0:00:02.640,0:00:07.671
In this video, we're going to[br]be looking at how we can use

0:00:07.671,0:00:10.767
logarithms to help simplify[br]certain functions before we

0:00:10.767,0:00:11.541
differentiate them.

0:00:12.720,0:00:17.160
To start off with, let's[br]just remind ourselves

0:00:17.160,0:00:21.045
about logarithms[br]themselves. So if we have

0:00:21.045,0:00:27.150
Y equals the log of X,[br]then this means log to

0:00:27.150,0:00:32.145
base E, and if we[br]differentiate that why by

0:00:32.145,0:00:36.030
DX is equal to one over X?

0:00:37.380,0:00:41.678
But of course. Inside this[br]logarithm here this might

0:00:41.678,0:00:47.880
not just be X, it might be a[br]function of X, so it might

0:00:47.880,0:00:52.310
be why equals the log of a[br]function of X.

0:00:53.550,0:00:57.834
So what happens when we[br]differentiate this? Well, we

0:00:57.834,0:01:03.546
have a rule that tells us why by[br]DX is equal to.

0:01:04.400,0:01:09.611
The derivative of F[br]of X. That's F dash

0:01:09.611,0:01:12.506
decks over F of X.

0:01:14.290,0:01:18.110
Now. We're going to be making[br]use these results.

0:01:18.910,0:01:24.470
But also the properties of[br]logarithms themselves so.

0:01:27.090,0:01:31.320
Let's take our first[br]example, why?

0:01:32.700,0:01:34.480
Equals the natural log.

0:01:36.440,0:01:42.533
3X to the 4th plus[br]Seven all raised to

0:01:42.533,0:01:44.564
the fifth power.

0:01:45.850,0:01:48.104
And we want to be able to

0:01:48.104,0:01:50.030
differentiate this. Well.

0:01:50.590,0:01:55.320
It looks very complicated in[br]here, but here we are raising.

0:01:56.020,0:02:01.684
3X to the power 4 + 7 to the[br]power five and one of the things

0:02:01.684,0:02:05.932
we do know about logarithms is[br]that if we are raising what's

0:02:05.932,0:02:06.994
inside the log.

0:02:07.870,0:02:12.690
To the power five, that's[br]exactly the same as multiplying

0:02:12.690,0:02:19.920
the log by 5. So In other words,[br]we can rewrite this as Y equals

0:02:19.920,0:02:27.632
5 times the log of three X to[br]the 4th plus 7 using our laws of

0:02:27.632,0:02:32.452
logarithms. And now this is now[br]much easier to differentiate

0:02:32.452,0:02:36.790
because this inside the log is[br]as very straightforward

0:02:36.790,0:02:42.860
function. So we can now[br]differentiate the why by DX is

0:02:42.860,0:02:45.470
equal to. 5.

0:02:46.280,0:02:51.296
Times by now, let's remember[br]what we do. This is the

0:02:51.296,0:02:54.944
function inside the log, so[br]we differentiate that.

0:02:56.500,0:03:04.420
So that gives us 12X cubed[br]and the derivative of the 70s

0:03:04.420,0:03:11.020
zero over the function itself 3X[br]to the 4th +7.

0:03:11.790,0:03:17.740
And if we just do a little bit[br]of simplifying five times by 12

0:03:17.740,0:03:23.690
gives us 60 X cubed over 3X to[br]the 4th plus Seven and what

0:03:23.690,0:03:27.940
looked like up here quieter[br]fearsome derivative that we were

0:03:27.940,0:03:33.040
going to have to do here, it's[br]turned out quite simply. So

0:03:33.040,0:03:35.590
let's have a look at another

0:03:35.590,0:03:41.210
example. Supposing we've got Y[br]equals the log of.

0:03:42.300,0:03:49.748
We've got 1 - 3 X divided[br]by 1 + 2 X. Now what

0:03:49.748,0:03:56.664
we've got here is the log of[br]a quotient, and there is a

0:03:56.664,0:03:59.324
formula for differentiating[br]a quotient.

0:04:00.500,0:04:05.373
That's going to lead to a very[br]very complicated expression, but

0:04:05.373,0:04:11.132
what we can do is make use of[br]the laws of logarithms because

0:04:11.132,0:04:16.891
the laws of logarithms tell us[br]that when we're doing the log of

0:04:16.891,0:04:22.207
a quotient than that is simply[br]the log of the numerator minus

0:04:22.207,0:04:26.637
the log of the denominator.[br]Because in logarithms we do

0:04:26.637,0:04:30.624
division by subtracting the logs[br]of the respective parts.

0:04:31.050,0:04:35.660
And now these two are both very[br]easy to differentiate.

