0:00:01.110,0:00:05.802
In this video, we're going to be[br]looking at how to differentiate

0:00:05.802,0:00:09.712
very simple functions from first[br]principles. So to begin with,

0:00:09.712,0:00:14.404
what we want to have a look at[br]is a straight line.

0:00:15.080,0:00:21.218
A straight line has a constant[br]gradient, or if we prefer its

0:00:21.218,0:00:27.914
rate of change of y with respect[br]to x is a constant.

0:00:29.080,0:00:33.433
If we take the straight line y[br]equals 3x + 2,

0:00:33.433,0:00:36.506
we can look at its gradient.

0:00:37.850,0:00:43.100
We take two points and look at[br]the change in y divided by the

0:00:43.100,0:00:44.225
change in x.

0:00:44.980,0:00:51.554
When x changes from -1 to 0, [br]y changes from -1 to 2,

0:00:51.554,0:00:53.674
and so the gradient is 3.

0:00:55.540,0:01:00.090
No matter which two points on[br]the line that we use, the value

0:01:00.090,0:01:02.190
of the gradient is always the

0:01:02.190,0:01:05.490
same. It is always 3.

0:01:07.430,0:01:13.646
Now we can see that in another[br]way, by looking at a table of values.

0:01:13.646,0:01:19.970
So let's take x ranging[br]in values from -3 up to 3.

0:01:19.970,0:01:26.674
And let's have a look at 3x.

0:01:26.680,0:01:33.112
Which is simply the values of x[br]we've got multiplied by 3.

0:01:33.830,0:01:38.012
And then add 2 to those values[br]to give us the function

0:01:38.012,0:01:42.296
y equals 3x + 2.

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Now as we look at this table[br]of values, what we see is that

0:01:53.536,0:02:01.648
for every unit increase in x we[br]always get 3 units increase in y.

0:02:01.648,0:02:07.764
So as we go from -3 to -2, [br]y goes from -7 to -4.

0:02:08.224,0:02:13.280
As we go from 0 to 1, y goes[br]from 2 to 5.

0:02:13.440,0:02:21.202
For every unit increase in x we[br]get the same increase in y of 3 units.

0:02:21.202,0:02:25.171
And so the gradient of this [br]straight line is 3 or

0:02:25.171,0:02:29.960
the rate of change of y with[br]respect to x is also 3.

0:02:31.190,0:02:36.110
Now, nice as it is that a straight[br]line has a constant gradient

0:02:36.110,0:02:40.620
and that y is changing at a [br]constant rate with respect to x,

0:02:40.620,0:02:45.160
this is not always the case. [br]So for instance when

0:02:45.160,0:02:48.733
you are in a car and you press[br]the accelerator, you watch the

0:02:48.733,0:02:53.509
needle on the speedometer. The[br]needle doesn't go up gradually,

0:02:53.509,0:02:57.940
it doesn't go up constantly. The[br]rate of increase of the speed varies.

0:02:59.596,0:03:05.400
Similarly the acceleration due to gravity,[br]now near the surface of the Earth,

0:03:05.400,0:03:09.500
we assume that it's constant.[br]But Newton actually showed that

0:03:09.500,0:03:14.772
it was proportional to 1 over[br]r squared, where r was the

0:03:14.772,0:03:19.338
distance that we were away from[br]the surface of the earth. So we

0:03:19.338,0:03:23.554
can see that as we move further[br]away from the surface of the

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Earth, the acceleration due to[br]gravity gets less, but it

0:03:27.624,0:03:32.488
doesn't change constantly. It[br]doesn't change uniformly.

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So to see what's happening we[br]need to look at some very simple curves.

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For a curve, we can see that if[br]we join any pair of points, we

0:03:44.790,0:03:46.182
get a straight line.

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But when we join different pairs[br]of points, we tend to get very

0:03:53.829,0:03:57.560
different straight lines with[br]very different gradients.

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We can see this again by looking[br]at very simple curve y equals x squared

0:04:06.910,0:04:12.110
and looking at a table[br]of values for it from -3 again

0:04:12.110,0:04:17.370
up to +3. So now let's calculate [br]the value of

0:04:17.370,0:04:22.089
the function y equals x squared[br]at each of these values.

