[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:02.64,0:00:07.67,Default,,0000,0000,0000,,In this video, we're going to\Nbe looking at how we can use
Dialogue: 0,0:00:07.67,0:00:10.77,Default,,0000,0000,0000,,logarithms to help simplify\Ncertain functions before we
Dialogue: 0,0:00:10.77,0:00:11.54,Default,,0000,0000,0000,,differentiate them.
Dialogue: 0,0:00:12.72,0:00:17.16,Default,,0000,0000,0000,,To start off with, let's\Njust remind ourselves
Dialogue: 0,0:00:17.16,0:00:21.04,Default,,0000,0000,0000,,about logarithms\Nthemselves. So if we have
Dialogue: 0,0:00:21.04,0:00:27.15,Default,,0000,0000,0000,,Y equals the log of X,\Nthen this means log to
Dialogue: 0,0:00:27.15,0:00:32.14,Default,,0000,0000,0000,,base E, and if we\Ndifferentiate that why by
Dialogue: 0,0:00:32.14,0:00:36.03,Default,,0000,0000,0000,,DX is equal to one over X?
Dialogue: 0,0:00:37.38,0:00:41.68,Default,,0000,0000,0000,,But of course. Inside this\Nlogarithm here this might
Dialogue: 0,0:00:41.68,0:00:47.88,Default,,0000,0000,0000,,not just be X, it might be a\Nfunction of X, so it might
Dialogue: 0,0:00:47.88,0:00:52.31,Default,,0000,0000,0000,,be why equals the log of a\Nfunction of X.
Dialogue: 0,0:00:53.55,0:00:57.83,Default,,0000,0000,0000,,So what happens when we\Ndifferentiate this? Well, we
Dialogue: 0,0:00:57.83,0:01:03.55,Default,,0000,0000,0000,,have a rule that tells us why by\NDX is equal to.
Dialogue: 0,0:01:04.40,0:01:09.61,Default,,0000,0000,0000,,The derivative of F\Nof X. That's F dash
Dialogue: 0,0:01:09.61,0:01:12.51,Default,,0000,0000,0000,,decks over F of X.
Dialogue: 0,0:01:14.29,0:01:18.11,Default,,0000,0000,0000,,Now. We're going to be making\Nuse these results.
Dialogue: 0,0:01:18.91,0:01:24.47,Default,,0000,0000,0000,,But also the properties of\Nlogarithms themselves so.
Dialogue: 0,0:01:27.09,0:01:31.32,Default,,0000,0000,0000,,Let's take our first\Nexample, why?
Dialogue: 0,0:01:32.70,0:01:34.48,Default,,0000,0000,0000,,Equals the natural log.
Dialogue: 0,0:01:36.44,0:01:42.53,Default,,0000,0000,0000,,3X to the 4th plus\NSeven all raised to
Dialogue: 0,0:01:42.53,0:01:44.56,Default,,0000,0000,0000,,the fifth power.
Dialogue: 0,0:01:45.85,0:01:48.10,Default,,0000,0000,0000,,And we want to be able to
Dialogue: 0,0:01:48.10,0:01:50.03,Default,,0000,0000,0000,,differentiate this. Well.
Dialogue: 0,0:01:50.59,0:01:55.32,Default,,0000,0000,0000,,It looks very complicated in\Nhere, but here we are raising.
Dialogue: 0,0:01:56.02,0:02:01.68,Default,,0000,0000,0000,,3X to the power 4 + 7 to the\Npower five and one of the things
Dialogue: 0,0:02:01.68,0:02:05.93,Default,,0000,0000,0000,,we do know about logarithms is\Nthat if we are raising what's
Dialogue: 0,0:02:05.93,0:02:06.99,Default,,0000,0000,0000,,inside the log.
