[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.41,0:00:07.27,Default,,0000,0000,0000,,In this video, you're going to\Nlearn how to differentiate the
Dialogue: 0,0:00:07.27,0:00:08.87,Default,,0000,0000,0000,,natural logarithm function.
Dialogue: 0,0:00:08.93,0:00:12.66,Default,,0000,0000,0000,,F of X is Ellen of
Dialogue: 0,0:00:12.66,0:00:15.92,Default,,0000,0000,0000,,X. And we're going to\Ndifferentiate this from first
Dialogue: 0,0:00:15.92,0:00:22.01,Default,,0000,0000,0000,,principles. And then we're going\Nto move to use the result from
Dialogue: 0,0:00:22.01,0:00:26.19,Default,,0000,0000,0000,,this to differentiate the\Nexponential function F of X
Dialogue: 0,0:00:26.19,0:00:32.69,Default,,0000,0000,0000,,equals E to the X, where you may\Nrecall that E is an irrational
Dialogue: 0,0:00:32.69,0:00:34.54,Default,,0000,0000,0000,,number. It's approximately equal
Dialogue: 0,0:00:34.54,0:00:41.18,Default,,0000,0000,0000,,to 2.718. And we call\Nthe the exponential constant.
Dialogue: 0,0:00:42.92,0:00:46.34,Default,,0000,0000,0000,,Now before we can start to do\Nthe differentiation of the
Dialogue: 0,0:00:46.34,0:00:49.14,Default,,0000,0000,0000,,natural logarithm, we need a\Nparticular result concerning the
Dialogue: 0,0:00:49.14,0:00:52.25,Default,,0000,0000,0000,,exponential constant. So I'm\Ngoing to derive that first. What
Dialogue: 0,0:00:52.25,0:00:56.92,Default,,0000,0000,0000,,I'd like you to like you to do\Nis I'd like you to think about
Dialogue: 0,0:00:56.92,0:01:03.53,Default,,0000,0000,0000,,this expression. One plus T all\Nraised to the power one over T.
Dialogue: 0,0:01:05.13,0:01:08.68,Default,,0000,0000,0000,,And what we're going to do is\Nwe're going to evaluate this
Dialogue: 0,0:01:08.68,0:01:09.87,Default,,0000,0000,0000,,expression for different values
Dialogue: 0,0:01:09.87,0:01:12.35,Default,,0000,0000,0000,,of tea. And see what happens.
Dialogue: 0,0:01:12.89,0:01:16.66,Default,,0000,0000,0000,,I'm going to drop a table of\Nvalues, so I'm going to have two
Dialogue: 0,0:01:16.66,0:01:23.16,Default,,0000,0000,0000,,columns. T values and then I'm\Ngoing to evaluate one plus T to
Dialogue: 0,0:01:23.16,0:01:25.38,Default,,0000,0000,0000,,the power one over T.
Dialogue: 0,0:01:26.09,0:01:29.47,Default,,0000,0000,0000,,And let's start with a very\Nsimple T value. Let's suppose T
Dialogue: 0,0:01:29.47,0:01:31.80,Default,,0000,0000,0000,,is one. Let's work this out.
Dialogue: 0,0:01:33.63,0:01:38.64,Default,,0000,0000,0000,,20 is one. This bracket becomes\N1 + 1 which is 2 and we want to
Dialogue: 0,0:01:38.64,0:01:43.02,Default,,0000,0000,0000,,raise it to the power one over\NT&T is one, so we're raising it
Dialogue: 0,0:01:43.02,0:01:45.21,Default,,0000,0000,0000,,to the power one over one, which
Dialogue: 0,0:01:45.21,0:01:50.62,Default,,0000,0000,0000,,is still 1. And two to the power\None is just two.
Dialogue: 0,0:01:52.28,0:01:56.22,Default,,0000,0000,0000,,Let's pick a different value for\NT and see what happens. And this
Dialogue: 0,0:01:56.22,0:01:59.86,Default,,0000,0000,0000,,time I'm going to choose a\Nsmaller value for T. I'm going
Dialogue: 0,0:01:59.86,0:02:01.98,Default,,0000,0000,0000,,to let TB not point, not one.
Dialogue: 0,0:02:03.29,0:02:04.42,Default,,0000,0000,0000,,Let's do the same thing.
Dialogue: 0,0:02:04.99,0:02:11.24,Default,,0000,0000,0000,,Put T in here will get one plus\NNorth Point N 1, which is one
Dialogue: 0,0:02:11.24,0:02:12.50,Default,,0000,0000,0000,,point N 1.
Dialogue: 0,0:02:13.34,0:02:14.79,Default,,0000,0000,0000,,And we're going to raise it to
Dialogue: 0,0:02:14.79,0:02:20.07,Default,,0000,0000,0000,,the power. 1 divided by T, which\Nis North Point N 1.
Dialogue: 0,0:02:20.69,0:02:25.53,Default,,0000,0000,0000,,And this is where you need\Na Calculator, so I want
Dialogue: 0,0:02:25.53,0:02:30.37,Default,,0000,0000,0000,,one point, not one raised\Nto the power 1 divided by
Dialogue: 0,0:02:30.37,0:02:32.13,Default,,0000,0000,0000,,nought point nought one.
Dialogue: 0,0:02:33.74,0:02:37.02,Default,,0000,0000,0000,,And that evaluates\Nto two point 705.
Dialogue: 0,0:02:38.28,0:02:42.02,Default,,0000,0000,0000,,So we see as TI has been made
Dialogue: 0,0:02:42.02,0:02:46.42,Default,,0000,0000,0000,,smaller. The value of this\Nexpression one plus T to the one
Dialogue: 0,0:02:46.42,0:02:48.26,Default,,0000,0000,0000,,over tease increased from 2 to
Dialogue: 0,0:02:48.26,0:02:53.55,Default,,0000,0000,0000,,just over 2.7. Now let's see\Nwhat happens when T gets even
Dialogue: 0,0:02:53.55,0:02:58.92,Default,,0000,0000,0000,,smaller and this time my value\Nfor T. I'm going to choose to be
Dialogue: 0,0:02:58.92,0:03:01.99,Default,,0000,0000,0000,,not .3 knots and 01 N point, not
Dialogue: 0,0:03:01.99,0:03:04.52,Default,,0000,0000,0000,,one. And let's evaluate this
Dialogue: 0,0:03:04.52,0:03:10.30,Default,,0000,0000,0000,,expression again. So I want one\Nplus T which is going to be one
Dialogue: 0,0:03:10.30,0:03:12.16,Default,,0000,0000,0000,,point not not not one.
