
In Differentiation, when
we differentiate A.

Function F of X.

So there's our function F of X
and this is a graph of it.

What do we actually doing? Well,
when we find the derivative were

actually finding the gradient of
a tangent. So when we

differentiate that's F dashed of
X, this represents the gradient.

Of. The curve.

Or since we have a curve
and the tangent at that point,

it is the gradient of.

The tangent so given.

A point where X is equal to

A. And therefore where?

The value of the function is F
of a, then F dashed at that

point is the gradient.

Of. The Tangent.

Where? X equals

a. And that's what we're going
to be using that the derivative.

The value of it at the point
where X equals a gives us the

gradient of the tangent, because
if we know the gradient of the

tangent and we know the point on
the curve, we can then find the

equation of this tangent because
it's just a straight line. So

let's have a look at some
examples. The first example will

take F of X is equal to X cubed.

Minus three X squared plus X
minus one, and what we want

is the equation of the Tangent.

To this curve at the point where
X equals 3.

Well, at the point where X
equals 3, so the first thing

we're going to find is where is
this point on the curve? So we

take X equals 3 and we'll find F

of three. So that's
three cubed minus 3

* 3 squared plus
3  1.

3 cubed is 27. Three squared is
9 and 3 nines are also 27.

So this builds the 27 
27 + 3  1 altogether.

We come up with two, so
the point that we're

actually interested in
on the curve is the .32.

Now the next thing we need is
the gradient. We've now got a

point on our line on our
tangent. We now need its

gradient, so we need to
differentiate F dashed of X. So

let's differentiate three X
squared for the derivative of X

cubed. Now we differentiate this
term and so that's minus six X.

Now this term, so the derivative
of X is.

One and now the final term, the
derivative of one is 0 because

it's a constant. We need the
gradient when X is equal to

three, so we can take that value
X equals 3 and substitute it

into our expression for the
gradient. And so we have 3 *

3 squared minus six times.

3 + 1.

3 squared is 9 and 3 nines

are 27. Minus 6 * 3
is minus 18 plus one, and so

this works out to be 27 takeaway
18. That's nine an add on one

that's 10. And so we've got a
point, and we've got a gradient,

so let's write those things
down. We've got our point.

Which is 3 two.

And we've got our gradient.

Which is 10. What we want is the
equation of the tangent at this

point on the curve with this
gradient. So that's the equation

of a straight line.

The equation of a straight line
that goes through a point X one

Y one is Y minus Y one over
X Minus X one is equal to M

the gradient. So this is
our X one, Y1. Let's just write

that in over the top X1Y one
and this is our gradient M.

So now will substitute these
numbers in so we have Y minus

two over X. Minus three is equal

to 10. Now will multiply it
by this X minus three here, so

we have Y minus two is equal to
10 times X minus three. Now all

we need to do is multiply out
the bracket Y minus two is 10 X

minus 30. Remembering to
multiply everything inside this

bracket by what's outside. And
now we just need to get this two

over here with this 30.

So we add two to both
sides. Y equals 10 X, now

minus 30. Adding onto is minus
28 and there's the equation of

our tangent. To the curve at
this point on the curve with

that gradient. OK, let's take
another example. This time.

Let's say that what we want
to be able to do is

looking at this curve Y equals
X cubed minus six X squared.

Plus X +3 and what we want
to find our what are the

tangents? What are the equations
of the tangents that are

parallel to the line Y equals X

+5? So here's our curve and what
we want are the equations of the

tangents parallel to this line.

Well, we've got to look at this
line and we've got to extract

some information from it.

Information that we can get from
it is what is its gradient,

because if the tangents have to
be parallel to this line then

they have to have the same
gradient. And what we can see is

that we've got One X here and
the standard equation for a

straight line is Y equals MX
plus. See where this M is the

gradient. So what we gain from
looking at this standard

equation and comparing it with
the straight line is that the

gradient of the straight line M
is equal to 1.

So now we know what the
gradients of the tangents have

to be. The gradients have to be
1, so how can we calculate that?

Well, we know that if we
differentiate this curve its

equation we will get an
expression for the gradients of

the tangents. Then we can put it
equal to 1 and solve an equation

that will give us the points or
at least will give us the X

coordinates of those points. So
let's do that. We know what. Why

is. Let's differentiate the why
by DX is.

Equal to. The derivative of X
cubed is 3 X squared. The

derivative of minus six X
squared is minus 12 X the

derivative of X is plus one, and
the derivative of three is 0

because it's a constant and we
want this expression to be equal

to 1. So we can take
one away from each side and we

have three X squared minus 12 X

equals 0. This is a quadratic
equation for X, which we can

solve, so let's do that.

Three X squared minus
12 X equals 0.

It's a quadratic. The first
question we've got to ask

ourselves is, does it factor
eyes? And if we look we can see

that we've got a common factor
in the numbers of three and a

common factor in the ex terms of
X, so we can take out three X as

a common factor.

3X times by something has to
give us three X squared, so

that's got to be X and three
X times by something has to

give us minus 12 eggs, so
that's got to be minus four,

and so that equals 0.