Dialogue: 0,0:02:07.87,0:02:12.69,Default,,0000,0000,0000,,To the power five, that's\Nexactly the same as multiplying
Dialogue: 0,0:02:12.69,0:02:19.92,Default,,0000,0000,0000,,the log by 5. So In other words,\Nwe can rewrite this as Y equals
Dialogue: 0,0:02:19.92,0:02:27.63,Default,,0000,0000,0000,,5 times the log of three X to\Nthe 4th plus 7 using our laws of
Dialogue: 0,0:02:27.63,0:02:32.45,Default,,0000,0000,0000,,logarithms. And now this is now\Nmuch easier to differentiate
Dialogue: 0,0:02:32.45,0:02:36.79,Default,,0000,0000,0000,,because this inside the log is\Nas very straightforward
Dialogue: 0,0:02:36.79,0:02:42.86,Default,,0000,0000,0000,,function. So we can now\Ndifferentiate the why by DX is
Dialogue: 0,0:02:42.86,0:02:45.47,Default,,0000,0000,0000,,equal to. 5.
Dialogue: 0,0:02:46.28,0:02:51.30,Default,,0000,0000,0000,,Times by now, let's remember\Nwhat we do. This is the
Dialogue: 0,0:02:51.30,0:02:54.94,Default,,0000,0000,0000,,function inside the log, so\Nwe differentiate that.
Dialogue: 0,0:02:56.50,0:03:04.42,Default,,0000,0000,0000,,So that gives us 12X cubed\Nand the derivative of the 70s
Dialogue: 0,0:03:04.42,0:03:11.02,Default,,0000,0000,0000,,zero over the function itself 3X\Nto the 4th +7.
Dialogue: 0,0:03:11.79,0:03:17.74,Default,,0000,0000,0000,,And if we just do a little bit\Nof simplifying five times by 12
Dialogue: 0,0:03:17.74,0:03:23.69,Default,,0000,0000,0000,,gives us 60 X cubed over 3X to\Nthe 4th plus Seven and what
Dialogue: 0,0:03:23.69,0:03:27.94,Default,,0000,0000,0000,,looked like up here quieter\Nfearsome derivative that we were
Dialogue: 0,0:03:27.94,0:03:33.04,Default,,0000,0000,0000,,going to have to do here, it's\Nturned out quite simply. So
Dialogue: 0,0:03:33.04,0:03:35.59,Default,,0000,0000,0000,,let's have a look at another
Dialogue: 0,0:03:35.59,0:03:41.21,Default,,0000,0000,0000,,example. Supposing we've got Y\Nequals the log of.
Dialogue: 0,0:03:42.30,0:03:49.75,Default,,0000,0000,0000,,We've got 1 - 3 X divided\Nby 1 + 2 X. Now what
Dialogue: 0,0:03:49.75,0:03:56.66,Default,,0000,0000,0000,,we've got here is the log of\Na quotient, and there is a
Dialogue: 0,0:03:56.66,0:03:59.32,Default,,0000,0000,0000,,formula for differentiating\Na quotient.
Dialogue: 0,0:04:00.50,0:04:05.37,Default,,0000,0000,0000,,That's going to lead to a very\Nvery complicated expression, but
Dialogue: 0,0:04:05.37,0:04:11.13,Default,,0000,0000,0000,,what we can do is make use of\Nthe laws of logarithms because
Dialogue: 0,0:04:11.13,0:04:16.89,Default,,0000,0000,0000,,the laws of logarithms tell us\Nthat when we're doing the log of
Dialogue: 0,0:04:16.89,0:04:22.21,Default,,0000,0000,0000,,a quotient than that is simply\Nthe log of the numerator minus
Dialogue: 0,0:04:22.21,0:04:26.64,Default,,0000,0000,0000,,the log of the denominator.\NBecause in logarithms we do
Dialogue: 0,0:04:26.64,0:04:30.62,Default,,0000,0000,0000,,division by subtracting the logs\Nof the respective parts.
Dialogue: 0,0:04:31.05,0:04:35.66,Default,,0000,0000,0000,,And now these two are both very\Neasy to differentiate.