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Now let's have a look how the[br]value of the function is

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changing as the value of[br]x changes. So if we have a

0:04:34.004,0:04:38.028
unit increase in x from -3 to[br]-2, we find that this time

0:04:38.028,0:04:43.866
the value of y is gone down[br]from 9 to 4, so it's gone down

0:04:43.866,0:04:49.596
by 5 units. Not a problem if[br]that decrease is going to be

0:04:49.596,0:04:53.050
sustained. But when we look at[br]the next unit.

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From -2 to -1, we find that it's [br]gone down from 4 to 1,

0:04:59.032,0:05:01.942
y has decreased by 3.

0:05:01.942,0:05:06.030
And in the next unit it is[br]decreased by 1, but then it

0:05:06.030,0:05:11.525
starts to increase again so we[br]can see that for a very, very

0:05:11.525,0:05:16.620
simple function like y equals X[br]squared, y is not changing

0:05:16.620,0:05:21.240
constantly with respect to x.[br]Now this is not a problem so

0:05:21.240,0:05:24.320
long as we can find a way of

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calculating it. To start that[br]process, let's just have a look

0:05:30.102,0:05:34.822
at a sketch of y equals x squared.

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And let me pick this[br]point on the curve.

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Now as we look at that look at[br]this very tiny little bit,

0:05:46.800,0:05:52.510
that's almost straight, and if I[br]was to home in and magnify, it

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would be almost more straight[br]and it would appear to be

0:05:57.300,0:06:02.340
straighter and straighter so as[br]we homed in closer and closer to

0:06:02.340,0:06:07.380
the point what we would see[br]would be much more like a

0:06:07.380,0:06:13.161
straight line. What straight[br]line might we be able to take

0:06:13.161,0:06:17.472
to represent that sort of[br]local straightness? And the

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answer would have to be a

0:06:20.346,0:06:23.208
tangent. Through the point.

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And that gives us what we need,[br]because this tangent is a

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straight line and we know that a[br]straight line has got a constant

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gradient, and so this is the[br]definition that we make.

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That the gradient

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of a curve

0:06:52.430,0:06:56.157
y equals a function of x

0:06:57.355,0:07:02.298
at a given point

0:07:03.260,0:07:06.008
is equal

0:07:07.432,0:07:11.137
to the gradient

0:07:12.740,0:07:15.618
of the tangent

0:07:17.830,0:07:20.560
to the curve

0:07:24.902,0:07:28.086
at that point.

0:07:30.386,0:07:32.742
So notice we haven't defined the

0:07:32.742,0:07:37.080
gradient of the curve globally,[br]so to speak,

0:07:37.080,0:07:39.480
we've said that at any particular

0:07:39.480,0:07:44.388
point, the gradient of that[br]curve is going to be the same

0:07:44.388,0:07:48.087
as the gradient of the[br]tangent. So now what we want

0:07:48.087,0:07:53.064
to see is how can we use this[br]definition to help us actually

0:07:53.064,0:07:57.067
calculate that gradient at any[br]particular point?

0:07:58.810,0:08:05.262
Let's take a fixed point P on[br]our curve and a sequence of

0:08:06.522,0:08:12.346
Points Q1, Q2 and so on. Getting[br]closer and closer to P.

0:08:14.100,0:08:19.833
We see that the lines from[br]P to each of the Qs gets

0:08:19.833,0:08:24.684
nearer and nearer to being[br]a tangent as the Qs come

0:08:24.684,0:08:26.889
nearer and nearer to P.

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So if we calculate the gradient[br]of one of these lines and let

0:08:34.745,0:08:40.530
the point Q Approach P along the[br]curve than the gradient of the

0:08:40.530,0:08:44.090
line should approach the[br]gradient of the Tangent.

0:08:45.730,0:08:49.090
And hence the gradient of[br]the curve.

0:08:50.320,0:08:56.326
So let's now try and set up[br]a calculation for the curve

0:08:56.326,0:09:03.120
y equals x squared at the point x[br]equals 3, so that this

0:09:03.120,0:09:06.192
calculation represents what[br]we've just seen.

0:09:06.730,0:09:10.018
So first of all, I'll[br]sketch the curve.

0:09:12.220,0:09:15.867
So there is y equals x squared.

0:09:16.560,0:09:19.080
Here will take our point.

0:09:19.650,0:09:24.993
Where x is equal to three. I'm[br]going to take a point Q,

0:09:27.410,0:09:32.132
which is near to P on the[br]curve and I'm going to join up

0:09:32.132,0:09:36.270
those two points with a straight line.

