br]and a delta x here, so let's

0:03:15.690,0:03:18.770
divide the numerator and the[br]denominator by delta x.

0:03:18.770,0:03:22.822
So this goes to 1,[br]this goes to 1.

0:03:22.822,0:03:26.350
And so this equals the limit--[br]I'll write smaller, because I'm

0:03:26.350,0:03:34.920
running out of space-- limit as[br]delta x approaches 0 of 1 over.

0:03:34.920,0:03:37.780
And of course we can only do[br]this assuming that delta--

0:03:37.780,0:03:40.220
well, we're dividing by delta[br]x to begin with, so we know

0:03:40.220,0:03:42.420
it's not 0, it's just[br]approaching zero.

0:03:42.420,0:03:50.320
So we get square root[br]of x plus delta x plus

0:03:50.320,0:03:51.860
the square root of x.

0:03:51.860,0:03:53.550
And now we can just[br]directly take the limit

0:03:53.550,0:03:54.410
as it approaches 0.

0:03:54.410,0:03:56.440
We can just set delta[br]x as equal to 0.

0:03:56.440,0:03:58.140
That's what it's approaching.

0:03:58.140,0:04:04.260
So then that equals one[br]over the square root of x.

0:04:04.260,0:04:06.790
Right, delta x is 0, so[br]we can ignore that.

0:04:06.790,0:04:09.120
We could take the limit[br]all the way to 0.

0:04:09.120,0:04:13.000
And then this is of course just[br]a square root of x here plus

0:04:13.000,0:04:17.160
the square root of x,[br]and that equals 1 over

0:04:17.160,0:04:19.350
2 square root of x.

0:04:19.350,0:04:24.890
And that equals 1/2x[br]to the negative 1/2.

0:04:24.890,0:04:28.900
So we just proved that x to the[br]1/2 power, the derivative of it

0:04:28.900,0:04:35.220
is 1/2x to the negative 1/2,[br]and so it is consistent with

0:04:35.220,0:04:41.700
the general property that the[br]derivative of-- oh I don't

0:04:41.700,0:04:50.850
know-- the derivative of x to[br]the n is equal to nx to the n

0:04:50.850,0:04:55.150
minus 1, even in this case[br]where the n was 1/2.

0:04:55.150,0:04:56.100
Well hopefully[br]that's satisfying.

0:04:56.100,0:04:58.960
I didn't prove it for all[br]fractions but this is a start.

0:04:58.960,0:05:01.120
This is a common one you[br]see, square root of x, and

0:05:01.120,0:05:03.770
it's hopefully not too[br]complicated for proof.

0:05:03.770,0:05:05.180
I will see you in[br]future videos.