Dialogue: 0,0:04:36.50,0:04:39.84,Default,,0000,0000,0000,,So we can have the why by DX
Dialogue: 0,0:04:39.84,0:04:45.93,Default,,0000,0000,0000,,equals. And this is a function\Nof X 1 - 3 X, so we
Dialogue: 0,0:04:45.93,0:04:47.40,Default,,0000,0000,0000,,differentiate it that's minus
Dialogue: 0,0:04:47.40,0:04:53.23,Default,,0000,0000,0000,,three. Over the function\Nitself, 1 - 3 X minus that
Dialogue: 0,0:04:53.23,0:04:58.48,Default,,0000,0000,0000,,minus sign, and now we do\Nthe same for this logarithm.
Dialogue: 0,0:04:58.48,0:05:02.77,Default,,0000,0000,0000,,Here's the function of X\Nwhich we differentiate, so
Dialogue: 0,0:05:02.77,0:05:08.97,Default,,0000,0000,0000,,the derivative of 1 + 2 X is\N2 over the function itself.
Dialogue: 0,0:05:08.97,0:05:10.88,Default,,0000,0000,0000,,1 + 2 X.
Dialogue: 0,0:05:12.07,0:05:17.79,Default,,0000,0000,0000,,Now that is a penalty to pay\Nfor getting away with such a
Dialogue: 0,0:05:17.79,0:05:19.99,Default,,0000,0000,0000,,straightforward\Ndifferentiation. What we have
Dialogue: 0,0:05:19.99,0:05:25.27,Default,,0000,0000,0000,,to do now is put these two\Ntogether. We have to bring
Dialogue: 0,0:05:25.27,0:05:29.67,Default,,0000,0000,0000,,them together all over the\Nsame denominator and this is
Dialogue: 0,0:05:29.67,0:05:32.75,Default,,0000,0000,0000,,our denominator. The product\Nof these two.
Dialogue: 0,0:05:34.93,0:05:39.77,Default,,0000,0000,0000,,So that's 1\N- 3 X times
Dialogue: 0,0:05:39.77,0:05:43.22,Default,,0000,0000,0000,,by 1 + 2 X.
Dialogue: 0,0:05:44.83,0:05:49.09,Default,,0000,0000,0000,,So we ask ourselves, how\Nmany times does this
Dialogue: 0,0:05:49.09,0:05:55.24,Default,,0000,0000,0000,,divide into this? And the\Nanswer is 1 + 2 X, so we
Dialogue: 0,0:05:55.24,0:06:01.38,Default,,0000,0000,0000,,have minus 3 * 1 + 2 X\Nminus two and then how
Dialogue: 0,0:06:01.38,0:06:06.12,Default,,0000,0000,0000,,many times does this go\Ninto this? And the answer
Dialogue: 0,0:06:06.12,0:06:12.26,Default,,0000,0000,0000,,is 1 - 3 X so it's 2\N* 1 - 3 X.
Dialogue: 0,0:06:13.42,0:06:20.35,Default,,0000,0000,0000,,Now we need to simplify this so\Nwe can keep the denominator as
Dialogue: 0,0:06:20.35,0:06:27.81,Default,,0000,0000,0000,,it is 1 - 3 X times\Nby 1 + 2 X and let's
Dialogue: 0,0:06:27.81,0:06:32.61,Default,,0000,0000,0000,,multiply out this bracket minus\Nthree times everything inside
Dialogue: 0,0:06:32.61,0:06:40.07,Default,,0000,0000,0000,,the bracket, so it's minus 3 -\N6 X and then minus two times
Dialogue: 0,0:06:40.07,0:06:43.80,Default,,0000,0000,0000,,everything inside the bracket,\Nso that's minus.
Dialogue: 0,0:06:43.89,0:06:46.73,Default,,0000,0000,0000,,2 + 6 X.
Dialogue: 0,0:06:47.95,0:06:54.08,Default,,0000,0000,0000,,Now we need to simplify this bit\Non top, and we've minus 6X plus
Dialogue: 0,0:06:54.08,0:07:01.09,Default,,0000,0000,0000,,6X, so that's no axis minus 3 -\N2 - 5. So we have minus 5.