Dialogue: 0,0:03:12.87,0:03:14.95,Default,,0000,0000,0000,,And I want to raise it to the
Dialogue: 0,0:03:14.95,0:03:21.47,Default,,0000,0000,0000,,power. One over T. That's one\Nover nought point nought nought
Dialogue: 0,0:03:21.47,0:03:27.42,Default,,0000,0000,0000,,nought one. And again we need\Na Calculator for this, so in
Dialogue: 0,0:03:27.42,0:03:31.94,Default,,0000,0000,0000,,the Calculator goes one point\Nnaut naut naut one I'm going
Dialogue: 0,0:03:31.94,0:03:36.46,Default,,0000,0000,0000,,to raise it to the power 1\Ndivided by nought. Point
Dialogue: 0,0:03:36.46,0:03:38.11,Default,,0000,0000,0000,,nought nought nought one.
Dialogue: 0,0:03:39.56,0:03:44.68,Default,,0000,0000,0000,,And this time I get the\Nvalue 2.718.
Dialogue: 0,0:03:45.92,0:03:48.49,Default,,0000,0000,0000,,Now, this is an exercise that\Nyou could continue for yourself.
Dialogue: 0,0:03:48.49,0:03:50.13,Default,,0000,0000,0000,,It would be a very useful thing
Dialogue: 0,0:03:50.13,0:03:55.07,Default,,0000,0000,0000,,for you to do. And see what\Nhappens as T continues to get
Dialogue: 0,0:03:55.07,0:03:59.63,Default,,0000,0000,0000,,smaller and smaller and smaller\Ncloser and closer to 0. And what
Dialogue: 0,0:03:59.63,0:04:03.81,Default,,0000,0000,0000,,we find is that as T gets\Nsmaller, this expression over
Dialogue: 0,0:04:03.81,0:04:09.13,Default,,0000,0000,0000,,here one plus T to the power one\Nover T gets closer and closer
Dialogue: 0,0:04:09.13,0:04:12.93,Default,,0000,0000,0000,,and closer, surprise, surprise.\NSo this value over here 2.718
Dialogue: 0,0:04:12.93,0:04:16.35,Default,,0000,0000,0000,,the exponential constant. Now\Nthat's a very important result
Dialogue: 0,0:04:16.35,0:04:20.15,Default,,0000,0000,0000,,that will need when we want to\Ndifferentiate the natural
Dialogue: 0,0:04:20.15,0:04:22.05,Default,,0000,0000,0000,,logarithm of X from first
Dialogue: 0,0:04:22.05,0:04:25.71,Default,,0000,0000,0000,,principles. I want to formally\Nwrite that down in a
Dialogue: 0,0:04:25.71,0:04:29.53,Default,,0000,0000,0000,,mathematical sort of way. What\Nwe've done is we've taken the
Dialogue: 0,0:04:29.53,0:04:34.04,Default,,0000,0000,0000,,expression one plus T raised to\Nthe power one over T, and we've
Dialogue: 0,0:04:34.04,0:04:37.86,Default,,0000,0000,0000,,evaluated this as TI gets closer\Nand closer to 0, and
Dialogue: 0,0:04:37.86,0:04:41.33,Default,,0000,0000,0000,,mathematically we write this\Nlike this. We right limb for
Dialogue: 0,0:04:41.33,0:04:44.74,Default,,0000,0000,0000,,limit. T tends to 0.
Dialogue: 0,0:04:45.35,0:04:47.22,Default,,0000,0000,0000,,Of this expression.
Dialogue: 0,0:04:47.84,0:04:51.85,Default,,0000,0000,0000,,Is E the exponential constant?\NIn other words, the limit as T
Dialogue: 0,0:04:51.85,0:04:56.52,Default,,0000,0000,0000,,tends to zero of one plus T\Nraised to the power one over T
Dialogue: 0,0:04:56.52,0:04:57.86,Default,,0000,0000,0000,,is the exponential constant.
Dialogue: 0,0:04:58.39,0:05:03.69,Default,,0000,0000,0000,,And that is a key result that\Nwill need very soon.
Dialogue: 0,0:05:04.95,0:05:10.01,Default,,0000,0000,0000,,We're now ready to do a bit of\Ndifferentiation from first
Dialogue: 0,0:05:10.01,0:05:15.07,Default,,0000,0000,0000,,principles, so I'm going to\Nremind you a little bit about
Dialogue: 0,0:05:15.07,0:05:18.29,Default,,0000,0000,0000,,the formula for differentiation\Nfrom first principles.
Dialogue: 0,0:05:18.85,0:05:20.57,Default,,0000,0000,0000,,What I'd like you to imagine.
Dialogue: 0,0:05:22.09,0:05:23.100,Default,,0000,0000,0000,,Is that we have a function?
Dialogue: 0,0:05:25.16,0:05:28.53,Default,,0000,0000,0000,,F of X.
Dialogue: 0,0:05:28.53,0:05:32.78,Default,,0000,0000,0000,,And what we're going to do is\Nwe're going to try to calculate
Dialogue: 0,0:05:32.78,0:05:36.05,Default,,0000,0000,0000,,the derivative of this function\Nat a particular point here.
Dialogue: 0,0:05:36.64,0:05:39.09,Default,,0000,0000,0000,,Let's call the point where\Ninterested in, let's call
Dialogue: 0,0:05:39.09,0:05:39.90,Default,,0000,0000,0000,,that point a.
Dialogue: 0,0:05:42.46,0:05:43.72,Default,,0000,0000,0000,,And A is the point.
Dialogue: 0,0:05:44.63,0:05:47.25,Default,,0000,0000,0000,,With X Coordinate X.
Dialogue: 0,0:05:49.12,0:05:55.58,Default,,0000,0000,0000,,When we try to calculate the\Nderivative of F of X at this
Dialogue: 0,0:05:55.58,0:05:59.81,Default,,0000,0000,0000,,point. That's when we try to\Ncalculate F, dash, devex. What
Dialogue: 0,0:05:59.81,0:06:03.15,Default,,0000,0000,0000,,we're doing is calculating the\Ngradient of the tangent to the
Dialogue: 0,0:06:03.15,0:06:06.80,Default,,0000,0000,0000,,curve at this point, so we're\Nworking out the gradient of the
Dialogue: 0,0:06:06.80,0:06:09.84,Default,,0000,0000,0000,,line, which is just touching the\Ncurve at this point.
Dialogue: 0,0:06:10.91,0:06:14.62,Default,,0000,0000,0000,,Now, the way we do this\Nmathematically is we move from
Dialogue: 0,0:06:14.62,0:06:18.66,Default,,0000,0000,0000,,our fixed point A to a nearby\Npoint a little bit further
Dialogue: 0,0:06:18.66,0:06:19.67,Default,,0000,0000,0000,,around the curve.
Dialogue: 0,0:06:20.25,0:06:23.40,Default,,0000,0000,0000,,Let's call this nearby point B.
Dialogue: 0,0:06:24.98,0:06:29.36,Default,,0000,0000,0000,,Now The X coordinate at B is\Njust a little bit more than the
Dialogue: 0,0:06:29.36,0:06:30.61,Default,,0000,0000,0000,,X coordinate at A.
