00:00Alright guys, so in this
00:03lesson I want to do
00:04some implicit differentiation. Now firstly
00:08what is it? Well imagine
00:11a circle. So this is
00:13the equation of a circle
00:14with radius 3 x squared
00:16plus y squared equals 9.
00:18Now what we're used to
00:19firstly implicit an explicit equation.
00:23So when you have an
00:24equation y equals f of
00:27x, this
00:28is someone called an explicit
00:30equation. Whereas something like this,
00:34x where plus y squared
00:34equals nine, is sometimes called
00:36an implicit equation. So it's
00:39basically where y is not
00:40expressed in terms of x
00:42on its own. So this
00:45is the equation of a
00:46circle. It is an implicit
00:48equation. And what I want
00:53to do is find the
00:55derivative
00:56So what I want is
00:58I want to be able
00:59to find the gradient of
01:02the curve at any point.
01:05So what I could do
01:06is, because remember a circle
01:10is not a function. So
01:11what I could do is
01:12I could say y squared
01:14equals 9 minus x squared,
01:18y equals the square root
01:20of 9 minus x squared,
01:21and then this is
01:24either plus or minus that
01:27and I can get two
01:31different derivatives and I can
01:33use, so say for example
01:35I wanted to get the
01:36gradient of the curve at,
01:39let's say at x equals
01:40two, it would be on
01:43my, it would be both
01:46this and
01:52Well, that's not great. You'd
01:54be both this and this
01:58depending if you wanted because
02:00when x is 2, the
02:02curve is both here and
02:04here, if that was straight.
02:07So you could do a
02:08twice, that would be called
02:10differentiating explicitly, but what I'm
02:13going to show you is
02:14a trick. Well, it's not
02:16a trick. It's a well
02:17-established rigor.
02:20your mathematical process, but whatever.
02:24Let's call it a trick
02:25for now. We are going
02:27to differentiate this in placidly.
02:30And what we do is,
02:30so we have x squared
02:32plus y squared equals nine.
02:35What we do is we
02:36get the derivative of the
02:37left and we get the
02:39derivative of the right. So
02:40we're doing the same thing
02:41to both sides of the
02:42equation. So I'm basically saying
02:44get d dx of
02:48this and get d dx
02:58of nine. Now we know
03:02the derivative of, we know
03:05the derivative of nine is
03:06zero and we know the
03:08derivative of x squared is
03:102x. So we're in the
03:12derivative of the left and
03:13the derivative of the right.
03:14The derivative of the left,
03:16this bit,
03:16and it's with respect to
03:18x. So the derivative of
03:19x squared is 2x, we
03:20know that. Now this is
03:22the, this is the tricky
03:24bit. How do I differentiate
03:25y squared with respect to
03:28x? Well, remember, y is
03:33a function of x. So,
03:38what we do is we
03:39use the chain rule and
03:41the chain rule is we
03:43differentiate the outside
03:44the outside function, which is
03:462 times y. And then
03:51we differentiate the inside function,
03:53which is y. And we
03:54multiplies with 2y times, what's
03:57the derivative of y? Well,
03:58let's just dy dx. And
04:03then the derivative of 9
04:04equals 0. Now, this bit,
04:08when you first see this,
04:09this is confusing. So let
04:10me go back to that.
04:12This is the, this is
04:12the,
04:12the only part that is
04:14new and that's confusing. And
04:16if you understand this, you'll
04:18be fine for the rest
04:18of it. Let me, let
04:21me, let me try and
04:22explain this with, with another
04:23example. Imagine instead of y
04:27squared, I want to differentiate
04:30sine x squared. So imagine,
04:36okay, this is the really
04:37explained, consider
04:40This is nothing to do
04:42with this question. I'm just
04:46using this example to help
04:47you understand it. Consider y
04:49equals sin x. Consider y
04:53equals sin of x. What's
04:58the derivative? What's the derivative?
05:01D dx of sin x.