Dialogue: 0,0:07:01.91,0:07:08.89,Default,,0000,0000,0000,,Over 1 - 3 X\Ntimes by 1 + 2
Dialogue: 0,0:07:08.89,0:07:12.03,Default,,0000,0000,0000,,X. But so long as we're good
Dialogue: 0,0:07:12.03,0:07:16.24,Default,,0000,0000,0000,,with the algebra. This\Ndifferentiation is well
Dialogue: 0,0:07:16.24,0:07:22.59,Default,,0000,0000,0000,,worth the simplifying as we\Nhave done here, so let's
Dialogue: 0,0:07:22.59,0:07:24.50,Default,,0000,0000,0000,,take another example.
Dialogue: 0,0:07:26.50,0:07:34.19,Default,,0000,0000,0000,,This time let's take Y equals\NX to the power sign X.
Dialogue: 0,0:07:35.08,0:07:35.49,Default,,0000,0000,0000,,Now.
Dialogue: 0,0:07:37.24,0:07:42.31,Default,,0000,0000,0000,,Real problems having the unknown\Nthe variable appear yet again in
Dialogue: 0,0:07:42.31,0:07:48.76,Default,,0000,0000,0000,,the power in the index, so this\Nsign X is the problem. If it
Dialogue: 0,0:07:48.76,0:07:54.30,Default,,0000,0000,0000,,wasn't in the index, we might be\Nable to differentiate it, but
Dialogue: 0,0:07:54.30,0:08:01.21,Default,,0000,0000,0000,,one way of getting it down, so\Nto speak out of the index is to
Dialogue: 0,0:08:01.21,0:08:07.20,Default,,0000,0000,0000,,take logs of both sides so we\Nhave log Y equals the logs.
Dialogue: 0,0:08:07.62,0:08:13.82,Default,,0000,0000,0000,,X to the power sign X\Nand if we remember.
Dialogue: 0,0:08:14.43,0:08:16.75,Default,,0000,0000,0000,,By using our log laws.
Dialogue: 0,0:08:17.33,0:08:23.63,Default,,0000,0000,0000,,When we raise the function\Ninside the log to a power, then
Dialogue: 0,0:08:23.63,0:08:28.88,Default,,0000,0000,0000,,that's just the same as\Nmultiplying the log of the
Dialogue: 0,0:08:28.88,0:08:30.98,Default,,0000,0000,0000,,function by the power.
Dialogue: 0,0:08:31.83,0:08:35.73,Default,,0000,0000,0000,,And now this is a\Nstraightforward product. It's
Dialogue: 0,0:08:35.73,0:08:38.65,Default,,0000,0000,0000,,sign X times by Log X.
Dialogue: 0,0:08:39.24,0:08:43.10,Default,,0000,0000,0000,,We know that we can\Ndifferentiate this. What
Dialogue: 0,0:08:43.10,0:08:48.40,Default,,0000,0000,0000,,about this at the left\Nhand side log Y? Well, if
Dialogue: 0,0:08:48.40,0:08:52.25,Default,,0000,0000,0000,,we're differentiating with\Nrespect to X, we can
Dialogue: 0,0:08:52.25,0:08:56.11,Default,,0000,0000,0000,,differentiate this log Y\Nimplicitly, and so the
Dialogue: 0,0:08:56.11,0:09:02.38,Default,,0000,0000,0000,,derivative of the log of Y\Nis one over Y times DY by
Dialogue: 0,0:09:02.38,0:09:03.82,Default,,0000,0000,0000,,the X equals.
Dialogue: 0,0:09:05.54,0:09:11.88,Default,,0000,0000,0000,,Now we need to differentiate\Nthis sign X times by Log X. It's
Dialogue: 0,0:09:11.88,0:09:18.23,Default,,0000,0000,0000,,a product you times by V and we\Nknow that the derivative of
Dialogue: 0,0:09:18.23,0:09:26.04,Default,,0000,0000,0000,,product is U times DV by DX Plus\NV Times du by DX. So let's write
Dialogue: 0,0:09:26.04,0:09:32.87,Default,,0000,0000,0000,,that down you which we said was\Nsign X times divided by DX and
Dialogue: 0,0:09:32.87,0:09:36.28,Default,,0000,0000,0000,,the derivative of log X is one.