Dialogue: 0,0:06:34.11,0:06:38.50,Default,,0000,0000,0000,,So the X coordinate here is X\Nplus a little bit more of X, and
Dialogue: 0,0:06:38.50,0:06:42.61,Default,,0000,0000,0000,,there's a little bit more of X\Nin here we write as Delta X.
Dialogue: 0,0:06:43.32,0:06:48.37,Default,,0000,0000,0000,,Delta X stands for a small\Nchange or small increment as we
Dialogue: 0,0:06:48.37,0:06:50.06,Default,,0000,0000,0000,,call it in X.
Dialogue: 0,0:06:50.99,0:06:52.94,Default,,0000,0000,0000,,So if this distance is Delta X.
Dialogue: 0,0:06:53.64,0:06:56.78,Default,,0000,0000,0000,,The X coordinate at B must be X.
Dialogue: 0,0:06:57.69,0:06:59.35,Default,,0000,0000,0000,,Plus Delta X.
Dialogue: 0,0:07:02.00,0:07:07.80,Default,,0000,0000,0000,,What about the Y coordinates at\NA&B? Well, at a The X coordinate
Dialogue: 0,0:07:07.80,0:07:13.25,Default,,0000,0000,0000,,is X. And a lies on the graph\Nof F of X.
Dialogue: 0,0:07:13.81,0:07:18.98,Default,,0000,0000,0000,,So the Y coordinate at a is just\Nthe function evaluated at X, so
Dialogue: 0,0:07:18.98,0:07:20.82,Default,,0000,0000,0000,,the Y coordinate at A.
Dialogue: 0,0:07:21.34,0:07:22.73,Default,,0000,0000,0000,,Is just F.
Dialogue: 0,0:07:23.26,0:07:25.55,Default,,0000,0000,0000,,Evaluated attacks or F of X?
Dialogue: 0,0:07:28.13,0:07:32.04,Default,,0000,0000,0000,,The Y coordinate at B in a\Nsimilar fashion is just the
Dialogue: 0,0:07:32.04,0:07:35.95,Default,,0000,0000,0000,,function evaluated at this X\Nvalue, which is X Plus Delta X.
Dialogue: 0,0:07:36.52,0:07:38.39,Default,,0000,0000,0000,,So the Y coordinate at B.
Dialogue: 0,0:07:40.71,0:07:45.51,Default,,0000,0000,0000,,Will be just the function\Nevaluated at X Plus Delta X.
Dialogue: 0,0:07:48.28,0:07:54.52,Default,,0000,0000,0000,,Knowing all these values F of XF\Nof X Plus Delta XX&X Plus Delta
Dialogue: 0,0:07:54.52,0:07:59.43,Default,,0000,0000,0000,,X, we can calculate the slope of\Nthe line through A&B.
Dialogue: 0,0:08:00.93,0:08:04.82,Default,,0000,0000,0000,,What's the slope of the line\Nthrough A and be? Well, we do
Dialogue: 0,0:08:04.82,0:08:07.81,Default,,0000,0000,0000,,that by forming a little right\Nangle triangle in here.
Dialogue: 0,0:08:08.82,0:08:12.53,Default,,0000,0000,0000,,And then the slope of the line
Dialogue: 0,0:08:12.53,0:08:15.37,Default,,0000,0000,0000,,through A&B. Is this distance?
Dialogue: 0,0:08:16.37,0:08:21.07,Default,,0000,0000,0000,,Divided by.\NThis distance.
Dialogue: 0,0:08:22.48,0:08:25.82,Default,,0000,0000,0000,,Now this distance is very\Nstraightforward to workout. It's
Dialogue: 0,0:08:25.82,0:08:29.53,Default,,0000,0000,0000,,the larger distance from the X\Naxis up to be.
Dialogue: 0,0:08:30.09,0:08:35.11,Default,,0000,0000,0000,,Subtract the smaller distance\Nfrom the X axis up to a. In
Dialogue: 0,0:08:35.11,0:08:40.54,Default,,0000,0000,0000,,other words, this distance in\Nhere is F of X Plus Delta X.
Dialogue: 0,0:08:41.23,0:08:43.63,Default,,0000,0000,0000,,Minus F of X.
Dialogue: 0,0:08:44.70,0:08:46.61,Default,,0000,0000,0000,,This distance in here.
Dialogue: 0,0:08:47.17,0:08:48.54,Default,,0000,0000,0000,,Is just Delta X.
Dialogue: 0,0:08:49.63,0:08:53.71,Default,,0000,0000,0000,,So now we're in a position to\Nwrite down a formula for the
Dialogue: 0,0:08:53.71,0:08:55.60,Default,,0000,0000,0000,,slope of the line through A&B.
Dialogue: 0,0:08:56.31,0:08:59.83,Default,,0000,0000,0000,,We say that the slope.
Dialogue: 0,0:08:59.83,0:09:02.46,Default,,0000,0000,0000,,As a baby.
Dialogue: 0,0:09:02.46,0:09:07.95,Default,,0000,0000,0000,,Is equal to this vertical\Ndistance which we've got here F
Dialogue: 0,0:09:07.95,0:09:10.44,Default,,0000,0000,0000,,of X plus Delta X.
Dialogue: 0,0:09:11.05,0:09:13.81,Default,,0000,0000,0000,,Minus F of X.
Dialogue: 0,0:09:14.32,0:09:16.84,Default,,0000,0000,0000,,All divided by the horizontal
Dialogue: 0,0:09:16.84,0:09:19.91,Default,,0000,0000,0000,,distance. Which is just Delta X.
Dialogue: 0,0:09:21.07,0:09:25.73,Default,,0000,0000,0000,,What's this that we found here?\NThis quantity is the slope.
Dialogue: 0,0:09:26.28,0:09:30.37,Default,,0000,0000,0000,,Of this line through A&B,\Nwhat we're really interested
Dialogue: 0,0:09:30.37,0:09:35.81,Default,,0000,0000,0000,,in is the slope of the\Ntangent at A and we achieve
Dialogue: 0,0:09:35.81,0:09:40.35,Default,,0000,0000,0000,,that by bringing B closer and\Ncloser into point A.
Dialogue: 0,0:09:41.77,0:09:46.62,Default,,0000,0000,0000,,What we do is we let be get\Ncloser and closer to A and as
Dialogue: 0,0:09:46.62,0:09:50.17,Default,,0000,0000,0000,,that happens the slope of this\Nline will become eventually the
Dialogue: 0,0:09:50.17,0:09:51.78,Default,,0000,0000,0000,,slope of the tangent today.
Dialogue: 0,0:09:53.25,0:09:56.65,Default,,0000,0000,0000,,Now as we bring B round\Nto a, what's happening is
Dialogue: 0,0:09:56.65,0:09:59.74,Default,,0000,0000,0000,,that Delta X is getting\Ncloser and closer to 0.