2 numbers multiplied together.
Give us 0, so one of them has

to be 0 or both of them have
to be 0. So we can say three

X equals 0 or X minus 4 equals
0, so therefore X is 0 or X

equals 4. Now having got these
two values of X, we want the

points. Where these tangents
were, what we've got now the X

coordinates. We now need the Y
coordinates and to do that we

need the equation of the curve
again. So let me just bring back

the page and it's X cubed minus
six X squared plus X +3.

So Y equals X cubed minus six
X squared plus X plus three and

first of all we want to point
X equals 0 and so Y equals

we put X equals 0.0 cubed. We
put X equals 0  6 *

0 squared plus 0 + 3 each
of these three terms is 0, so

we. End up with three.

X equals

4. Why is
equal to 4 cubed?

Minus 6 * 4
squared plus 4 +

3. Equals.
Well, 4 cubed is 4 times

by 4 times by 4 four
416 and four times by 16

is 64 takeaway. Now we've got
6 times by 16 and so

that's 96 + 4 + 3.

Equals 64 takeaway. 96
is minus 32, and

then we're adding on
another Seven, so that's

minus 25. So our
two points are 03.

And 4  25 and those are the
two points where the gradients

of the tangent are equal to 1
and so where the tangents are

parallel to the line that we
started out with. That's Y

equals X +5.

Let's take another example
and this time I want to

introduce the word normal.
What's a normal? Well, we

tend to think of the word
normal in English as mean the

same everything's alright,
each usual. But in

mathematics, the word normal
has a very specific meaning.

It means perpendicular.

Or at right angles.

At right
angles.

So if I have a curve.

Let's say that's my curve and I
have a tangent at that point

there. Then what's the normal to
the curve while the normal is at

right angles to the curve, so
it's also at right angles to the

tangent, so there's the Tangent.

To the curve and this is
the normal to the curve normal

because it's at right angles
perpendicular to the Tangent.

OK.

If we can find the equation
of a tangent, we can surely

find the equation of a
normal, but there is one

little piece of information
that we need. That is, if we

have two lines at right
angles, what's the

relationship between their
gradients?

So if I take 2 lines there,
right angles to each other,

and let's say that this line
has gradient.

M1 and let's say
this line has gradient.

M2 And
the relationship between these

two gradients, because they are
at right angles, is that M1

times by M2 is equal to
minus one, and that's for lines.

At right angles.

And so, since tangent normal at
right angles, we can use this

relationship. We can calculate
the gradient of the tangent and

use this to find the gradient of

the normal. Now let's have a
look at that practice.

Let's take the curve Y equals
X plus one over X.

At the point where X equals 2,
let's ask ourselves what's the

equation of the tangent at the
point where X equals 2. What's

the equation of the normal? So
first of all, let's establish

what the point is, X equals 2, Y
equals 2 plus one over 2, which

is 2 1/2.

But thinking ahead, I think I
would prefer to have that as an

improper fraction, as five over
2. That's because I'm going to

have to do some algebra with it
later, and I'd rather keep it is

5 over 2. Then keep it as 2 1/2.

OK, next we want the gradient at
this point X equals 2.

So let me just write down.

The equation of the curve Y
equals X plus and in order to

differentiate, be ready to
differentiate the one over X.

I'm going to write it as X to
the power minus one.

And now we can differentiate it
DY by DX is equal to the

derivative of X is one plus
differentiate this we multiply

by the minus one and we take one
away from the minus one to give

us a power of minus two. So

that's one. Plus and A minus
gives us a minus there one over

X to the minus 2 means one over

X squared. X is equal

to 2. And so my
gradient D why by DX is

equal to 1  1 over 4,
two squared is 4 and one

take away a quarter leaves
me with three quarters.

So we have got a .25
over 2 and we've got a

gradient of 3/4 and so we
can find the equation of the

Tangent. So let's just list
what we know. We've got the

tangents at the point.

And this is 2, five over 2 and
we know that the gradient at

that point is 3/4.

So the standard equation for a
straight line Y minus Y one over

X Minus X one is equal to
M the gradient. This is X one

and This is why one. So now
we can substitute these things

into here, Y minus five over 2
over X minus two is equal to

3/4. Let me multiply both
sides by X minus 2.

And then let's
multiply both sides

by this form.

Now if I
multiply out, the

brackets that I've
got for, why?

Now 4 times by minus five over 2
four times by the five is 20

divided by the two is 10 and
then the minus sign minus 10

equals 3 X minus six and it will
be nice to be able to get all

these numbers together. So I'm
going to add 10 to each side

will give me 4 Y equals 3X plus

4. So the equation of
the curve is 4 Y

equals 3X plus 4.

Now we need to find the
equation of normal to the curve.

Let's say the gradient of
the normal.

Is M2 and that the gradient
of the tangent?

Is M1.
And let's recall that normal

and tangent are perpendicular.

And the thing that we said about
2 lines that were perpendicular

at right angles to each other
was that if we multiplied their

gradients together, the answer
we got was minus one.

Now we do know what the value of
M1 is. We do know the gradient

of the tangent is 3/4. So if we
put that into here 3/4 times by

M2 is equal to minus one.