0:04:36.500,0:04:39.836
So we can have the why by DX

0:04:39.836,0:04:45.927
equals. And this is a function[br]of X 1 - 3 X, so we

0:04:45.927,0:04:47.403
differentiate it that's minus

0:04:47.403,0:04:53.230
three. Over the function[br]itself, 1 - 3 X minus that

0:04:53.230,0:04:58.477
minus sign, and now we do[br]the same for this logarithm.

0:04:58.477,0:05:02.770
Here's the function of X[br]which we differentiate, so

0:05:02.770,0:05:08.971
the derivative of 1 + 2 X is[br]2 over the function itself.

0:05:08.971,0:05:10.879
1 + 2 X.

0:05:12.070,0:05:17.790
Now that is a penalty to pay[br]for getting away with such a

0:05:17.790,0:05:19.990
straightforward[br]differentiation. What we have

0:05:19.990,0:05:25.270
to do now is put these two[br]together. We have to bring

0:05:25.270,0:05:29.670
them together all over the[br]same denominator and this is

0:05:29.670,0:05:32.750
our denominator. The product[br]of these two.

0:05:34.930,0:05:39.767
So that's 1[br]- 3 X times

0:05:39.767,0:05:43.222
by 1 + 2 X.

0:05:44.830,0:05:49.087
So we ask ourselves, how[br]many times does this

0:05:49.087,0:05:55.236
divide into this? And the[br]answer is 1 + 2 X, so we

0:05:55.236,0:06:01.385
have minus 3 * 1 + 2 X[br]minus two and then how

0:06:01.385,0:06:06.115
many times does this go[br]into this? And the answer

0:06:06.115,0:06:12.264
is 1 - 3 X so it's 2[br]* 1 - 3 X.

0:06:13.420,0:06:20.349
Now we need to simplify this so[br]we can keep the denominator as

0:06:20.349,0:06:27.811
it is 1 - 3 X times[br]by 1 + 2 X and let's

0:06:27.811,0:06:32.608
multiply out this bracket minus[br]three times everything inside

0:06:32.608,0:06:40.070
the bracket, so it's minus 3 -[br]6 X and then minus two times

0:06:40.070,0:06:43.801
everything inside the bracket,[br]so that's minus.

0:06:43.890,0:06:46.730
2 + 6 X.

0:06:47.950,0:06:54.082
Now we need to simplify this bit[br]on top, and we've minus 6X plus

0:06:54.082,0:07:01.090
6X, so that's no axis minus 3 -[br]2 - 5. So we have minus 5.

0:07:01.910,0:07:08.890
Over 1 - 3 X[br]times by 1 + 2

0:07:08.890,0:07:12.030
X. But so long as we're good

0:07:12.030,0:07:16.240
with the algebra. This[br]differentiation is well

0:07:16.240,0:07:22.590
worth the simplifying as we[br]have done here, so let's

0:07:22.590,0:07:24.495
take another example.

0:07:26.500,0:07:34.192
This time let's take Y equals[br]X to the power sign X.

0:07:35.080,0:07:35.490
Now.

0:07:37.240,0:07:42.311
Real problems having the unknown[br]the variable appear yet again in

0:07:42.311,0:07:48.765
the power in the index, so this[br]sign X is the problem. If it

0:07:48.765,0:07:54.297
wasn't in the index, we might be[br]able to differentiate it, but

0:07:54.297,0:08:01.212
one way of getting it down, so[br]to speak out of the index is to

0:08:01.212,0:08:07.205
take logs of both sides so we[br]have log Y equals the logs.

0:08:07.620,0:08:13.820
X to the power sign X[br]and if we remember.

0:08:14.430,0:08:16.750
By using our log laws.

0:08:17.330,0:08:23.630
When we raise the function[br]inside the log to a power, then

0:08:23.630,0:08:28.880
that's just the same as[br]multiplying the log of the

0:08:28.880,0:08:30.980
function by the power.

0:08:31.830,0:08:35.726
And now this is a[br]straightforward product. It's

0:08:35.726,0:08:38.648
sign X times by Log X.

0:08:39.240,0:08:43.096
We know that we can[br]differentiate this. What

0:08:43.096,0:08:48.398
about this at the left[br]hand side log Y? Well, if

0:08:48.398,0:08:52.254
we're differentiating with[br]respect to X, we can

0:08:52.254,0:08:56.110
differentiate this log Y[br]implicitly, and so the

0:08:56.110,0:09:02.376
derivative of the log of Y[br]is one over Y times DY by

0:09:02.376,0:09:03.822
the X equals.

0:09:05.540,0:09:11.884
Now we need to differentiate[br]this sign X times by Log X. It's

0:09:11.884,0:09:18.228
a product you times by V and we[br]know that the derivative of

0:09:18.228,0:09:26.036
product is U times DV by DX Plus[br]V Times du by DX. So let's write

0:09:26.036,0:09:32.868
that down you which we said was[br]sign X times divided by DX and

0:09:32.868,0:09:36.284
the derivative of log X is one.

