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In this video, we're going to be
looking at how to differentiate

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very simple functions from first
principles. So to begin with,

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what we want to have a look at
is a straight line.

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A straight line has a constant
gradient, or if we prefer its

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rate of change of y with respect
to x is a constant.

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If we take the straight line y
equals 3x + 2,

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we can look at its gradient.

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We take two points and look at
the change in y divided by the

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change in x.

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When x changes from -1 to 0, 
y changes from -1 to 2,

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and so the gradient is 3.

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No matter which two points on
the line that we use, the value

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of the gradient is always the

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same. It is always 3.

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Now we can see that in another
way, by looking at a table of values.

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So let's take x ranging
in values from -3 up to 3.

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And let's have a look at 3x.

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Which is simply the values of x
we've got multiplied by 3.

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And then add 2 to those values
to give us the function

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y equals 3x + 2.

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

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for every unit increase in x we
always get 3 units increase in y.

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So as we go from -3 to -2, 
y goes from -7 to -4.

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As we go from 0 to 1, y goes
from 2 to 5.

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For every unit increase in x we
get the same increase in y of 3 units.

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And so the gradient of this 
straight line is 3 or

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the rate of change of y with
respect to x is also 3.

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Now, nice as it is that a straight
line has a constant gradient

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and that y is changing at a 
constant rate with respect to x,

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this is not always the case. 
So for instance when

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you are in a car and you press
the accelerator, you watch the

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needle on the speedometer. The
needle doesn't go up gradually,

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it doesn't go up constantly. The
rate of increase of the speed varies.

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Similarly the acceleration due to gravity,
now near the surface of the Earth,

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we assume that it's constant.
But Newton actually showed that

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it was proportional to 1 over
r squared, where r was the

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distance that we were away from
the surface of the earth. So we

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can see that as we move further
away from the surface of the

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Earth, the acceleration due to
gravity gets less, but it

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doesn't change constantly. It
doesn't change uniformly.

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

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

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get a straight line.

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

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different straight lines with
very different gradients.

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

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and looking at a table
of values for it from -3 again

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up to +3. So now let's calculate 
the value of

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the function y equals x squared
at each of these values.

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

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

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unit increase in x from -3 to
-2, we find that this time

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the value of y is gone down
from 9 to 4, so it's gone down

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by 5 units. Not a problem if
that decrease is going to be

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sustained. But when we look at
the next unit.

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

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y has decreased by 3.

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And in the next unit it is
decreased by 1, but then it

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starts to increase again so we
can see that for a very, very

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simple function like y equals X
squared, y is not changing

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constantly with respect to x.
Now this is not a problem so

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long as we can find a way of

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

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at a sketch of y equals x squared.

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And let me pick this
point on the curve.

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

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that's almost straight, and if I
was to home in and magnify, it

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would be almost more straight
and it would appear to be

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straighter and straighter so as
we homed in closer and closer to

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the point what we would see
would be much more like a

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straight line. What straight
line might we be able to take

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to represent that sort of
local straightness? And the

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answer would have to be a

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tangent. Through the point.

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

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straight line and we know that a
straight line has got a constant

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

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That the gradient

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of a curve

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y equals a function of x

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at a given point

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is equal

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to the gradient

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of the tangent

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to the curve

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at that point.

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So notice we haven't defined the

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gradient of the curve globally,
so to speak,

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we've said that at any particular

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point, the gradient of that
curve is going to be the same

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as the gradient of the
tangent. So now what we want

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to see is how can we use this
definition to help us actually

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calculate that gradient at any
particular point?

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Let's take a fixed point P on
our curve and a sequence of

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Points Q1, Q2 and so on. Getting
closer and closer to P.

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We see that the lines from
P to each of the Qs gets

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nearer and nearer to being
a tangent as the Qs come

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nearer and nearer to P.

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

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the point Q Approach P along the
curve than the gradient of the

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line should approach the
gradient of the Tangent.

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And hence the gradient of
the curve.

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So let's now try and set up
a calculation for the curve

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y equals x squared at the point x
equals 3, so that this

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calculation represents what
we've just seen.