So if we multiply it by the four
and divide by the three we get

M2 is equal to minus four over
three and so now we know the

gradient of the normal. We also
know the point on the curve that

still hasn't changed. That still
the .25 over 2.

So we now want the equation of
the normal. Let's just write

down what we know. We know the
point that's 25 over 2 an we

know the gradient that's minus
four over 3.

So our standard equation for
straight line Y minus Y one over

X Minus X one is equal to the

gradient. And this is the
Point X one Y one. So

now we substitute that in Y
minus five over 2 over X.

Minus two is equal to minus
four over 3.

Let's multiply up by X
minus two and by three.

So we have three times Y minus
five over 2 is equal to minus

four times X minus 2.

Multiply out the brackets 3 Y
minus 15 over 2 is equal

to minus four X +8.

Now let's get things together.
It's awkward having this minus

4X here, so we'll add 4X to
each side, and we'll add 15 over

2 to each side, so will have
3 Y plus 4X is equal to

8 + 15 over 2. Now eight
is 16 over 2. So here I'll

have 30 one over 2.

So we'll have 3 Y plus 4X is
31 over 2 and this tools and

awkward thing. So let's multiply
throughout by two in order to

get rid of it. 6 Y plus 8X
equals 31 and that's the

equation of our normal.

Now when we've got tangents and
normals because the normally

sort of inside the curve if you
like or passing through the

curve, sometimes a normal can
actually meet the curve again,

and we might be interested to
know where it meets that curve.

So. Question we're going to ask
is if we have the curve XY

equals 4. And we look at
the point, X equals 2.

Find the equation of the normal.
Where does the normal meet the

curve? Again? If it does, well,
let's have a look at a picture

why equals 4 over X? What does
this look like as a curve?

OK, if we have a very large
value of X, say 104, / 100 is

very small, so we got a little
bit of curve down here. If we

have a very small value of X,
say point nor one.

Then 4 divided by Point N 1.1 is
100, so 4 / 100 is 400 very

large, so we're very small
values of X. We've got a bit of

curve there and we can join it
up like that.

Now if we take negative values
of X, the same thing happens,

except we get negative values of
Y and so the curve looks like

that. Notice we do not allow X
to be 0 because we cannot divide

by zero, so there is a hole in
this curve. There is a gap here

at X equals 0. Well, here's the
point. Let's say X equals 2.

That's the point on the curve.

There is the tangent to the
curve, and there's the normal to

the curve, and as we can see,
this normal goes through there.

So the question is where is it
where is that point?

So first of all, let's establish
what this point is up here so we

know that X is equal to two and
Y is equal to 4 over 2 equals

2. So our point is the .22.

Next, we want the gradient of
the tangent in order that we can

find the gradient of the normal.

So here we have Y equals 4 over
X, which is 4 times X to the

minus one. So I'll differentiate
that the why by DX equals minus

4 multiplied by the minus one
and then taking one of the index

that gets it to be minus two. So
we have minus four over X

squared. X is equal

to 2. And so DY
by the X is minus 4

over 4 is minus one.

So what have I got? I've got
the .22 and I've got the

gradient of the tangent and
remember tangent and normal

are right angles.

So for two lines at right
angles, M1 times by M2 is minus

one. I know the value of this,
it's minus one times by N 2

equals minus one. The only
number M2 can be there is one.

So now I have got for the normal
I want its equation. I've got

the point that it goes through
which is 2 two and I've got its

gradient which is one.

So I can write down the standard
equation of a straight line Y

minus Y one over X Minus X one
is equal to M.

This is my X One Y1.

And I can substitute those in so
I have Y minus two over X, minus

two is equal to 1.

Multiply it by X minus two, so
we have Y minus two is equal

to X minus two, and so why
is equal to X?

And so now we have the equation
of the normal Y equals X and

we have the equation of the
curve XY equals 4.

Now. Where does this line meet
this curve? That's what we're

asking ourselves. Where does the
normal intersect the curve?

At the points of intersection,
both of these equations are true

at the same time, so that means
I can take the value of Y, which

is X and substituted into here.
So therefore I have X times by X

equals 4 at these points. In
other words, X squared is equal

to 4. If I take the square root,
that will give me the value of

X. We might think that's just

two. But remember, when you take
a square root you get a plus and

A minus. So we have X equals 2
or minus 2.

And so if X equals 2, then why
must be equal to X? So that

tells us Y equals 2, or and if
we take minus two, Y is equal to

X. That gives us minus two and
so our two points are two 2 and

minus 2  2.

Those are the two points where
the normal meets the curve.

Notice this is the first point
that we started off with.

And indeed, when we're doing
this kind of question, the point

where we started off with always
is going to be a part of the

solution. So we've dealt with
applications to tangents and

normals. We've seen that in
order to find the gradient of

the tangent you differentiate,
substituting the value of X and

that gives you the gradient of
the curve and hence the gradient

of the tangent and the other
relationship that we found was

that a normal was perpendicular
to the tangent and that the

product result. Of multiplying
two gradients together, where

the two lines are perpendicular
was minus one. That's an

important relationship when
we're looking at 2 lines that

are perpendicular, as is the
case for tangent and normal.