0:09:37.570,0:09:41.154
So there's my point, P as[br]my point Q and I'm going to take

0:09:41.154,0:09:45.455
a triangle formed by dropping a[br]perpendicular down to reach the

0:09:45.455,0:09:49.756
horizontal through P. So that will[br]give me a little right angle

0:09:49.756,0:09:53.275
triangle with the right angle at[br]the point are.

0:09:54.550,0:10:01.165
Now I'm going to take[br]small values of PR.

0:10:01.760,0:10:05.336
So that I get the coordinate

0:10:05.336,0:10:14.751
of Q. So let's line that up. [br]PR will be first of all 0.1,

0:10:15.780,0:10:21.890
then 0.01, 0.001

0:10:21.890,0:10:26.778
and finally 0.0001.

0:10:27.550,0:10:35.056
So what will be the x coordinate[br]of Q, x coordinate of Q?

0:10:36.016,0:10:41.720
Well, it will just be 3 + whatever PR is

0:10:41.720,0:10:51.711
so that will be 3.1, 3.01, 3.001 [br]and 3.0001.

0:10:52.221,0:10:59.639
Now I've got the x coordinate of Q, [br]let's calculate the y coordinate of Q.

0:11:01.019,0:11:04.951
So that means we've got to[br]square each of these,

0:11:04.951,0:11:09.944
so our first result will be 9.61 [br]when we square that.

0:11:10.814,0:11:15.956
Next is 9.0601.

0:11:16.323,0:11:23.207
Next 9.006001

0:11:23.207,0:11:30.219
and the next 9.00060001.

0:11:32.099,0:11:37.034
So what's the change in y[br]for each of these?

0:11:37.034,0:11:39.735
In other words, what's QR?

0:11:43.060,0:11:49.575
We know that the base for where[br]P is is if x is equal 3 at P,

0:11:49.575,0:11:54.706
y must be equal to 9. [br]So QR, the increase in y

0:11:54.706,0:11:59.432
we can get by taking 9 away [br]from each of these.

0:11:59.432,0:12:16.023
That's 0.61, 0.601, 0.006001,[br]0.00060001.

0:12:17.773,0:12:21.519
So now we need to [br]calculate the gradient of

0:12:21.519,0:12:27.140
each of these little segments[br]PQ. So what is that gradient?

0:12:27.140,0:12:35.423
will it be the increase in y, which is [br]QR, over PR, which is the increase in x.

0:12:35.830,0:12:44.072
So we've got 0.61 divided[br]by 0.1. So to do the first one

0:12:44.072,0:12:51.290
I'll write it down 0.61 divided [br]by 0.1, that is 6.1.

0:12:52.740,0:13:00.231
If we do the same calculation[br]here, the answer is 6.01.

0:13:00.761,0:13:06.566
Do the same calculation here [br]and it's 6.001.

0:13:07.146,0:13:13.952
And do the same calculation here [br]and it's 6.0001.

0:13:15.160,0:13:19.597
So let's just have a look what's[br]happening here. This gradient

0:13:19.597,0:13:29.400
seems to be getting nearer and[br]nearer to 6. And 6, perhaps by coincidence

0:13:29.849,0:13:38.366
is 2 times by 3, where 3 is the x value[br]at this point, P.

0:13:38.950,0:13:44.923
So we need to do this[br]calculation again, but we need

0:13:44.923,0:13:47.638
to do it in general.

0:13:48.750,0:13:52.710
So let's set that situation.

0:13:53.690,0:13:56.398
We take our axes.

0:13:58.470,0:14:04.698
And we plot our curve[br]y equals x squared.

0:14:05.410,0:14:14.278
Take my point P on the curve [br]and P is (x,y).

0:14:15.180,0:14:20.691
And I take a point Q a little[br]further along the curve.

0:14:21.810,0:14:24.526
How far along the curve is it?

0:14:25.220,0:14:31.336
Well, I'm going to increase x by[br]a very small amount.

0:14:32.680,0:14:36.118
So we got the same triangle.

0:14:37.110,0:14:39.360
As we had in the calculation.

0:14:39.970,0:14:42.838
So there's R.

0:14:42.840,0:14:49.425
And I am increasing x, to get to R,[br]by a very small amount and

0:14:49.425,0:14:54.254
that small amount is going to be[br]written as delta x.

0:14:54.960,0:15:01.504
Now I cannot emphasize enough[br]that delta x is not delta times by x.

0:15:01.504,0:15:08.822
It is a single symbol on its own, [br]and it represents a small change in x.