Dialogue: 0,0:09:36.45,0:09:37.83,Default,,0000,0000,0000,,Over X.
Dialogue: 0,0:09:39.05,0:09:42.54,Default,,0000,0000,0000,,Close the which is log
Dialogue: 0,0:09:42.54,0:09:47.34,Default,,0000,0000,0000,,X. Times the\Nderivative of U and
Dialogue: 0,0:09:47.34,0:09:51.45,Default,,0000,0000,0000,,you re sign X its\Nderivative is
Dialogue: 0,0:09:51.45,0:09:53.22,Default,,0000,0000,0000,,therefore cause X.
Dialogue: 0,0:09:56.36,0:09:59.79,Default,,0000,0000,0000,,Now we can do a little bit of\Nsimplifying. Here we can put it
Dialogue: 0,0:09:59.79,0:10:05.18,Default,,0000,0000,0000,,all. Over this denominator of X,\Nand so we have sine X.
Dialogue: 0,0:10:07.02,0:10:11.67,Default,,0000,0000,0000,,Plus and if we're putting\Neverything over this denominator
Dialogue: 0,0:10:11.67,0:10:17.36,Default,,0000,0000,0000,,of X, we need to multiply\Nanything that's not over the
Dialogue: 0,0:10:17.36,0:10:24.08,Default,,0000,0000,0000,,denominator of X already by X,\Nso that's X Times Log X times
Dialogue: 0,0:10:24.08,0:10:26.67,Default,,0000,0000,0000,,cause X all over X.
Dialogue: 0,0:10:27.70,0:10:33.02,Default,,0000,0000,0000,,But this side with one over Y\Nfrom what we're really after is
Dialogue: 0,0:10:33.02,0:10:34.65,Default,,0000,0000,0000,,DY by The X.
Dialogue: 0,0:10:35.83,0:10:41.90,Default,,0000,0000,0000,,And so we need to multiply up by\Nwhy, but we know what? Why is
Dialogue: 0,0:10:41.90,0:10:44.74,Default,,0000,0000,0000,,its X to the power sign X?
Dialogue: 0,0:10:46.39,0:10:53.17,Default,,0000,0000,0000,,So that's sine XX\Nlog X cause X.
Dialogue: 0,0:10:54.03,0:11:00.76,Default,,0000,0000,0000,,OX and we're going to multiply\Nup by Y, and that's X sign
Dialogue: 0,0:11:00.76,0:11:06.46,Default,,0000,0000,0000,,X to the power sign X, and\Nso we finished the
Dialogue: 0,0:11:06.46,0:11:10.09,Default,,0000,0000,0000,,differentiation and again,\Nsomething looked impossible at
Dialogue: 0,0:11:10.09,0:11:16.82,Default,,0000,0000,0000,,the beginning by making use of\Nthe laws of logs, we have been
Dialogue: 0,0:11:16.82,0:11:18.38,Default,,0000,0000,0000,,able to differentiate.
Dialogue: 0,0:11:21.52,0:11:28.64,Default,,0000,0000,0000,,So let's take a look another\Nexample, this time a more
Dialogue: 0,0:11:28.64,0:11:33.17,Default,,0000,0000,0000,,complicated one. Quotient one\Nreally is quite.
Dialogue: 0,0:11:34.05,0:11:35.12,Default,,0000,0000,0000,,Complex.
Dialogue: 0,0:11:38.67,0:11:44.21,Default,,0000,0000,0000,,Now again, we could do this as a\Nquotient you over V.