Dialogue: 0,0:10:00.80,0:10:06.68,Default,,0000,0000,0000,,So what we want to do is we want\Nto workout this expression in
Dialogue: 0,0:10:06.68,0:10:12.14,Default,,0000,0000,0000,,the limit as Delta X approaches\N0. So we want to workout the
Dialogue: 0,0:10:12.14,0:10:16.34,Default,,0000,0000,0000,,limit as Delta X tends to zero\Nof all this.
Dialogue: 0,0:10:16.40,0:10:22.18,Default,,0000,0000,0000,,When we\Nwork all
Dialogue: 0,0:10:22.18,0:10:28.66,Default,,0000,0000,0000,,that out. That\Nwill give us a value for the
Dialogue: 0,0:10:28.66,0:10:35.02,Default,,0000,0000,0000,,derivative. Of the function at\Npoint a, so that will define F
Dialogue: 0,0:10:35.02,0:10:36.51,Default,,0000,0000,0000,,dash of X.
Dialogue: 0,0:10:37.63,0:10:41.91,Default,,0000,0000,0000,,That is a very important result\Nthat will need as well when we
Dialogue: 0,0:10:41.91,0:10:44.54,Default,,0000,0000,0000,,come to differentiate the\Nnatural logarithm from first
Dialogue: 0,0:10:44.54,0:10:47.17,Default,,0000,0000,0000,,principles. Now I've rushed\Nthrough this rather quickly.
Dialogue: 0,0:10:47.17,0:10:49.80,Default,,0000,0000,0000,,There's a whole unit on\Ndifferentiation from first
Dialogue: 0,0:10:49.80,0:10:53.75,Default,,0000,0000,0000,,principles, and if you want a\Ngreat deal more detail, I would
Dialogue: 0,0:10:53.75,0:10:56.05,Default,,0000,0000,0000,,go back and look at that video.
Dialogue: 0,0:10:56.85,0:11:03.83,Default,,0000,0000,0000,,OK, but now into the real\Nbusiness of the video. We want
Dialogue: 0,0:11:03.83,0:11:10.24,Default,,0000,0000,0000,,to learn how to differentiate\Nthis function F of X is
Dialogue: 0,0:11:10.24,0:11:13.73,Default,,0000,0000,0000,,the natural logarithm Ln of X.
Dialogue: 0,0:11:14.25,0:11:18.26,Default,,0000,0000,0000,,Now to do that will need lots of\Ndifferent things. Lots of
Dialogue: 0,0:11:18.26,0:11:20.93,Default,,0000,0000,0000,,results from different branches\Nof mathematics, so let's
Dialogue: 0,0:11:20.93,0:11:22.60,Default,,0000,0000,0000,,summarize what will need first.
Dialogue: 0,0:11:23.15,0:11:27.74,Default,,0000,0000,0000,,One thing will need is the\Nformula for the derivative that
Dialogue: 0,0:11:27.74,0:11:34.41,Default,,0000,0000,0000,,we just wrote down F dash of X\Nis the limit as Delta X tends to
Dialogue: 0,0:11:34.41,0:11:37.74,Default,,0000,0000,0000,,zero of F of X Plus Delta X.
Dialogue: 0,0:11:38.37,0:11:41.18,Default,,0000,0000,0000,,Minus F of X.
Dialogue: 0,0:11:41.18,0:11:42.68,Default,,0000,0000,0000,,All over Delta X.
Dialogue: 0,0:11:43.28,0:11:45.51,Default,,0000,0000,0000,,That's the formula for\Ndifferentiation from first
Dialogue: 0,0:11:45.51,0:11:49.00,Default,,0000,0000,0000,,principles and will apply that\Nto this function in a minute.
Dialogue: 0,0:11:49.99,0:11:54.15,Default,,0000,0000,0000,,Will also need the result that I\Ngave you right at the beginning
Dialogue: 0,0:11:54.15,0:11:57.99,Default,,0000,0000,0000,,of the video, which was that the\Nexponential constant E can be
Dialogue: 0,0:11:57.99,0:11:59.27,Default,,0000,0000,0000,,written as the limit.
Dialogue: 0,0:11:59.83,0:12:02.05,Default,,0000,0000,0000,,As T tends to 0.
Dialogue: 0,0:12:02.83,0:12:07.95,Default,,0000,0000,0000,,Of one plus T or raised to the\Npower one over T, so that's
Dialogue: 0,0:12:07.95,0:12:09.78,Default,,0000,0000,0000,,another result that will need.
Dialogue: 0,0:12:10.33,0:12:15.03,Default,,0000,0000,0000,,We need a bit more actually. We\Nneed something about logarithms.
Dialogue: 0,0:12:15.03,0:12:22.29,Default,,0000,0000,0000,,You need to know that the log of\NA minus the log of B is the log
Dialogue: 0,0:12:22.29,0:12:27.41,Default,,0000,0000,0000,,of a divided by B. That's one of\Nthe laws of logarithms.
Dialogue: 0,0:12:28.11,0:12:33.03,Default,,0000,0000,0000,,And another law of logarithms\Nthat will need is that if you
Dialogue: 0,0:12:33.03,0:12:37.13,Default,,0000,0000,0000,,have a multiple M times a\Nlogarithm, then that multiple
Dialogue: 0,0:12:37.13,0:12:43.28,Default,,0000,0000,0000,,can come inside as a power. So\Nwe can write M log a as the
Dialogue: 0,0:12:43.28,0:12:44.51,Default,,0000,0000,0000,,logarithm of A.
Dialogue: 0,0:12:45.15,0:12:47.17,Default,,0000,0000,0000,,Bring it in as a power to the
Dialogue: 0,0:12:47.17,0:12:51.57,Default,,0000,0000,0000,,power N. These are two laws of\Nlogarithms, and if you're not
Dialogue: 0,0:12:51.57,0:12:55.33,Default,,0000,0000,0000,,familiar with those then you\Nshould go back and look at the
Dialogue: 0,0:12:55.33,0:12:56.89,Default,,0000,0000,0000,,unit on laws of logarithms.
Dialogue: 0,0:12:57.97,0:13:03.55,Default,,0000,0000,0000,,1 final result we need to know\Nthat the natural logarithm of
Dialogue: 0,0:13:03.55,0:13:08.20,Default,,0000,0000,0000,,the exponential constant E is\None and again you'll find
Dialogue: 0,0:13:08.20,0:13:09.60,Default,,0000,0000,0000,,information about that.
Dialogue: 0,0:13:10.12,0:13:14.37,Default,,0000,0000,0000,,On the unit, on logarithms. So\Nall those things will need as we
Dialogue: 0,0:13:14.37,0:13:15.68,Default,,0000,0000,0000,,start to differentiate this
Dialogue: 0,0:13:15.68,0:13:21.16,Default,,0000,0000,0000,,function here. OK, let's dive\Nin. Here's our function F of X
Dialogue: 0,0:13:21.16,0:13:22.32,Default,,0000,0000,0000,,is log X.