0:15:09.682,0:15:15.674
What's that resulted in? Its pushed the [br]point further along the curve to Q,

0:15:15.674,0:15:21.072
and so we've got a change in y[br]of delta y

0:15:21.480,0:15:34.464
So that the coordinates of Q are now[br]x + delta x and y + delta y.

0:15:35.670,0:15:39.376
Now. We've got the[br]diagram labeled up.

0:15:40.790,0:15:46.334
And so what we need to be able[br]to do is do some calculation.

0:15:46.334,0:15:49.827
What do we know? We know that[br]this is the graph of the curve

0:15:49.827,0:15:58.630
y equals x squared.[br]So here at the point Q we know that

0:15:58.630,0:16:06.896
y + delta y is equal to [br]x + delta x all squared.

0:16:07.930,0:16:18.785
This we can multiply out. It will[br]give us x squared plus 2x delta x,

0:16:18.785,0:16:22.948
and I'm going to put that[br]in a bracket to emphasize the

0:16:22.948,0:16:26.356
fact that delta x is a [br]single symbol,

0:16:27.133,0:16:32.703
plus delta x all squared.

0:16:34.560,0:16:40.620
But what else do we know? Will we[br]know at this point P that y is

0:16:40.620,0:16:43.360
equal to x squared.

0:16:44.460,0:16:51.300
So y + delta y is equal to this but [br]the y part is equal to x squared.

0:16:51.300,0:16:58.877
So that tells us therefore that the [br]change in y, delta y, is this bit here:

0:16:59.600,0:17:07.141
2x delta x plus delta x all squared.

0:17:07.901,0:17:12.886
So let's now calculate the gradient of[br]this line PQ.

0:17:13.808,0:17:18.210
The gradient of PQ,

0:17:20.210,0:17:23.440
well, it's QR over PR

0:17:23.440,0:17:26.120
or the change in y,

0:17:29.150,0:17:35.676
delta y, over the change in x.

0:17:35.676,0:17:39.237
And we've got expressions[br]for both of these:

0:17:39.237,0:17:44.538
delta x is delta x and delta y is this.

0:17:48.600,0:17:56.931
So we've got delta y over delta x[br]is equal to...

0:17:57.132,0:18:06.352
Let's remember that this was 2x[br]delta x plus delta x all squared

0:18:06.352,0:18:10.652
and that was all over delta x.

0:18:10.652,0:18:14.032
Now, delta x is a common factor here.

0:18:14.032,0:18:21.407
Let's take it out as a common factor: [br]2x + delta x

0:18:21.407,0:18:29.842
times by delta x; all over delta x. [br]Now delta x is a small positive

0:18:29.842,0:18:35.366
amount of x, so we can[br]cancel it out there and there.

0:18:35.366,0:18:44.504
So we have delta y over delta x[br]is equal to 2x + delta x.

0:18:45.944,0:18:48.472
Now remember what we're doing.

0:18:48.472,0:18:54.191
We've calculated the gradient of [br]this line PQ and we're going to let

0:18:54.191,0:18:59.670
Q come nearer and nearer to P [br]along the curve.

0:18:59.670,0:19:04.395
So we're letting delta x[br]tend to 0,

0:19:04.460,0:19:11.695
until this secant or cord, PQ, [br]becomes the tangent.

0:19:13.872,0:19:17.960
And so we can say that the gradient

0:19:20.411,0:19:21.610
of

0:19:22.165,0:19:23.849
the tangent

0:19:26.050,0:19:30.870
at P is equal to...

0:19:31.580,0:19:35.071
Now we have to use some[br]mathematical language.

0:19:35.071,0:19:40.382
What happens to this expression as[br]delta x approaches 0?

0:19:40.950,0:19:43.835
Well, it would seem that if[br]delta x got smaller and smaller

0:19:43.835,0:19:46.400
and smaller and smaller[br]and smaller, it got to zero.

0:19:46.400,0:19:51.975
This would become 2x. We are[br]approaching a limit of 2x and so

0:19:51.975,0:19:55.440
we use some mathematical[br]language to help us say that. So

0:19:55.440,0:19:57.935
we say that this is the limit,

0:19:59.460,0:20:02.996
as delta x tends to 0,

0:20:04.343,0:20:07.629
of delta y over delta x

0:20:08.219,0:20:10.090
is equal to

0:20:10.090,0:20:14.371
the limit as delta x tends to 0

0:20:14.701,0:20:18.006
of 2x + delta x

0:20:18.786,0:20:25.016
And if delta x is going off to 0,[br]this is just 2x and notice that

0:20:25.016,0:20:28.260
agrees with what we had before.[br]When we said that the gradient

0:20:28.260,0:20:35.852
was 2 times by 3 equal 6 at[br]the point where x is equal to 3.