05:04The derivative of sin x
05:06squared. What we do is,
05:08we differentiate the outside function,
05:11which is the sin of
05:12x squared, so it becomes
05:13two times sin of x.
05:17And then we multiply it
05:18by the derivative of the
05:19inside function, so it's the
05:21derivative of sin x, which
05:22is cos x. So it
05:23becomes two sin x cos
05:25x, or two times y
05:29times dy dx. This is
05:33purely the chain rule. And
05:35if you think of y
05:36that
05:36I use sign x here
05:39because it's nice when you
05:41explain it. Then you have
05:43to multiply it by the
05:45derivative. So hopefully this makes
05:48sense and if this makes
05:49sense, then this makes sense.
05:52Okay, fine. Now, once I
05:55have done that, I can
05:58get the derivative. I can
06:00say dy dx. dy dx
06:04is equal to
06:04negative 2x over negative 2x
06:11divided by 2y, which is
06:14actually equal to negative x
06:16over y. And that is
06:19my derivative. And it's a
06:22derivative in terms of x
06:23and y. So now when
06:25you're getting the derivative when
06:27x is 2, well, what
06:29you end up substituting into
06:32my derivative
06:33is my x value and
06:35my y value. Hence, I
06:37get there will only be
06:39one solution. It will give
06:40me a depending which one
06:42you want, but it will
06:43give me that solution because
06:45that gradient, because that I
06:48will sub in two for
06:49x and well, what is
06:52it? The square root of
06:54five for y. And if
06:56I wanted this, the gradient
06:57of this of the graph
07:00at this point,
07:01I'd sewn to a negative
07:02root 5. Okay, so that's
07:05kind of the theory of
07:07implicit differentiation. I want to
07:09do an example. This is,
07:13it's pretty much taken from
07:16a pass paper question. I've
07:18altered it very slightly just
07:20the first part I actually
07:22want us to get the
07:23derivative. Okay, so what we're
07:25going to do is, we're
07:27going to differentiate the left
07:29differentiate the right. So the
07:30derivative of y cubed, similarly
07:33similar to how I differentiated
07:35y squared, I'm going to
07:37do three times y squared,
07:41three times y squared times
07:43dy dx. Same thing, derivative
07:48of y cubed is three
07:50y squared and then I
07:51multiply by the derivative of
07:52y, which is just dy
07:53dx. Minus. Okay.
07:57Now this is a product.
08:00I have 3x times y
08:03squared. This is the part
08:07of these questions that always
08:09causes lots of problems because
08:11there's a minus here to
08:12begin with. There's a 3
08:14here and I've got my
08:17product with x and y
08:18squared. So I'm going to
08:20leave the minus there and
08:21I'm going to put a
08:22bracket here.
08:25My u and you could
08:26actually take out the, well,
08:28why don't we do that
08:28actually? I can take out
08:30the three as well. I
08:30take out minus three and
08:32I'm gonna do my product
08:33rule with x and y
08:34squared. Now I'm gonna do
08:36the product rule over here
08:37to the right, like I
08:38always do. So u is
08:40equal to x, v is
08:43equal to y squared. d
08:45u dx equals one, d
08:49v dx equals, okay.
08:53Any more space here? It
08:55would be very, very, very
08:56careful. dV dx equals 2y
09:00dy dx. That's like my
09:03circle example above. 2y dy
09:07dx. You don't even need
09:08that dot there. Okay, now
09:11in the brackets I'm going
09:12to put u times dV
09:14dx x times dV dx,
09:17which is 2y. So if
09:19you don't mind I'm going
09:20to put the two firsts.
09:212 x y dy dx.
09:27That's just this times this.
09:29Plus v times d dx,
09:31which is just y squared
09:33times 1, which is just
09:35y squared. Close my bracket.
09:40Plus, what's the derivative of
09:42x cubed? Well, that's easy.
09:44It's just 3 x squared.
09:47Now here we have another,
09:48let me just move
09:49this bit over here. We
09:53have another situation here where
09:58we have a common mistake
10:01that occurs. You have to
10:04differentiate the right as well,
10:07so we differentiate the left,
10:08and we differentiate the right.