0:09:36.450,0:09:37.830
Over X.

0:09:39.050,0:09:42.540
Close the which is log

0:09:42.540,0:09:47.338
X. Times the[br]derivative of U and

0:09:47.338,0:09:51.454
you re sign X its[br]derivative is

0:09:51.454,0:09:53.218
therefore cause X.

0:09:56.360,0:09:59.790
Now we can do a little bit of[br]simplifying. Here we can put it

0:09:59.790,0:10:05.180
all. Over this denominator of X,[br]and so we have sine X.

0:10:07.020,0:10:11.673
Plus and if we're putting[br]everything over this denominator

0:10:11.673,0:10:17.360
of X, we need to multiply[br]anything that's not over the

0:10:17.360,0:10:24.081
denominator of X already by X,[br]so that's X Times Log X times

0:10:24.081,0:10:26.666
cause X all over X.

0:10:27.700,0:10:33.017
But this side with one over Y[br]from what we're really after is

0:10:33.017,0:10:34.653
DY by The X.

0:10:35.830,0:10:41.905
And so we need to multiply up by[br]why, but we know what? Why is

0:10:41.905,0:10:44.740
its X to the power sign X?

0:10:46.390,0:10:53.174
So that's sine XX[br]log X cause X.

0:10:54.030,0:11:00.764
OX and we're going to multiply[br]up by Y, and that's X sign

0:11:00.764,0:11:06.462
X to the power sign X, and[br]so we finished the

0:11:06.462,0:11:10.088
differentiation and again,[br]something looked impossible at

0:11:10.088,0:11:16.822
the beginning by making use of[br]the laws of logs, we have been

0:11:16.822,0:11:18.376
able to differentiate.

0:11:21.520,0:11:28.637
So let's take a look another[br]example, this time a more

0:11:28.637,0:11:33.166
complicated one. Quotient one[br]really is quite.

0:11:34.050,0:11:35.120
Complex.

0:11:38.670,0:11:44.214
Now again, we could do this as a[br]quotient you over V.

0:11:44.800,0:11:49.545
And use the formula which if[br]we remember is VDU by DX minus

0:11:49.545,0:11:53.560
UDV by the X all over V[br]squared. But simply having

0:11:53.560,0:11:57.210
said it, that's quite[br]frightening. So if we have a

0:11:57.210,0:12:02.685
look at this one of the ways[br]we can do it is to take logs

0:12:02.685,0:12:03.415
both sides.

0:12:10.580,0:12:14.396
And then having taken logs of

0:12:14.396,0:12:20.340
both sides. We can make use of[br]our laws of logarithms to

0:12:20.340,0:12:21.549
simplify this side.

0:12:22.850,0:12:27.050
So let's begin by remembering[br]that if we have a quotient when

0:12:27.050,0:12:30.900
we do the logarithm, then that's[br]the log of the numerator.

0:12:33.980,0:12:38.948
Minus the log of[br]the denominator.

0:12:42.960,0:12:46.710
But we can simplify this one[br]even further because we're

0:12:46.710,0:12:52.335
raising 1 - 2 X to the power[br]three, and in terms of our laws

0:12:52.335,0:12:57.585
of logs, that's exactly the same[br]as multiplying the log of 1 - 2

0:12:57.585,0:12:58.710
X by three.

0:13:00.200,0:13:05.260
Minus now here we've got a[br]square root and the square

0:13:05.260,0:13:10.320
root is the same as raising[br]something to the power 1/2

0:13:10.320,0:13:16.300
so that this is exactly the[br]same as taking a half of the

0:13:16.300,0:13:19.060
log of one plus X squared.

0:13:20.950,0:13:27.060
Now what we see is that this[br]that we've got here is much

0:13:27.060,0:13:28.470
simpler to differentiate.

0:13:29.350,0:13:36.266
So we can do that. The[br]derivative of the log of Y doing

0:13:36.266,0:13:43.714
it implicitly is one over YDY by[br]X equals 3 Times Now we want

0:13:43.714,0:13:50.630
to now differentiate the log of[br]1 - 2 X. So the function

0:13:50.630,0:13:56.482
is 1 - 2 X. So we[br]differentiate that it's minus

0:13:56.482,0:13:59.674
two over 1 - 2 X.

0:14:00.500,0:14:02.130
Minus 1/2.

0:14:04.080,0:14:08.535
Differentiating the log of[br]one plus X squared, the

0:14:08.535,0:14:13.485
function is one plus X[br]squared, so its derivative is

0:14:13.485,0:14:16.455
2X over one plus X squared.