0:20:37.270,0:20:40.720
Now we could do this[br]calculation in just the same

0:20:40.720,0:20:43.825
way for all kinds of[br]different curves, but it

0:20:43.825,0:20:48.310
would become a bit of a bind[br]if we had to keep writing

0:20:48.310,0:20:52.450
this down all the time. So we[br]have a special symbol for

0:20:52.450,0:20:53.830
this phrase, the limit.

0:20:55.530,0:21:02.240
as delta x tends to 0 of[br]delta y over delta x.

0:21:03.420,0:21:11.466
And we write that as dy by[br]dx and we say d y by d x

0:21:11.466,0:21:17.072
and that's a symbol all on its[br]own as a unit which stands for

0:21:17.072,0:21:25.298
the limit as delta x tends to 0[br]of delta y over delta x,

0:21:26.418,0:21:29.622
and in the case we've just

0:21:29.622,0:21:38.453
looked at, y equals X squared, [br]dy by dx we've just seen is equal to 2x.

0:21:44.060,0:21:48.990
Now. Another video in the[br]sequence will show how this is

0:21:48.990,0:21:54.170
done for y equals sine x and the[br]processes that we go through.

0:22:00.211,0:22:07.203
We often use function notation[br]y equals f of x.

0:22:08.523,0:22:10.868
So in function notation,

0:22:10.868,0:22:16.204
our point P, which[br]was the point x, y,

0:22:16.204,0:22:20.206
becomes the point[br]x, f of x.

0:22:21.240,0:22:32.492
Our point Q, which remember [br]was x + delta x, y + delta y

0:22:33.152,0:22:39.194
Well, this remains as x + delta x [br]for the x coordinate,

0:22:39.194,0:22:46.958
but the y + delta y. Well that's the[br]function of x + delta x.

0:22:49.240,0:22:54.070
And so if we want to have a look[br]at what we've done before, if

0:22:54.070,0:22:55.680
you remember, then we take

0:22:56.790,0:23:04.336
the change in y, that's delta y.[br]So the change in y is:

0:23:04.336,0:23:12.033
what we had here, at Q, [br]minus, what we had at P.

0:23:12.202,0:23:24.772
So that's f of x + delta x minus f of x, [br]and so our gradient delta y over delta x

0:23:24.772,0:23:34.786
is f of x + delta x minus f of x [br]all over delta x.

0:23:37.830,0:23:40.460
So dy by dx

0:23:44.070,0:23:49.770
is the limit as delta x tends to 0

0:23:50.780,0:23:55.817
of the change in y divided[br]by the change in x that gave rise to it.

0:23:57.300,0:23:58.980
Which is the limit

0:23:59.520,0:24:03.150
as delta x tends to 0

0:24:04.350,0:24:14.154
of f of x + delta x[br]minus f of x all over delta x

0:24:14.154,0:24:17.338
So, if we are using the language[br]of functions. That's the

0:24:17.338,0:24:20.916
language that we use and[br]sometimes for functions when

0:24:20.916,0:24:24.932
we're differentiating them[br]instead of using the symbol

0:24:24.932,0:24:30.454
dy by dx, sometimes use[br]the symbol a dash to indicate

0:24:30.454,0:24:33.968
that we've differentiated[br]with respect to x.

0:24:35.240,0:24:39.321
Let's just take one[br]example using this

0:24:39.321,0:24:45.925
notation, f of x is equal to the [br]function 1 over x.

0:24:48.240,0:24:51.815
Have a sketch of this curve for[br]positive values of x.

0:24:52.950,0:24:57.306
There it is coming down like[br]that. So if I take

0:24:57.850,0:25:03.817
a point P and I increase[br]the value of x by a small

0:25:03.817,0:25:09.325
amount to give me a second[br]point Q on the curve. Then

0:25:09.325,0:25:12.079
I can see that in fact,

0:25:13.120,0:25:15.840
there's my delta x.

0:25:16.930,0:25:22.494
And there's delta y, so for all[br]delta x still means a small

0:25:22.494,0:25:27.630
positive increasing x, delta y[br]just means the change in y and

0:25:27.630,0:25:32.766
we can see that this is actually[br]a decrease. Let me complete.