10:09The common mistake is to
10:10put down 27. That is
10:12wrong. You have to differentiate
10:1427, and the derivative of
10:1627 is
10:170. Okay. Now I'm going
10:23to multiply all this out.
10:25But if I actually look
10:26at this, look, these three
10:27cancels. Nice. Let's get, let's
10:29cancel the three. Three, three,
10:31three. I'm dividing everything by
10:33three, dividing cross by three.
10:34So now I have y
10:35squared, dy dx, careful with
10:38this minus, minus 2xy dy
10:43dx.
10:45minus y squared plus x
10:51squared equals zero. Now the
10:55first question is find dy
10:56dx. How do I find
10:58dy dx? Well, the terms
11:01that have dy dx in
11:02them are the first two
11:03terms. So I'm going to
11:05leave them here. So it's
11:06y squared dy dx minus
11:112xy.
11:142i dx equals, I'm gonna
11:17add y squared to both
11:20sides. So y squared on
11:21the right, and I'm gonna
11:23subtract x squared. Then, I'm
11:27gonna take out my dy
11:29dx, because I'm trying to
11:32find dy dx on its
11:33own. So I factor out
11:34dy dx. I get y
11:36squared minus 2xy equals y
11:41squared.
11:41minus x squared and then
11:44finally, dy dx equals y
11:48squared minus x squared over
11:53y squared minus 2 x
11:57y. Done. This is the
12:01derivative. Now this is the
12:03type of question I call
12:05familiar difficult. Definitely I'm not
12:08going to say that's an
12:09easy question
12:09But I like that question
12:11and I would like to
12:13see that in an exam
12:14because if that if that
12:16question comes up in an
12:17exam I'm thinking at least
12:19well I know how to
12:21do it. Yes, like there's
12:23plenty places for me to
12:24make mistakes, but it's the
12:26type of question that I've
12:27seen before and I know
12:29how to do it. Part
12:32B also is not is
12:34not too bad once once
12:35you've done this. So it's
12:36just finding the coordinates at
12:37the points and the
12:38graph where dy dx equals
12:39zero. So this slight change
12:42I made to the past
12:43paper question was I actually
12:44said to find this. They
12:46just said this, find the
12:47coordinates on the graph where
12:48dy dx was zero. And
12:49if they do that, it's
12:51a little bit easier because
12:52you can just, even at
12:53this point here, you can
12:55just sub zero for this
12:57and zero for this. And
12:59you just left with, well,
13:02you just left with, with,
13:06with,
13:06minus y squared plus x
13:08squared equals zero and solve
13:10from there. But I could
13:12either do that or now
13:14that I have dy dx
13:14I can just do this
13:15as well. So it says
13:16find the coordinates of the
13:18points in the graph where
13:20dy dx equals zero. So
13:22let's do part B down
13:24here, part B. dy dx
13:27equals zero, that means, look,
13:29this is my dy dx
13:30equals zero, that means the
13:33numerator, y squared,
13:34square minus x squared has
13:35to equal zero. That means
13:37y squared equals x squared
13:40and then y is equal
13:43to the square root of
13:45x squared, which is either
13:47plus or minus x. Okay.
13:53Now, am I finished? No,
13:57I need to find the
13:57coordinates. What am I going
14:00to do? Well, this
14:02is my original curve. So
14:06let's bring that down here.
14:12Okay, this is my original
14:13curve. I need to find
14:16the coordinates where y is
14:23equal to x and negative
14:24x and it satisfies this.
14:26So what I'm going to
14:27do is I'm going to
14:28sub in I am going
14:30sub in x for y.
14:33So when y equals x,
14:36let's do this one first.
14:38When y equals x. So
14:39I'm going to sub in
14:39x. So now I have
14:41x cubed instead of y
14:42cubed, x cubed, minus 3
14:45x times x squared plus
14:50x cubed equals 27. This
14:57is actually x cubed.