Dialogue: 0,0:11:44.80,0:11:49.54,Default,,0000,0000,0000,,And use the formula which if\Nwe remember is VDU by DX minus
Dialogue: 0,0:11:49.54,0:11:53.56,Default,,0000,0000,0000,,UDV by the X all over V\Nsquared. But simply having
Dialogue: 0,0:11:53.56,0:11:57.21,Default,,0000,0000,0000,,said it, that's quite\Nfrightening. So if we have a
Dialogue: 0,0:11:57.21,0:12:02.68,Default,,0000,0000,0000,,look at this one of the ways\Nwe can do it is to take logs
Dialogue: 0,0:12:02.68,0:12:03.42,Default,,0000,0000,0000,,both sides.
Dialogue: 0,0:12:10.58,0:12:14.40,Default,,0000,0000,0000,,And then having taken logs of
Dialogue: 0,0:12:14.40,0:12:20.34,Default,,0000,0000,0000,,both sides. We can make use of\Nour laws of logarithms to
Dialogue: 0,0:12:20.34,0:12:21.55,Default,,0000,0000,0000,,simplify this side.
Dialogue: 0,0:12:22.85,0:12:27.05,Default,,0000,0000,0000,,So let's begin by remembering\Nthat if we have a quotient when
Dialogue: 0,0:12:27.05,0:12:30.90,Default,,0000,0000,0000,,we do the logarithm, then that's\Nthe log of the numerator.
Dialogue: 0,0:12:33.98,0:12:38.95,Default,,0000,0000,0000,,Minus the log of\Nthe denominator.
Dialogue: 0,0:12:42.96,0:12:46.71,Default,,0000,0000,0000,,But we can simplify this one\Neven further because we're
Dialogue: 0,0:12:46.71,0:12:52.34,Default,,0000,0000,0000,,raising 1 - 2 X to the power\Nthree, and in terms of our laws
Dialogue: 0,0:12:52.34,0:12:57.58,Default,,0000,0000,0000,,of logs, that's exactly the same\Nas multiplying the log of 1 - 2
Dialogue: 0,0:12:57.58,0:12:58.71,Default,,0000,0000,0000,,X by three.
Dialogue: 0,0:13:00.20,0:13:05.26,Default,,0000,0000,0000,,Minus now here we've got a\Nsquare root and the square
Dialogue: 0,0:13:05.26,0:13:10.32,Default,,0000,0000,0000,,root is the same as raising\Nsomething to the power 1/2
Dialogue: 0,0:13:10.32,0:13:16.30,Default,,0000,0000,0000,,so that this is exactly the\Nsame as taking a half of the
Dialogue: 0,0:13:16.30,0:13:19.06,Default,,0000,0000,0000,,log of one plus X squared.
Dialogue: 0,0:13:20.95,0:13:27.06,Default,,0000,0000,0000,,Now what we see is that this\Nthat we've got here is much
Dialogue: 0,0:13:27.06,0:13:28.47,Default,,0000,0000,0000,,simpler to differentiate.
Dialogue: 0,0:13:29.35,0:13:36.27,Default,,0000,0000,0000,,So we can do that. The\Nderivative of the log of Y doing
Dialogue: 0,0:13:36.27,0:13:43.71,Default,,0000,0000,0000,,it implicitly is one over YDY by\NX equals 3 Times Now we want
Dialogue: 0,0:13:43.71,0:13:50.63,Default,,0000,0000,0000,,to now differentiate the log of\N1 - 2 X. So the function
Dialogue: 0,0:13:50.63,0:13:56.48,Default,,0000,0000,0000,,is 1 - 2 X. So we\Ndifferentiate that it's minus
Dialogue: 0,0:13:56.48,0:13:59.67,Default,,0000,0000,0000,,two over 1 - 2 X.
Dialogue: 0,0:14:00.50,0:14:02.13,Default,,0000,0000,0000,,Minus 1/2.
Dialogue: 0,0:14:04.08,0:14:08.54,Default,,0000,0000,0000,,Differentiating the log of\None plus X squared, the
Dialogue: 0,0:14:08.54,0:14:13.48,Default,,0000,0000,0000,,function is one plus X\Nsquared, so its derivative is
Dialogue: 0,0:14:13.48,0:14:16.46,Default,,0000,0000,0000,,2X over one plus X squared.