0:14:17.640,0:14:22.860
Again, there's a penalty to be[br]paid. We really ought to put

0:14:22.860,0:14:26.340
these back together again, so[br]it's all one.

0:14:27.010,0:14:31.930
So let's do that and let's just[br]simplify a little bit here.

0:14:31.930,0:14:35.620
There's a two will cancel there[br]with a 2.

0:14:36.250,0:14:36.740
So.

0:14:38.740,0:14:41.740
We've got one over Y.

0:14:42.540,0:14:47.034
DY by the X[br]is equal to.

0:14:49.760,0:14:55.808
We've got 3 times by minus two[br]over 1 - 2 X, so that's going to

0:14:55.808,0:14:56.942
give us minus.

0:14:57.500,0:14:58.530
6.

0:14:59.610,0:15:03.756
All over[br]1 - 2 X.

0:15:06.410,0:15:10.793
We've got minus X over one[br]plus X squared.

0:15:18.200,0:15:23.708
Now we need to bring these[br]two terms together so we look

0:15:23.708,0:15:28.757
for a common denominator that[br]must be the product of the

0:15:28.757,0:15:29.675
two denominators.

0:15:32.320,0:15:39.250
And 1 - 2 X goes into this[br]one plus X squared times minus

0:15:39.250,0:15:46.675
six times by one plus X squared[br]minus X times by well one plus X

0:15:46.675,0:15:53.605
squared goes into the whole of[br]this 1 - 2 X times. So we

0:15:53.605,0:15:58.555
needed multiply the minus X by 1[br]- 2 X.

0:15:59.890,0:16:05.500
Multiply out the bracket minus 6[br]- 6 X squared.

0:16:06.090,0:16:09.914
Minus X +2 X

0:16:09.914,0:16:17.037
squared. All over 1[br]- 2 X one plus X

0:16:17.037,0:16:17.650
squared.

0:16:20.840,0:16:22.670
We can simplify the top.

0:16:23.400,0:16:30.848
Minus 6 minus X minus four X[br]squared all over 1 - 2 X

0:16:30.848,0:16:37.764
one plus X squared. And now[br]finally we want DY by the X

0:16:37.764,0:16:44.148
equals, but it's going to be[br]this. Let's take that minus sign

0:16:44.148,0:16:46.808
out so we have minus.

0:16:47.550,0:16:53.605
6 plus X plus[br]four X squared.

0:16:54.530,0:17:02.390
All over 1 - 2 X[br]times by one plus X squared.

0:17:03.680,0:17:10.050
And we need to multiply both[br]sides by Y and let's just recall

0:17:10.050,0:17:15.440
what why was. Why was 1 - 2[br]X or cubed?

0:17:16.300,0:17:23.250
So that's 1 - 2[br]X or cubed all over

0:17:23.250,0:17:26.725
one plus X squared square

0:17:26.725,0:17:31.690
root off. And clearly is a[br]little bit more simplified, can

0:17:31.690,0:17:37.270
be done here, because we've got[br]a 1 - 2 X here and a 1 - 2 X

0:17:37.270,0:17:40.680
cubed there so we can cancel[br]that with one of those.

0:17:41.430,0:17:45.668
Here we've got a one plus X[br]squared times by the square root

0:17:45.668,0:17:49.906
of 1 plus X squared, and if we[br]follow our laws of indices,

0:17:49.906,0:17:54.470
that's to the power one. And[br]this is to the power half. So we

0:17:54.470,0:17:58.382
add them together. That gives us[br]to the power three over 2.

0:17:59.800,0:18:06.019
So do you? Why by[br]DX finally will be.

0:18:09.850,0:18:11.500
Let's take this top line.

0:18:13.740,0:18:20.476
Minus 6 plus X[br]plus four X squared.

0:18:23.650,0:18:24.940
Times by.

0:18:26.320,0:18:28.850
1 - 2 X squared.

0:18:30.410,0:18:35.946
1 - 2 X or[br]squared all over.

0:18:37.210,0:18:41.890
Just check on that denominator[br]one plus X squared to the power

0:18:41.890,0:18:47.740
one times by one plus X squared[br]to the power a half and the one

0:18:47.740,0:18:53.980
and add it to the half. Gives us[br]1 1/2 or three over 2, so that's

0:18:53.980,0:18:58.270
one plus X squared raised to the[br]power three over 2.

0:19:00.940,0:19:05.321
So what we've seen in this[br]video is that we can use the

0:19:05.321,0:19:08.354
laws of logarithms to help us[br]simplify certain functions

0:19:08.354,0:19:10.039
before we come to[br]differentiating.

0:19:11.070,0:19:14.254
That can make the whole[br]process of differentiation

0:19:14.254,0:19:15.448
so much simpler.