Dialogue: 0,0:13:22.91,0:13:27.33,Default,,0000,0000,0000,,We want to put this this\Nfunction log X into this formula
Dialogue: 0,0:13:27.33,0:13:30.64,Default,,0000,0000,0000,,for differentiation from first\Nprinciples. To do that, we're
Dialogue: 0,0:13:30.64,0:13:34.69,Default,,0000,0000,0000,,going to have to evaluate F at X\NPlus Delta X.
Dialogue: 0,0:13:35.19,0:13:37.36,Default,,0000,0000,0000,,Let's do that.
Dialogue: 0,0:13:37.42,0:13:42.94,Default,,0000,0000,0000,,I remember this function. The\Nlogarithm takes the log of the
Dialogue: 0,0:13:42.94,0:13:49.97,Default,,0000,0000,0000,,input, so when the input is X\NPlus Delta X, the output will be
Dialogue: 0,0:13:49.97,0:13:53.48,Default,,0000,0000,0000,,the logarithm of X Plus Delta X.
Dialogue: 0,0:13:54.74,0:13:56.48,Default,,0000,0000,0000,,So now we've got F of X Plus
Dialogue: 0,0:13:56.48,0:13:58.76,Default,,0000,0000,0000,,Delta X. And we've got F of X.
Dialogue: 0,0:13:59.37,0:14:01.77,Default,,0000,0000,0000,,We can plug them into the\Ntop line of the formula.
Dialogue: 0,0:14:03.44,0:14:06.46,Default,,0000,0000,0000,,F of X Plus Delta X.
Dialogue: 0,0:14:07.00,0:14:09.26,Default,,0000,0000,0000,,Minus F of X.
Dialogue: 0,0:14:09.81,0:14:11.74,Default,,0000,0000,0000,,What will that be?
Dialogue: 0,0:14:12.45,0:14:16.24,Default,,0000,0000,0000,,It will be the logarithm of X\NPlus Delta X.
Dialogue: 0,0:14:16.24,0:14:23.44,Default,,0000,0000,0000,,Subtract F of X so\Nit'll be. Subtract the logarithm
Dialogue: 0,0:14:23.44,0:14:27.26,Default,,0000,0000,0000,,of X. Ah, with
Dialogue: 0,0:14:27.26,0:14:30.95,Default,,0000,0000,0000,,a logarithm. Minus a
Dialogue: 0,0:14:30.95,0:14:35.66,Default,,0000,0000,0000,,logarithm. And this is where I'm\Ngoing to use this law of
Dialogue: 0,0:14:35.66,0:14:39.32,Default,,0000,0000,0000,,logarithms here. Which says that\Nif you've got the log of a
Dialogue: 0,0:14:39.32,0:14:41.58,Default,,0000,0000,0000,,quantity minus the log of\Nanother quantity, that's the
Dialogue: 0,0:14:41.58,0:14:44.84,Default,,0000,0000,0000,,same as the log of the first\Nterm divided by the second term.
Dialogue: 0,0:14:45.49,0:14:49.76,Default,,0000,0000,0000,,So In other words, we can\Nsimplify these two logarithms to
Dialogue: 0,0:14:49.76,0:14:54.80,Default,,0000,0000,0000,,a single logarithm. The Ln of X\NPlus Delta X. That's the first
Dialogue: 0,0:14:54.80,0:14:58.68,Default,,0000,0000,0000,,one divided by the second one,\Nwhich is just X.
Dialogue: 0,0:15:00.99,0:15:05.01,Default,,0000,0000,0000,,And because I know where this is\Ngoing, I want to simplify this a
Dialogue: 0,0:15:05.01,0:15:07.88,Default,,0000,0000,0000,,little bit more. I'm going to\Nrecognize that within this
Dialogue: 0,0:15:07.88,0:15:10.74,Default,,0000,0000,0000,,bracket here. We've X divided by
Dialogue: 0,0:15:10.74,0:15:13.04,Default,,0000,0000,0000,,X. Which is just one.
Dialogue: 0,0:15:13.73,0:15:17.25,Default,,0000,0000,0000,,And we've Delta X divided
Dialogue: 0,0:15:17.25,0:15:23.13,Default,,0000,0000,0000,,by X. OK, so just a\Nbit of tidying up there.
Dialogue: 0,0:15:24.18,0:15:28.80,Default,,0000,0000,0000,,Also, I'm keeping in mind that\NI've got this result here about
Dialogue: 0,0:15:28.80,0:15:32.26,Default,,0000,0000,0000,,the exponential constant and in\Nthis limiting process here
Dialogue: 0,0:15:32.26,0:15:35.73,Default,,0000,0000,0000,,you'll remember we have the\Nquantity one plus T.
Dialogue: 0,0:15:36.66,0:15:40.34,Default,,0000,0000,0000,,In order that I can use this\Nresult, I'm going to make a
Dialogue: 0,0:15:40.34,0:15:43.74,Default,,0000,0000,0000,,substitution in here and I'm\Ngoing to write Delta X over X.
Dialogue: 0,0:15:44.92,0:15:47.37,Default,,0000,0000,0000,,As simply T so I can start to
Dialogue: 0,0:15:47.37,0:15:54.79,Default,,0000,0000,0000,,use this result. So I'm going to\Nlet Delta X over X equal T.
Dialogue: 0,0:15:56.04,0:15:58.45,Default,,0000,0000,0000,,So I'll have a log of\None plus T.
Dialogue: 0,0:16:00.80,0:16:03.53,Default,,0000,0000,0000,,We still going to use the\Nformula for differentiation from
Dialogue: 0,0:16:03.53,0:16:07.35,Default,,0000,0000,0000,,first principles. Let's go to\Nthat. We want F dash of X is the
Dialogue: 0,0:16:07.35,0:16:10.08,Default,,0000,0000,0000,,limit of all this quantity.\NLet's take this quantity and
Dialogue: 0,0:16:10.08,0:16:11.45,Default,,0000,0000,0000,,divide it by Delta X.
Dialogue: 0,0:16:12.27,0:16:14.54,Default,,0000,0000,0000,,So F of X Plus Delta X.
Dialogue: 0,0:16:15.25,0:16:19.10,Default,,0000,0000,0000,,Minus F of X divided by Delta
Dialogue: 0,0:16:19.10,0:16:25.33,Default,,0000,0000,0000,,X. Well, it's the top line. Is\Nthis quantity in here the
Dialogue: 0,0:16:25.33,0:16:31.29,Default,,0000,0000,0000,,logarithm? And we want to divide\Nit all by Delta X, but I've made
Dialogue: 0,0:16:31.29,0:16:36.25,Default,,0000,0000,0000,,a substitution here. I've let\NDelta X over X equal T. In other
Dialogue: 0,0:16:36.25,0:16:41.96,Default,,0000,0000,0000,,words, Delta X is X times T. So\Ninstead of the Delta X here I'm
Dialogue: 0,0:16:41.96,0:16:46.53,Default,,0000,0000,0000,,going to write this as one over\NX times T down there.