0:25:34.520,0:25:41.460
This triangle by drawing in[br]that secant, PQ and let's

0:25:41.460,0:25:45.624
have these coordinates,[br]this is x, y.

0:25:47.190,0:25:52.270
And this is. x + delta x

0:25:52.270,0:25:59.150
and then f of x + delta x.

0:26:01.140,0:26:09.280
So let's have a look what we've[br]got: delta y is the change in y

0:26:09.280,0:26:17.519
and so it's f of x + delta x [br]minus f of x.

0:26:22.800,0:26:27.000
And so delta y over delta x is

0:26:27.000,0:26:33.930
f of x + delta x minus f of x, [br]all divided by delta x.

0:26:37.940,0:26:43.430
And this is the gradient of[br]this line PQ.

0:26:43.990,0:26:51.005
OK, what is f of x + delta x[br]well, f of x is 1 over x so

0:26:51.005,0:26:55.430
this is 1 over [br]x + delta x.

0:26:55.980,0:27:00.957
minus f of x, which is just[br]1 over x,

0:27:00.957,0:27:03.706
all divided by

0:27:03.706,0:27:05.282
delta x.

0:27:05.970,0:27:13.277
Here we have two fractions and we are[br]subtracting them, so we need a

0:27:13.277,0:27:17.390
common denominator and that[br]common denominator will be

0:27:17.390,0:27:20.345
x times, x + delta x.

0:27:22.400,0:27:26.770
So I've multiplied this by, this[br]x + delta x, I have

0:27:26.770,0:27:30.680
multiplied it by x, so I must [br]multiply the 1 by x.

0:27:31.410,0:27:36.975
Minus, and I've multiplied the x[br]by x + delta x, so I must

0:27:36.975,0:27:44.005
multiply the 1 by x + delta x[br]and then this is all

0:27:44.005,0:27:47.497
divided by delta x.

0:27:49.027,0:27:51.610
So let's look at the top, x - delta x,[br]minus delta x.

0:27:51.610,0:27:58.175
So I've got x - x that goes out.

0:27:59.040,0:28:04.930
And I got on that line there - delta x.

0:28:06.170,0:28:11.186
All over, well, I'm dividing by[br]that, and by that, so effectively

0:28:11.186,0:28:13.694
I'm dividing by both of them.

0:28:19.107,0:28:21.605
So there we've got them both[br]written down.

0:28:22.095,0:28:25.893
Now delta x is a small positive quantity,

0:28:26.613,0:28:33.262
so I can divide top and bottom there[br]by delta x, canceling delta x out.

0:28:36.382,0:28:50.358
So delta y over delta x is equal to [br]-1 over, x + delta x, times by x.

0:28:50.358,0:28:54.395
Let me just multiply that[br]denominator out so I

0:28:54.395,0:29:00.904
have x squared plus [br]delta x times by x.

0:29:03.060,0:29:13.140
dy by dx is equal to the limit[br]as delta x tends to 0 of

0:29:13.140,0:29:16.290
delta y over delta x.

0:29:16.830,0:29:21.980
Which will be the limit[br]as delta x tends to 0

0:29:23.160,0:29:30.780
of -1 over x squared plus[br]delta x times by x.

0:29:32.010,0:29:36.030
Equals... Let's have a look at[br]what's happening in this denominator.

0:29:36.030,0:29:46.860
As delta x becomes very, very small[br]x times by delta x is also very very small

0:29:46.860,0:29:50.548
and very much smaller than x squared.

0:29:50.978,0:29:57.138
So as delta x goes off to 0, this term[br]here goes away to 0 and

0:29:57.138,0:29:59.839
just leaves us with the x squared.

0:29:59.839,0:30:05.134
So our limit is -1 over x squared.

0:30:06.800,0:30:11.486
And that's our calculation done[br]in function notation to find the

0:30:11.486,0:30:18.684
derivative, as we call dy by dx,[br]or the gradient of the curve at a point.

0:30:19.570,0:30:23.840
Now, the calculations that we've[br]done or all quite complicated.

0:30:24.860,0:30:27.900
You need to do one or two[br]to practice the method and

0:30:27.900,0:30:32.681
to see what's going on with[br]it - to see how it's actually working.

0:30:32.681,0:30:38.279
But, to do most of your differentiation,[br]there are a set of rules that you

0:30:38.279,0:30:43.290
can follow that will help you do the[br]differentiation much more quickly.