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So first of all, I'll
sketch the curve.

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So there is y equals x squared.

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Here will take our point.

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Where x is equal to three. I'm
going to take a point Q,

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which is near to P on the
curve and I'm going to join up

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those two points with a straight line.

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So there's my point, P as
my point Q and I'm going to take

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a triangle formed by dropping a
perpendicular down to reach the

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horizontal through P. So that will
give me a little right angle

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triangle with the right angle at
the point are.

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Now I'm going to take
small values of PR.

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So that I get the coordinate

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of Q. So let's line that up. 
PR will be first of all 0.1,

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then 0.01, 0.001

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and finally 0.0001.

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So what will be the x coordinate
of Q, x coordinate of Q?

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Well, it will just be 3 + whatever PR is

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so that will be 3.1, 3.01, 3.001 
and 3.0001.

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Now I've got the x coordinate of Q, 
let's calculate the y coordinate of Q.

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So that means we've got to
square each of these,

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so our first result will be 9.61 
when we square that.

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Next is 9.0601.

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Next 9.006001

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and the next 9.00060001.

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So what's the change in y
for each of these?

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In other words, what's QR?

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We know that the base for where
P is is if x is equal 3 at P,

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y must be equal to 9. 
So QR, the increase in y

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we can get by taking 9 away 
from each of these.

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That's 0.61, 0.601, 0.006001,
0.00060001.

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So now we need to 
calculate the gradient of

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each of these little segments
PQ. So what is that gradient?

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will it be the increase in y, which is 
QR, over PR, which is the increase in x.

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So we've got 0.61 divided
by 0.1. So to do the first one

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I'll write it down 0.61 divided 
by 0.1, that is 6.1.

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If we do the same calculation
here, the answer is 6.01.

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Do the same calculation here 
and it's 6.001.

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And do the same calculation here 
and it's 6.0001.

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So let's just have a look what's
happening here. This gradient

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seems to be getting nearer and
nearer to 6. And 6, perhaps by coincidence

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is 2 times by 3, where 3 is the x value
at this point, P.

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So we need to do this
calculation again, but we need

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to do it in general.

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So let's set that situation.

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We take our axes.

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And we plot our curve
y equals x squared.

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Take my point P on the curve 
and P is (x,y).

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And I take a point Q a little
further along the curve.

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How far along the curve is it?

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Well, I'm going to increase x by
a very small amount.

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So we got the same triangle.

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As we had in the calculation.

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So there's R.

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And I am increasing x, to get to R,
by a very small amount and

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that small amount is going to be
written as delta x.

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Now I cannot emphasize enough
that delta x is not delta times by x.

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It is a single symbol on its own, 
and it represents a small change in x.

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What's that resulted in? Its pushed the 
point further along the curve to Q,

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and so we've got a change in y
of delta y

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So that the coordinates of Q are now
x + delta x and y + delta y.

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Now. We've got the
diagram labeled up.

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And so what we need to be able
to do is do some calculation.

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What do we know? We know that
this is the graph of the curve

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y equals x squared.
So here at the point Q we know that

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y + delta y is equal to 
x + delta x all squared.

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This we can multiply out. It will
give us x squared plus 2x delta x,

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and I'm going to put that
in a bracket to emphasize the

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fact that delta x is a 
single symbol,

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plus delta x all squared.

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But what else do we know? Will we
know at this point P that y is

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equal to x squared.

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So y + delta y is equal to this but 
the y part is equal to x squared.

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So that tells us therefore that the 
change in y, delta y, is this bit here:

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2x delta x plus delta x all squared.

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So let's now calculate the gradient of
this line PQ.

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The gradient of PQ,

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well, it's QR over PR

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or the change in y,

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delta y, over the change in x.

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And we've got expressions
for both of these:

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delta x is delta x and delta y is this.

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So we've got delta y over delta x
is equal to...

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Let's remember that this was 2x
delta x plus delta x all squared

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and that was all over delta x.

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Now, delta x is a common factor here.