Dialogue: 0,0:16:48.36,0:16:51.19,Default,,0000,0000,0000,,OK, so far so this is the\Nexpression for our derivative.
Dialogue: 0,0:16:51.19,0:16:52.99,Default,,0000,0000,0000,,The only thing we have to do.
Dialogue: 0,0:16:53.73,0:16:57.68,Default,,0000,0000,0000,,Is take the limit as Delta\NX tends to 0. Let me write
Dialogue: 0,0:16:57.68,0:16:58.90,Default,,0000,0000,0000,,that expression down again.
Dialogue: 0,0:17:19.02,0:17:23.56,Default,,0000,0000,0000,,So that's where we've got to and\Nwhat we want to do is let Delta
Dialogue: 0,0:17:23.56,0:17:25.08,Default,,0000,0000,0000,,X 10 to 0 now.
Dialogue: 0,0:17:25.83,0:17:29.90,Default,,0000,0000,0000,,Remember that Delta X we've said\Nwas X times T.
Dialogue: 0,0:17:30.77,0:17:34.32,Default,,0000,0000,0000,,Remember also that X is a fixed\Npoint, but remember that from
Dialogue: 0,0:17:34.32,0:17:38.17,Default,,0000,0000,0000,,right back at the beginning when\Nwe came up with our formula for
Dialogue: 0,0:17:38.17,0:17:39.35,Default,,0000,0000,0000,,different station from first
Dialogue: 0,0:17:39.35,0:17:43.86,Default,,0000,0000,0000,,principles. The point a was a\Nfixed point. It was believed
Dialogue: 0,0:17:43.86,0:17:47.92,Default,,0000,0000,0000,,that was the variable point. So\Na was fixed, X was fixed.
Dialogue: 0,0:17:48.27,0:17:53.08,Default,,0000,0000,0000,,Now if Delta X tends to note an\Nex is fixed.
Dialogue: 0,0:17:54.17,0:17:57.88,Default,,0000,0000,0000,,That means that team must tend\Nto not as well. So in this
Dialogue: 0,0:17:57.88,0:17:59.87,Default,,0000,0000,0000,,process over here, we must let T
Dialogue: 0,0:17:59.87,0:18:06.48,Default,,0000,0000,0000,,10 to 0. So\Nwhat we really
Dialogue: 0,0:18:06.48,0:18:10.26,Default,,0000,0000,0000,,want is the
Dialogue: 0,0:18:10.26,0:18:13.95,Default,,0000,0000,0000,,limit. As TI tends to
Dialogue: 0,0:18:13.95,0:18:17.28,Default,,0000,0000,0000,,0. Of one over XT.
Dialogue: 0,0:18:17.83,0:18:22.63,Default,,0000,0000,0000,,Natural logarithm of 1 plus T\Nand when we work this out,
Dialogue: 0,0:18:22.63,0:18:27.03,Default,,0000,0000,0000,,that's actually going to give us\NF dash devex, the derivative
Dialogue: 0,0:18:27.03,0:18:28.63,Default,,0000,0000,0000,,that we're looking for.
Dialogue: 0,0:18:29.87,0:18:32.19,Default,,0000,0000,0000,,Now we're nearly finished. I\Njust want to tidy this up a
Dialogue: 0,0:18:32.19,0:18:34.70,Default,,0000,0000,0000,,little bit more, take the limit,\Nand then will be will be all
Dialogue: 0,0:18:34.70,0:18:39.61,Default,,0000,0000,0000,,done. You remember one of the\Nlaws of logarithms that I just
Dialogue: 0,0:18:39.61,0:18:42.04,Default,,0000,0000,0000,,wrote down was that M Log A?
Dialogue: 0,0:18:42.96,0:18:46.02,Default,,0000,0000,0000,,Was logged A to the power N.
Dialogue: 0,0:18:46.59,0:18:50.29,Default,,0000,0000,0000,,The power. The power can be\Nwritten outside or a number
Dialogue: 0,0:18:50.29,0:18:51.78,Default,,0000,0000,0000,,outside can go inside as a
Dialogue: 0,0:18:51.78,0:18:55.12,Default,,0000,0000,0000,,power. Well, with the one over T
Dialogue: 0,0:18:55.12,0:19:00.26,Default,,0000,0000,0000,,outside here. And that one over\NT I'm going to take inside the
Dialogue: 0,0:19:00.26,0:19:04.16,Default,,0000,0000,0000,,logarithm as a power. So I'm\Ngoing to write this as the
Dialogue: 0,0:19:04.16,0:19:07.16,Default,,0000,0000,0000,,limit. As TI tends to 0.
Dialogue: 0,0:19:07.67,0:19:11.73,Default,,0000,0000,0000,,Of one over X, I'm going to\Nleave that bit outside, but
Dialogue: 0,0:19:11.73,0:19:15.44,Default,,0000,0000,0000,,the one over T bit will go\Ninside and have logarithm
Dialogue: 0,0:19:15.44,0:19:19.84,Default,,0000,0000,0000,,one plus T and one over T\Ninside as a power and give
Dialogue: 0,0:19:19.84,0:19:22.20,Default,,0000,0000,0000,,me a one over T up there.
Dialogue: 0,0:19:23.69,0:19:28.10,Default,,0000,0000,0000,,Now this is looking much more\Nlike that result that we had at
Dialogue: 0,0:19:28.10,0:19:31.83,Default,,0000,0000,0000,,the very beginning. Remember the\Nresult we had at the beginning
Dialogue: 0,0:19:31.83,0:19:35.89,Default,,0000,0000,0000,,about the exponential constant.\NIt said that E was the limit as
Dialogue: 0,0:19:35.89,0:19:40.64,Default,,0000,0000,0000,,T tends to zero of one plus T to\Nthe power one over T.
Dialogue: 0,0:19:41.39,0:19:45.47,Default,,0000,0000,0000,,Now you see I've got the one\Nplus T to the one over T in here
Dialogue: 0,0:19:45.47,0:19:47.51,Default,,0000,0000,0000,,I've got a limit as T tends to
Dialogue: 0,0:19:47.51,0:19:52.65,Default,,0000,0000,0000,,zero there. The X is fixed, So\Nwhat we can do is we can leave
Dialogue: 0,0:19:52.65,0:19:57.30,Default,,0000,0000,0000,,the one over X there. We want to\Ntake the limit as T tends to
Dialogue: 0,0:19:57.30,0:20:00.40,Default,,0000,0000,0000,,zero of this expression in here,\Nwhich is the logarithm.
Dialogue: 0,0:20:01.11,0:20:03.12,Default,,0000,0000,0000,,Of one plus T to the one over T.