Dialogue: 0,0:14:17.64,0:14:22.86,Default,,0000,0000,0000,,Again, there's a penalty to be\Npaid. We really ought to put
Dialogue: 0,0:14:22.86,0:14:26.34,Default,,0000,0000,0000,,these back together again, so\Nit's all one.
Dialogue: 0,0:14:27.01,0:14:31.93,Default,,0000,0000,0000,,So let's do that and let's just\Nsimplify a little bit here.
Dialogue: 0,0:14:31.93,0:14:35.62,Default,,0000,0000,0000,,There's a two will cancel there\Nwith a 2.
Dialogue: 0,0:14:36.25,0:14:36.74,Default,,0000,0000,0000,,So.
Dialogue: 0,0:14:38.74,0:14:41.74,Default,,0000,0000,0000,,We've got one over Y.
Dialogue: 0,0:14:42.54,0:14:47.03,Default,,0000,0000,0000,,DY by the X\Nis equal to.
Dialogue: 0,0:14:49.76,0:14:55.81,Default,,0000,0000,0000,,We've got 3 times by minus two\Nover 1 - 2 X, so that's going to
Dialogue: 0,0:14:55.81,0:14:56.94,Default,,0000,0000,0000,,give us minus.
Dialogue: 0,0:14:57.50,0:14:58.53,Default,,0000,0000,0000,,6.
Dialogue: 0,0:14:59.61,0:15:03.76,Default,,0000,0000,0000,,All over\N1 - 2 X.
Dialogue: 0,0:15:06.41,0:15:10.79,Default,,0000,0000,0000,,We've got minus X over one\Nplus X squared.
Dialogue: 0,0:15:18.20,0:15:23.71,Default,,0000,0000,0000,,Now we need to bring these\Ntwo terms together so we look
Dialogue: 0,0:15:23.71,0:15:28.76,Default,,0000,0000,0000,,for a common denominator that\Nmust be the product of the
Dialogue: 0,0:15:28.76,0:15:29.68,Default,,0000,0000,0000,,two denominators.
Dialogue: 0,0:15:32.32,0:15:39.25,Default,,0000,0000,0000,,And 1 - 2 X goes into this\None plus X squared times minus
Dialogue: 0,0:15:39.25,0:15:46.68,Default,,0000,0000,0000,,six times by one plus X squared\Nminus X times by well one plus X
Dialogue: 0,0:15:46.68,0:15:53.60,Default,,0000,0000,0000,,squared goes into the whole of\Nthis 1 - 2 X times. So we
Dialogue: 0,0:15:53.60,0:15:58.56,Default,,0000,0000,0000,,needed multiply the minus X by 1\N- 2 X.
Dialogue: 0,0:15:59.89,0:16:05.50,Default,,0000,0000,0000,,Multiply out the bracket minus 6\N- 6 X squared.
Dialogue: 0,0:16:06.09,0:16:09.91,Default,,0000,0000,0000,,Minus X +2 X
Dialogue: 0,0:16:09.91,0:16:17.04,Default,,0000,0000,0000,,squared. All over 1\N- 2 X one plus X
Dialogue: 0,0:16:17.04,0:16:17.65,Default,,0000,0000,0000,,squared.
Dialogue: 0,0:16:20.84,0:16:22.67,Default,,0000,0000,0000,,We can simplify the top.
Dialogue: 0,0:16:23.40,0:16:30.85,Default,,0000,0000,0000,,Minus 6 minus X minus four X\Nsquared all over 1 - 2 X
Dialogue: 0,0:16:30.85,0:16:37.76,Default,,0000,0000,0000,,one plus X squared. And now\Nfinally we want DY by the X
Dialogue: 0,0:16:37.76,0:16:44.15,Default,,0000,0000,0000,,equals, but it's going to be\Nthis. Let's take that minus sign
Dialogue: 0,0:16:44.15,0:16:46.81,Default,,0000,0000,0000,,out so we have minus.
Dialogue: 0,0:16:47.55,0:16:53.60,Default,,0000,0000,0000,,6 plus X plus\Nfour X squared.
