www.mathcentre.ac.uk/.../8.1%20Differentiation%201st%20principles%20Tangent.mp4

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

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

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

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

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

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

we can look at its gradient.

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

change in x.

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

and so the gradient is 3.

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

of the gradient is always the

same. It is always 3.

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

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

And let's have a look at 3x.

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

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

y equals 3x + 2.

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

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

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

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

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

And so the gradient of this
straight line is 3 or

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

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

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

this is not always the case.
So for instance when

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

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

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

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

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

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

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

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

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

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

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

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

get a straight line.

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

different straight lines with
very different gradients.

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

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

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

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

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

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

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

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

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

sustained. But when we look at
the next unit.

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

y has decreased by 3.

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

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

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

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

long as we can find a way of

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

at a sketch of y equals x squared.

And let me pick this
point on the curve.

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

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

would be almost more straight
and it would appear to be

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

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

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

to represent that sort of
local straightness? And the

answer would have to be a

tangent. Through the point.

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

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

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

That the gradient

of a curve

y equals a function of x

at a given point

is equal

to the gradient

of the tangent

to the curve

at that point.

So notice we haven't defined the

gradient of the curve globally,
so to speak,

we've said that at any particular

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

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

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

calculate that gradient at any
particular point?

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

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

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

nearer and nearer to being
a tangent as the Qs come

nearer and nearer to P.

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

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

line should approach the
gradient of the Tangent.

And hence the gradient of
the curve.

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

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

calculation represents what
we've just seen.

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

So there is y equals x squared.

Here will take our point.

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

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

those two points with a straight line.

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

a triangle formed by dropping a
perpendicular down to reach the

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

triangle with the right angle at
the point are.

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

So that I get the coordinate

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

then 0.01, 0.001

and finally 0.0001.

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

Well, it will just be 3 + whatever PR is

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

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

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

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

Next is 9.0601.

Next 9.006001

and the next 9.00060001.

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

In other words, what's QR?

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

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

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

That's 0.61, 0.601, 0.006001,
0.00060001.

So now we need to
calculate the gradient of

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

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

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

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

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

Do the same calculation here
and it's 6.001.

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

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

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

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

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

to do it in general.

So let's set that situation.

We take our axes.

And we plot our curve
y equals x squared.

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

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

How far along the curve is it?

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

So we got the same triangle.

As we had in the calculation.

So there's R.

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

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

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

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

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

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

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

Now. We've got the
diagram labeled up.

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

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

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

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

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

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

fact that delta x is a
single symbol,

plus delta x all squared.

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

equal to x squared.

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

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

2x delta x plus delta x all squared.

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

The gradient of PQ,

well, it's QR over PR

or the change in y,

delta y, over the change in x.

And we've got expressions
for both of these:

delta x is delta x and delta y is this.

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

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

and that was all over delta x.

Now, delta x is a common factor here.

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

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

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

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

Now remember what we're doing.

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

Q come nearer and nearer to P
along the curve.

So we're letting delta x
tend to 0,

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

And so we can say that the gradient

of

the tangent

at P is equal to...

Now we have to use some
mathematical language.

What happens to this expression as
delta x approaches 0?

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

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

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

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

we say that this is the limit,

as delta x tends to 0,

of delta y over delta x

is equal to

the limit as delta x tends to 0

of 2x + delta x

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

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

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

Now we could do this
calculation in just the same

way for all kinds of
different curves, but it

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

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

this phrase, the limit.

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

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

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

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

and in the case we've just

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

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

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

We often use function notation
y equals f of x.

So in function notation,

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

becomes the point
x, f of x.

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

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

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

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

you remember, then we take

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

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

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

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

So dy by dx

is the limit as delta x tends to 0

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

Which is the limit

as delta x tends to 0

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

So, if we are using the language
of functions. That's the

language that we use and
sometimes for functions when

we're differentiating them
instead of using the symbol

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

that we've differentiated
with respect to x.

Let's just take one
example using this

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

Have a sketch of this curve for
positive values of x.

There it is coming down like
that. So if I take

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

amount to give me a second
point Q on the curve. Then

I can see that in fact,

there's my delta x.

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

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

we can see that this is actually
a decrease. Let me complete.

This triangle by drawing in
that secant, PQ and let's

have these coordinates,
this is x, y.

And this is. x + delta x

and then f of x + delta x.

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

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

And so delta y over delta x is

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

And this is the gradient of
this line PQ.

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

this is 1 over
x + delta x.

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

all divided by

delta x.

Here we have two fractions and we are
subtracting them, so we need a

common denominator and that
common denominator will be

x times, x + delta x.

So I've multiplied this by, this
x + delta x, I have

multiplied it by x, so I must
multiply the 1 by x.

Minus, and I've multiplied the x
by x + delta x, so I must

multiply the 1 by x + delta x
and then this is all

divided by delta x.

So let's look at the top, x - delta x,
minus delta x.

So I've got x - x that goes out.

And I got on that line there - delta x.

All over, well, I'm dividing by
that, and by that, so effectively

I'm dividing by both of them.

So there we've got them both
written down.

Now delta x is a small positive quantity,

so I can divide top and bottom there
by delta x, canceling delta x out.

So delta y over delta x is equal to
-1 over, x + delta x, times by x.

Let me just multiply that
denominator out so I

have x squared plus
delta x times by x.

dy by dx is equal to the limit
as delta x tends to 0 of

delta y over delta x.

Which will be the limit
as delta x tends to 0

of -1 over x squared plus
delta x times by x.

Equals... Let's have a look at
what's happening in this denominator.

As delta x becomes very, very small
x times by delta x is also very very small

and very much smaller than x squared.

So as delta x goes off to 0, this term
here goes away to 0 and

just leaves us with the x squared.

So our limit is -1 over x squared.

And that's our calculation done
in function notation to find the

derivative, as we call dy by dx,
or the gradient of the curve at a point.

Now, the calculations that we've
done or all quite complicated.

You need to do one or two
to practice the method and

to see what's going on with
it - to see how it's actually working.

But, to do most of your differentiation,
there are a set of rules that you

can follow that will help you do the
differentiation much more quickly.