Dialogue: 0,0:20:03.67,0:20:08.19,Default,,0000,0000,0000,,But we've already said that as T\Ntends to 0, this quantity just
Dialogue: 0,0:20:08.19,0:20:09.59,Default,,0000,0000,0000,,tends to the exponential
Dialogue: 0,0:20:09.59,0:20:13.79,Default,,0000,0000,0000,,constant E. So we end up with\Njust simply Ln of E.
Dialogue: 0,0:20:15.21,0:20:19.43,Default,,0000,0000,0000,,Final result we had was that the\Nnatural logarithm of the
Dialogue: 0,0:20:19.43,0:20:20.97,Default,,0000,0000,0000,,exponential constant was one.
Dialogue: 0,0:20:21.67,0:20:26.27,Default,,0000,0000,0000,,So this bit in here is going to\Nsimplify to just one.
Dialogue: 0,0:20:27.76,0:20:31.73,Default,,0000,0000,0000,,In other words, we've deduced\Nthat the derivative F Dash Devex
Dialogue: 0,0:20:31.73,0:20:36.06,Default,,0000,0000,0000,,is simply just one over X Times\None or one over X.
Dialogue: 0,0:20:37.18,0:20:40.98,Default,,0000,0000,0000,,It's a very important result.\NLet me summarize it for you.
Dialogue: 0,0:20:41.93,0:20:45.88,Default,,0000,0000,0000,,We're saying that if F of X is\Nthe natural logarithm of X.
Dialogue: 0,0:20:46.46,0:20:52.92,Default,,0000,0000,0000,,Then F dash to vex. The\Nderivative is simply one over X
Dialogue: 0,0:20:52.92,0:20:57.22,Default,,0000,0000,0000,,and we've proved that using\Ndifferentiation from first
Dialogue: 0,0:20:57.22,0:21:01.83,Default,,0000,0000,0000,,principles. I'm well aware that\Nto do that we require lots of
Dialogue: 0,0:21:01.83,0:21:04.52,Default,,0000,0000,0000,,different bits of mathematics\Nfrom lots of different areas.
Dialogue: 0,0:21:04.52,0:21:06.31,Default,,0000,0000,0000,,Differentiation from first\Nprinciples, properties of
Dialogue: 0,0:21:06.31,0:21:10.50,Default,,0000,0000,0000,,logarithms, and so on, so you\Nmay need to look back at lots of
Dialogue: 0,0:21:10.50,0:21:14.39,Default,,0000,0000,0000,,other bits of work that you've\Ndone in the past in order to
Dialogue: 0,0:21:14.39,0:21:17.38,Default,,0000,0000,0000,,pull all this together, but\Nwe've got a very important
Dialogue: 0,0:21:17.38,0:21:20.96,Default,,0000,0000,0000,,result now that we can use in\Nlots of situations. What we've
Dialogue: 0,0:21:20.96,0:21:24.55,Default,,0000,0000,0000,,shown is that if you want to\Ndifferentiate with respect to X
Dialogue: 0,0:21:24.55,0:21:27.54,Default,,0000,0000,0000,,and natural logarithm function,\NEllen of X, the result that
Dialogue: 0,0:21:27.54,0:21:29.34,Default,,0000,0000,0000,,you'll get is one over X.
Dialogue: 0,0:21:31.49,0:21:36.09,Default,,0000,0000,0000,,We can use this result whatever\Nthe letter is in here. So for
Dialogue: 0,0:21:36.09,0:21:38.57,Default,,0000,0000,0000,,example, if we had D by DY.
Dialogue: 0,0:21:39.12,0:21:40.52,Default,,0000,0000,0000,,Of Allen of Y.
Dialogue: 0,0:21:41.06,0:21:43.26,Default,,0000,0000,0000,,The result will be one over Y.
Dialogue: 0,0:21:44.07,0:21:49.12,Default,,0000,0000,0000,,If we had D by DTLN of T,\Nthe result will be one over
Dialogue: 0,0:21:49.12,0:21:52.37,Default,,0000,0000,0000,,T. In other words, when you\Ndifferentiate the natural
Dialogue: 0,0:21:52.37,0:21:55.26,Default,,0000,0000,0000,,logarithm function, you'll\Nget one over whatever the
Dialogue: 0,0:21:55.26,0:21:55.98,Default,,0000,0000,0000,,variable is.
Dialogue: 0,0:21:57.08,0:21:58.59,Default,,0000,0000,0000,,OK, very important result.
Dialogue: 0,0:21:59.25,0:22:04.79,Default,,0000,0000,0000,,We're going to use that result\Nnow to move on and differentiate
Dialogue: 0,0:22:04.79,0:22:06.18,Default,,0000,0000,0000,,the exponential function.
Dialogue: 0,0:22:06.18,0:22:11.76,Default,,0000,0000,0000,,So now I'm going to look at the\Nfunction Y is a function of X,
Dialogue: 0,0:22:11.76,0:22:14.36,Default,,0000,0000,0000,,is equal to Y to the X.
Dialogue: 0,0:22:14.94,0:22:19.83,Default,,0000,0000,0000,,Or just simply Y equals E\Nto the X and we're going to
Dialogue: 0,0:22:19.83,0:22:22.46,Default,,0000,0000,0000,,learn how to differentiate\Nthis exponential function.
Dialogue: 0,0:22:23.99,0:22:26.41,Default,,0000,0000,0000,,Let me remind you something
Dialogue: 0,0:22:26.41,0:22:31.59,Default,,0000,0000,0000,,about logarithms. If A equals B\Nto the power, see.
Dialogue: 0,0:22:32.88,0:22:34.76,Default,,0000,0000,0000,,An alternative way of writing
Dialogue: 0,0:22:34.76,0:22:40.96,Default,,0000,0000,0000,,this expression. Is that the\Nlogarithm to base B of A?
Dialogue: 0,0:22:41.56,0:22:46.78,Default,,0000,0000,0000,,Is C. This is something that's\Nbeen discussed at great length
Dialogue: 0,0:22:46.78,0:22:51.64,Default,,0000,0000,0000,,in the unit on logarithms, but\Nit's a result that will need
Dialogue: 0,0:22:51.64,0:22:54.88,Default,,0000,0000,0000,,here in order to differentiate\Nthe exponential function.
Dialogue: 0,0:22:56.52,0:23:00.31,Default,,0000,0000,0000,,What I want to do is take\Nthis Y equals E to the power
Dialogue: 0,0:23:00.31,0:23:02.75,Default,,0000,0000,0000,,X and write it in its log\Nrhythmic form.
Dialogue: 0,0:23:03.99,0:23:09.11,Default,,0000,0000,0000,,And the log rhythmic form using\Nthe results over here is that
Dialogue: 0,0:23:09.11,0:23:11.25,Default,,0000,0000,0000,,the log to base E.