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Let's take it out as a common factor: 
2x + delta x

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times by delta x; all over delta x. 
Now delta x is a small positive

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amount of x, so we can
cancel it out there and there.

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So we have delta y over delta x
is equal to 2x + delta x.

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Now remember what we're doing.

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We've calculated the gradient of 
this line PQ and we're going to let

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Q come nearer and nearer to P 
along the curve.

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So we're letting delta x
tend to 0,

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until this secant or cord, PQ, 
becomes the tangent.

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And so we can say that the gradient

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of

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the tangent

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at P is equal to...

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Now we have to use some
mathematical language.

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What happens to this expression as
delta x approaches 0?

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Well, it would seem that if
delta x got smaller and smaller

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and smaller and smaller
and smaller, it got to zero.

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This would become 2x. We are
approaching a limit of 2x and so

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we use some mathematical
language to help us say that. So

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we say that this is the limit,

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as delta x tends to 0,

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of delta y over delta x

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is equal to

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the limit as delta x tends to 0

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of 2x + delta x

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And if delta x is going off to 0,
this is just 2x and notice that

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agrees with what we had before.
When we said that the gradient

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was 2 times by 3 equal 6 at
the point where x is equal to 3.

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Now we could do this
calculation in just the same

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way for all kinds of
different curves, but it

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would become a bit of a bind
if we had to keep writing

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this down all the time. So we
have a special symbol for

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this phrase, the limit.

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as delta x tends to 0 of
delta y over delta x.

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And we write that as dy by
dx and we say d y by d x

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and that's a symbol all on its
own as a unit which stands for

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the limit as delta x tends to 0
of delta y over delta x,

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and in the case we've just

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looked at, y equals X squared, 
dy by dx we've just seen is equal to 2x.

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Now. Another video in the
sequence will show how this is

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done for y equals sine x and the
processes that we go through.

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We often use function notation
y equals f of x.

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So in function notation,

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our point P, which
was the point x, y,

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becomes the point
x, f of x.

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Our point Q, which remember 
was x + delta x, y + delta y

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Well, this remains as x + delta x 
for the x coordinate,

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but the y + delta y. Well that's the
function of x + delta x.

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And so if we want to have a look
at what we've done before, if

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you remember, then we take

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the change in y, that's delta y.
So the change in y is:

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what we had here, at Q, 
minus, what we had at P.

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So that's f of x + delta x minus f of x, 
and so our gradient delta y over delta x

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is f of x + delta x minus f of x 
all over delta x.

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00:23:37,830 --> 00:23:40,460
So dy by dx

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00:23:44,070 --> 00:23:49,770
is the limit as delta x tends to 0

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of the change in y divided
by the change in x that gave rise to it.

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00:23:57,300 --> 00:23:58,980
Which is the limit

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as delta x tends to 0

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of f of x + delta x
minus f of x all over delta x

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00:24:14,154 --> 00:24:17,338
So, if we are using the language
of functions. That's the

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language that we use and
sometimes for functions when

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we're differentiating them
instead of using the symbol

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dy by dx, sometimes use
the symbol a dash to indicate

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that we've differentiated
with respect to x.

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00:24:35,240 --> 00:24:39,321
Let's just take one
example using this

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notation, f of x is equal to the 
function 1 over x.

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00:24:48,240 --> 00:24:51,815
Have a sketch of this curve for
positive values of x.

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00:24:52,950 --> 00:24:57,306
There it is coming down like
that. So if I take

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a point P and I increase
the value of x by a small

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00:25:03,817 --> 00:25:09,325
amount to give me a second
point Q on the curve. Then

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00:25:09,325 --> 00:25:12,079
I can see that in fact,

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there's my delta x.

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And there's delta y, so for all
delta x still means a small

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positive increasing x, delta y
just means the change in y and

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00:25:27,630 --> 00:25:32,766
we can see that this is actually
a decrease. Let me complete.

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00:25:34,520 --> 00:25:41,460
This triangle by drawing in
that secant, PQ and let's

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00:25:41,460 --> 00:25:45,624
have these coordinates,
this is x, y.

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And this is. x + delta x

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00:25:52,270 --> 00:25:59,150
and then f of x + delta x.