14:58is x cubed minus 3x
15:03cubed plus x cubed which
15:05is minus x cubed equals
15:1027 x cubed equals negative
15:1727 x is equal to
15:22the cubed root of negative
15:2627 x
15:267 was the cube root
15:28of negative 27, but it's
15:29negative 3. So when x
15:31is equal to negative 3,
15:36y is equal to therefore,
15:40y is equal to negative
15:433. So my point is
15:47negative 3, negative 3. My
15:52second one is when
15:54Because remember, x, y can
15:55be positive x or negative
15:57x. So when y equals
15:59negative x, there's seven negative
16:02x here. So I have
16:04negative x. Don't forget my
16:05brackets cubed minus 3x negative
16:10x squared plus x cubed
16:15equals 27. Negative x cubed
16:20is negative x cubed.
16:22minus 3, this becomes positive,
16:26so it's still minus, so
16:28it's minus 3x cubed plus
16:33x cubed equals 27, those
16:40cancel. I'm left with, I'm
16:44gonna divide by negative 3
16:45here, so I've x cubed
16:47equals negative
16:509, excuse is equal to
16:53the negative 9. Therefore, x
16:55is equal to the cube
16:59root, the cubed root of
17:02negative 9. This is a
17:06paper one question and I
17:07don't know if a calculator.
17:08So I'm just going to
17:08have to leave it as
17:09x is equal to the
17:10cube root of 9. Therefore,
17:15why is equal to negative?
17:18negative negative, the cube root
17:21of negative nine. So negative,
17:25the cube root of negative
17:26nine. And the point is
17:29therefore the point is the
17:33cube root of negative nine
17:36comma negative the cube root
17:40of negative nine, which equals
17:44the cube root of negative
17:46nine, which equals the cube
17:46root of negative nine,
17:46cubed root of negative nine,
17:50comma, the cubed root of
17:54nine. The reason for that
17:56is, because the cubed root
17:57of negative nine will give
17:58me a negative number, and
18:00then it's minus that negative
18:02number. So it's a positive
18:03number, which is the same
18:04as the cubed root of
18:05nine. And as I don't
18:06have a calculator, I can't
18:09simplify that any further unless
18:10you know what the cubed
18:11root of nine is. I
18:13don't.
18:14So that is implicit differentiation.
18:17Hopefully you'll see that it's
18:20actually easier to do it
18:21that way than to imagine.
18:23Imagine me trying to write
18:27this equation out implicitly and
18:30then try to figure out
18:32the derivative of that function.
18:37I want to just show
18:38you actually though when I
18:39graph it, if I graph
18:42this,
18:42What I get, well look,
18:47this is the relation, you
18:49can see it's clearly not
18:50a function, because it doesn't
18:52pass the vertical line test.
18:54But where is the gradient,
18:56where is the derivative equal
18:57to zero, where is the
18:58gradient equal to zero? Well,
19:01hopefully you can see it,
19:02it is here, and here.
19:07Now what are these points?
19:08Well,
19:11My solution where the derivative
19:18equals zero was at negative
19:20three, negative three. So there
19:22should be one at y
19:24equals negative three. So right
19:26here, all right where x
19:32is negative three and y
19:33is negative three and yes,
19:34you can see it there.
19:36I can't actually click on
19:37that there, but I can't.
19:39Well, negative 3, negative 3,
19:42negative 3, that this is
19:44a tangent to my curve.
19:48You can clearly see it.
19:51Okay. And then this one
19:54is at, well this was
19:56my y equals the cubed
20:00root. I can find the
20:02cubed root here.
20:07root of 9, remember became
20:12positive, and there we go,
20:13same thing. You can see
20:16there that we have tangent
20:26when y is equal to
20:27the cube root of 9
20:29and x is negative, the
20:32cube root of the cube
20:35root
20:35negative 9, which is the
20:37same as negative the cube
20:38root of 9. Okay, that's
20:40it. That's implicit differentiation. Hopefully
20:42that made sense. Any questions,
20:46as always, let me know.