Dialogue: 0,0:23:11.85,0:23:17.51,Default,,0000,0000,0000,,Of Y is X, so that's just\Nthe log arhythmic form of
Dialogue: 0,0:23:17.51,0:23:19.40,Default,,0000,0000,0000,,the same expression here.
Dialogue: 0,0:23:20.87,0:23:25.24,Default,,0000,0000,0000,,Now log base E of X is\Nwhat we call Ln of X
Dialogue: 0,0:23:25.24,0:23:26.25,Default,,0000,0000,0000,,the natural logarithm.
Dialogue: 0,0:23:27.38,0:23:32.09,Default,,0000,0000,0000,,What I want to do is deal with\Nthis equivalent form instead of
Dialogue: 0,0:23:32.09,0:23:33.17,Default,,0000,0000,0000,,the original form.
Dialogue: 0,0:23:34.02,0:23:37.72,Default,,0000,0000,0000,,To differentiate this, what I\Nwant to do is differentiate both
Dialogue: 0,0:23:37.72,0:23:39.40,Default,,0000,0000,0000,,sides with respect to X.
Dialogue: 0,0:23:40.34,0:23:44.12,Default,,0000,0000,0000,,So I want deep IDX of the left
Dialogue: 0,0:23:44.12,0:23:51.07,Default,,0000,0000,0000,,hand side. Will equal D by\NDX of the right hand
Dialogue: 0,0:23:51.07,0:23:54.73,Default,,0000,0000,0000,,side. Now this is easy if you\Ndifferentiate X with respect to
Dialogue: 0,0:23:54.73,0:23:56.09,Default,,0000,0000,0000,,X, you just get one.
Dialogue: 0,0:23:57.33,0:23:59.34,Default,,0000,0000,0000,,The left hand side needs a bit
Dialogue: 0,0:23:59.34,0:24:03.29,Default,,0000,0000,0000,,more thought. What we've got\Nhere is that why is a function
Dialogue: 0,0:24:03.29,0:24:05.12,Default,,0000,0000,0000,,of XY is E to the X?
Dialogue: 0,0:24:05.81,0:24:08.58,Default,,0000,0000,0000,,And we're taking another\Nfunction of it. We want the
Dialogue: 0,0:24:08.58,0:24:12.84,Default,,0000,0000,0000,,logarithm of Y. So we've got a\Nfunction of a function we're
Dialogue: 0,0:24:12.84,0:24:15.33,Default,,0000,0000,0000,,trying to differentiate a\Nfunction of a function.
Dialogue: 0,0:24:16.70,0:24:19.29,Default,,0000,0000,0000,,So we need a rule for\Ndifferentiating functions of
Dialogue: 0,0:24:19.29,0:24:22.46,Default,,0000,0000,0000,,functions, and that's called\Na chain rule or the rule for
Dialogue: 0,0:24:22.46,0:24:23.90,Default,,0000,0000,0000,,differentiating function of\Na function.
Dialogue: 0,0:24:25.29,0:24:29.27,Default,,0000,0000,0000,,The rule says that if you want\Nto differentiate this function
Dialogue: 0,0:24:29.27,0:24:33.25,Default,,0000,0000,0000,,of a function with respect to X,\Nyou differentiate it with
Dialogue: 0,0:24:33.25,0:24:34.34,Default,,0000,0000,0000,,respect to Y.
Dialogue: 0,0:24:34.36,0:24:38.38,Default,,0000,0000,0000,,And then multiply by\NDYDX.
Dialogue: 0,0:24:40.11,0:24:43.22,Default,,0000,0000,0000,,So that's the function of a\Nfunction rule applied there.
Dialogue: 0,0:24:43.73,0:24:47.03,Default,,0000,0000,0000,,And if you've not met that\Nbefore, there's a unit on
Dialogue: 0,0:24:47.03,0:24:49.73,Default,,0000,0000,0000,,differentiating function of a\Nfunction using the chain rule.
Dialogue: 0,0:24:50.27,0:24:53.85,Default,,0000,0000,0000,,On the right hand\Nside we have one.
Dialogue: 0,0:24:55.80,0:24:59.30,Default,,0000,0000,0000,,Nearly finished all we have to\Ndo now is differentiate log Y
Dialogue: 0,0:24:59.30,0:25:03.10,Default,,0000,0000,0000,,with respect to Y, but we've\Njust done a lot of that. We've
Dialogue: 0,0:25:03.10,0:25:06.31,Default,,0000,0000,0000,,just realized improved that if\Nyou want to differentiate, log Y
Dialogue: 0,0:25:06.31,0:25:10.11,Default,,0000,0000,0000,,with respect to Y. You'll just\Nget one over Y, so following on
Dialogue: 0,0:25:10.11,0:25:12.15,Default,,0000,0000,0000,,from here will get one over Y.
Dialogue: 0,0:25:12.80,0:25:16.09,Default,,0000,0000,0000,,Multiplied by DYDX.
Dialogue: 0,0:25:16.09,0:25:18.66,Default,,0000,0000,0000,,And that's got to equal 1.
Dialogue: 0,0:25:20.03,0:25:23.51,Default,,0000,0000,0000,,So rearranging all this dyd X.
Dialogue: 0,0:25:24.01,0:25:28.37,Default,,0000,0000,0000,,Will equal just why times one\Nwhich is just Y.
Dialogue: 0,0:25:30.82,0:25:33.86,Default,,0000,0000,0000,,But why is E to the X?
Dialogue: 0,0:25:34.82,0:25:42.08,Default,,0000,0000,0000,,So we've shown that dyd X\Nis E to the X.
Dialogue: 0,0:25:42.09,0:25:46.75,Default,,0000,0000,0000,,So we've another very important\Nresult now, and the result is
Dialogue: 0,0:25:46.75,0:25:50.57,Default,,0000,0000,0000,,that if Y is E to the power X.
Dialogue: 0,0:25:51.31,0:25:53.96,Default,,0000,0000,0000,,Then the derivative dyd X.
Dialogue: 0,0:25:54.49,0:25:58.82,Default,,0000,0000,0000,,Is the same function E to the\Npower X, and that's a result
Dialogue: 0,0:25:58.82,0:26:01.82,Default,,0000,0000,0000,,which is well worth\Nremembering. It will crop up
Dialogue: 0,0:26:01.82,0:26:05.15,Default,,0000,0000,0000,,over and over again in lots of\Ndifferent calculations. Lots
Dialogue: 0,0:26:05.15,0:26:07.81,Default,,0000,0000,0000,,of different applications. In\Nfact the exponential function
Dialogue: 0,0:26:07.81,0:26:11.47,Default,,0000,0000,0000,,or multiples of it is the only\Nfunction which when you
Dialogue: 0,0:26:11.47,0:26:14.80,Default,,0000,0000,0000,,differentiate it, it gives you\Nthe same function that you
Dialogue: 0,0:26:14.80,0:26:17.13,Default,,0000,0000,0000,,started with, so that's\Nsomething worth remembering.