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So let's have a look what we've
got: delta y is the change in y

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and so it's f of x + delta x 
minus f of x.

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00:26:22,800 --> 00:26:27,000
And so delta y over delta x is

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f of x + delta x minus f of x, 
all divided by delta x.

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00:26:37,940 --> 00:26:43,430
And this is the gradient of
this line PQ.

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OK, what is f of x + delta x
well, f of x is 1 over x so

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this is 1 over 
x + delta x.

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minus f of x, which is just
1 over x,

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all divided by

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00:27:03,706 --> 00:27:05,282
delta x.

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00:27:05,970 --> 00:27:13,277
Here we have two fractions and we are
subtracting them, so we need a

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common denominator and that
common denominator will be

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00:27:17,390 --> 00:27:20,345
x times, x + delta x.

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00:27:22,400 --> 00:27:26,770
So I've multiplied this by, this
x + delta x, I have

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00:27:26,770 --> 00:27:30,680
multiplied it by x, so I must 
multiply the 1 by x.

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00:27:31,410 --> 00:27:36,975
Minus, and I've multiplied the x
by x + delta x, so I must

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00:27:36,975 --> 00:27:44,005
multiply the 1 by x + delta x
and then this is all

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00:27:44,005 --> 00:27:47,497
divided by delta x.

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00:27:49,027 --> 00:27:51,610
So let's look at the top, x - delta x,
minus delta x.

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00:27:51,610 --> 00:27:58,175
So I've got x - x that goes out.

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00:27:59,040 --> 00:28:04,930
And I got on that line there - delta x.

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00:28:06,170 --> 00:28:11,186
All over, well, I'm dividing by
that, and by that, so effectively

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00:28:11,186 --> 00:28:13,694
I'm dividing by both of them.

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00:28:19,107 --> 00:28:21,605
So there we've got them both
written down.

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00:28:22,095 --> 00:28:25,893
Now delta x is a small positive quantity,

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00:28:26,613 --> 00:28:33,262
so I can divide top and bottom there
by delta x, canceling delta x out.

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00:28:36,382 --> 00:28:50,358
So delta y over delta x is equal to 
-1 over, x + delta x, times by x.

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00:28:50,358 --> 00:28:54,395
Let me just multiply that
denominator out so I

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00:28:54,395 --> 00:29:00,904
have x squared plus 
delta x times by x.

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00:29:03,060 --> 00:29:13,140
dy by dx is equal to the limit
as delta x tends to 0 of

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00:29:13,140 --> 00:29:16,290
delta y over delta x.

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00:29:16,830 --> 00:29:21,980
Which will be the limit
as delta x tends to 0

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00:29:23,160 --> 00:29:30,780
of -1 over x squared plus
delta x times by x.

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00:29:32,010 --> 00:29:36,030
Equals... Let's have a look at
what's happening in this denominator.

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00:29:36,030 --> 00:29:46,860
As delta x becomes very, very small
x times by delta x is also very very small

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00:29:46,860 --> 00:29:50,548
and very much smaller than x squared.

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00:29:50,978 --> 00:29:57,138
So as delta x goes off to 0, this term
here goes away to 0 and

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00:29:57,138 --> 00:29:59,839
just leaves us with the x squared.

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00:29:59,839 --> 00:30:05,134
So our limit is -1 over x squared.

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00:30:06,800 --> 00:30:11,486
And that's our calculation done
in function notation to find the

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00:30:11,486 --> 00:30:18,684
derivative, as we call dy by dx,
or the gradient of the curve at a point.

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00:30:19,570 --> 00:30:23,840
Now, the calculations that we've
done or all quite complicated.

313
00:30:24,860 --> 00:30:27,900
You need to do one or two
to practice the method and

314
00:30:27,900 --> 00:30:32,681
to see what's going on with
it - to see how it's actually working.

315
00:30:32,681 --> 00:30:38,279
But, to do most of your differentiation,
there are a set of rules that you

316
00:30:38,279 --> 00:30:43,290
can follow that will help you do the
differentiation much more quickly.