Dialogue: 0,0:16:54.53,0:17:02.39,Default,,0000,0000,0000,,All over 1 - 2 X\Ntimes by one plus X squared.
Dialogue: 0,0:17:03.68,0:17:10.05,Default,,0000,0000,0000,,And we need to multiply both\Nsides by Y and let's just recall
Dialogue: 0,0:17:10.05,0:17:15.44,Default,,0000,0000,0000,,what why was. Why was 1 - 2\NX or cubed?
Dialogue: 0,0:17:16.30,0:17:23.25,Default,,0000,0000,0000,,So that's 1 - 2\NX or cubed all over
Dialogue: 0,0:17:23.25,0:17:26.72,Default,,0000,0000,0000,,one plus X squared square
Dialogue: 0,0:17:26.72,0:17:31.69,Default,,0000,0000,0000,,root off. And clearly is a\Nlittle bit more simplified, can
Dialogue: 0,0:17:31.69,0:17:37.27,Default,,0000,0000,0000,,be done here, because we've got\Na 1 - 2 X here and a 1 - 2 X
Dialogue: 0,0:17:37.27,0:17:40.68,Default,,0000,0000,0000,,cubed there so we can cancel\Nthat with one of those.
Dialogue: 0,0:17:41.43,0:17:45.67,Default,,0000,0000,0000,,Here we've got a one plus X\Nsquared times by the square root
Dialogue: 0,0:17:45.67,0:17:49.91,Default,,0000,0000,0000,,of 1 plus X squared, and if we\Nfollow our laws of indices,
Dialogue: 0,0:17:49.91,0:17:54.47,Default,,0000,0000,0000,,that's to the power one. And\Nthis is to the power half. So we
Dialogue: 0,0:17:54.47,0:17:58.38,Default,,0000,0000,0000,,add them together. That gives us\Nto the power three over 2.
Dialogue: 0,0:17:59.80,0:18:06.02,Default,,0000,0000,0000,,So do you? Why by\NDX finally will be.
Dialogue: 0,0:18:09.85,0:18:11.50,Default,,0000,0000,0000,,Let's take this top line.
Dialogue: 0,0:18:13.74,0:18:20.48,Default,,0000,0000,0000,,Minus 6 plus X\Nplus four X squared.
Dialogue: 0,0:18:23.65,0:18:24.94,Default,,0000,0000,0000,,Times by.
Dialogue: 0,0:18:26.32,0:18:28.85,Default,,0000,0000,0000,,1 - 2 X squared.
Dialogue: 0,0:18:30.41,0:18:35.95,Default,,0000,0000,0000,,1 - 2 X or\Nsquared all over.
Dialogue: 0,0:18:37.21,0:18:41.89,Default,,0000,0000,0000,,Just check on that denominator\None plus X squared to the power
Dialogue: 0,0:18:41.89,0:18:47.74,Default,,0000,0000,0000,,one times by one plus X squared\Nto the power a half and the one
Dialogue: 0,0:18:47.74,0:18:53.98,Default,,0000,0000,0000,,and add it to the half. Gives us\N1 1/2 or three over 2, so that's
Dialogue: 0,0:18:53.98,0:18:58.27,Default,,0000,0000,0000,,one plus X squared raised to the\Npower three over 2.
Dialogue: 0,0:19:00.94,0:19:05.32,Default,,0000,0000,0000,,So what we've seen in this\Nvideo is that we can use the
Dialogue: 0,0:19:05.32,0:19:08.35,Default,,0000,0000,0000,,laws of logarithms to help us\Nsimplify certain functions
Dialogue: 0,0:19:08.35,0:19:10.04,Default,,0000,0000,0000,,before we come to\Ndifferentiating.
Dialogue: 0,0:19:11.07,0:19:14.25,Default,,0000,0000,0000,,That can make the whole\Nprocess of differentiation
Dialogue: 0,0:19:14.25,0:19:15.45,Default,,0000,0000,0000,,so much simpler.