www.mathcentre.ac.uk/.../8.9%20Inplicit%20Differentiation.mp4

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

differentiate. Functions of Y
with respect to X.

Let's begin by looking at an
equation like this.

X squared
Plus Y

squared. Minus

4X. +5 Y
minus 8 equals

0. Now here the X
is and wiser all tangled

together and the wise Y squared

and Y. It will be quite
difficult to rearrange this, so

it said Y equals a function of

X. We could perhaps given values
of X workout what values of why

were and thereby draw graph.

But nevertheless, these things
are intimately connected, and

differentiating something like
this is going to be much harder,

so it would seem than if it just
said why equals some function of

X. But that's what we're going
to have a look at in this video.

How can we differentiate a
function and equation that looks

like this where the X and wiser
all tangled up together?

We're going to use something
known as the chain rule or

function of a function. It's got
those two names. Chain rule an

function of a function.

I'm going to try and stick with
using the name chain rule, but

you may know it as function of a

function. There is a video which
covers this particular rule

explicitly. So I'm just going to

revise it. So let's take
an example such as Y

equals. 5 + 2
X to the 10th power. And

let's say I want to
differentiate this. Then the

chain rule says that if I

port. You
equals 5

+ 2

X. Then

Why will be equal to you to
the power 10?

And we can form
the why by DX

by doing DY by
du times DU by

DX and it's this.

That is the chain rule.

Or some of you may know it
function of a function.

So this is what we've got to do.
We've got to work out the why by

DU and workout. Do you buy the
X? So I'm going to turn the

page. I'm going to virtually
write this out again in order to

make sure that I've got the
space in which to do it.

So we begin with Y equals

5. Plus 2X to the

power 10. We put
you equals 5 +

2 X and then
Y is equal to

U to the power

10. DY by the X
is given by the Y by

EU times. Do you buy the
X and we can workout each

of these two divided by DU&U
by DX. So let's have a

look at that why by EU
is the derivative of U to

the power 10 with respect to

you. Which is 10 U to the power
9. Remember we multiply by the

index and take one of the index
to give us nine and so that's

10. 5 + 2 X
to the Power 9 replacing EU

by 5 + 2 X.

And we can calculate you
by The X.

That's the derivative of 5
+ 2 X.

With respect to X, and that's
just two because the derivative

of five 5 is a constant stats
zero, the derivative of two X is

just two. So now we know what
divided by du is. It's this and

we know what do you buy the
access. It's this so we can

write those in.

10 Times 5
+ 2 X to the

9th. Times by two and
of course, the two times by 10

gives us 20.

So we've used our
chain rule in order

to be able to
differentiate this function.

So let me just
write our chain rule

down here DY by
X is equal to

Y by DU times
du by X.

Now. Let's suppose that we

had zed. And said was a
function of Why?

Then D zed by the
X would be equal to

D zed. Bye.

DY times DY

by X. Using
our chain rule again.

Let's take an example.
Let's say that zed is

equal to Y squared.

Then these Ed by the
X would be equal to.

The derivative of Y squared
with respect to Y.

Times DY by the X
and the derivative of Y

squared with respect to Y
is equal to two Y

times DY by X.

Notice what we seem to have done
is just differentiated. Why

square root respect to Y and
multiplied by DY by the X? And

I'll draw attention to that
again later on. For now, let's

have a look at some examples.
Begin with this one Y squared.

Plus X cubed.

Minus Y cubed plus 6.

Equals 3 Y. We want to
differentiate this with respect

to X, so we're looking for the
derivative of Y squared with

respect to X.

Derivative of X cubed
with respect to X.

Derivative of Y cubed with
respect to X.

Derivative of the six with
respect to X and the derivative

of three Y with respect to X.
Not remember how we said we

would do this? We said we will
take the derivative of Y squared

with respect to Y.

And multiply by DY by the X that
was our chain rule.

Plus this is straightforward,
don't need to worry about this.

This is the derivative of X
cubed with respect to X. We know

that is 3 X squared multiplied
by the index and take one away

minus. And here again, we've got
to apply our chain rule. So

we've got the derivative of Y
cubed with respect to Y.

Times by DY. By

The X. Plus

Now the derivative of six well
six is just a constant, so we

know its derivative is 0.

Equals and again we want the
derivative of three Y, so our

chain rule tells us this is the
derivative of three Y with

respect to Y times DY by DX.

Now we can go back and work out
each of these derivatives with

respect to Y, so that will give

us 2Y. DYIDX
plus three

X squared.
Minus the derivative of Y Cube

with respect to Y is 3 Y
squared divided by DX. We can

ignore the zero.

And the derivative of three Y
with respect to Y is 3 times

divided by DX.

Now. If we get together all the
terms that have a DY by the X in

them. Then, having done that, we
can sort out what divided by DX

actually is, so I'm going to
gather together all the terms in

do I buy DX over on this side of
the equation so I can keep .3 X

squared there on its own. So I
have three X squared equals.

Now here on this side I've
got 3D Y by X.

I'm going to take this away from
each side, so that's minus two

Y. DY by X and I'm going
to add this term to both sides

plus three Y squared DY by The
X. what I can see here is that

I've got a common factor of the
why by DX that I can take out,

so that's what we're going to do
next, so will have.

Three X squared equals.

Bracket. And taking out
that common factor of the why

by DX. So let's just go back
and have a lot. What do I

buy? DX was multiplying. It
was multiplying A3.

It was multiplying a minus two Y
and it was multiplying A plus

three Y squared, so it was
multiplying a 3 - 2 Y and

a plus. 3 Y squared. So now
we can get divided by DX on

its own if we divide throughout
by this expression sode, why by

DX is equal to three X squared
and dividing throughout dividing

both sides by 3 - 2 Y
plus three Y squared.

And then we've got our
expression for DY by DX.

That was a reasonably
straightforward example. The

work that many complications and
it followed very directly from

our first look at this. So now
let's look at a slightly more

complicated example, one where
in fact we've got other

functions. Of X&Y. So in this
case will start up with a sign

Y where we've got the Axis and
the wise actually combined

together, so we've got X
squared times by Y cubed.

Minus calls X and let's say
equals 2 Yi. Want to be

able to differentiate this with
respect to X, so that's the

derivative of sine Y with
respect to X plus the derivative

of X squared Y cubed.

With respect to X minus the
derivative of Cos X with respect

to X equals the derivative of
two Y with respect to X.

Now let's remember what our
function of a function rule

tells us that this is done
as the derivative of sine Y

with respect to Y times by
DY by DX.

This one. Bit of a problem
'cause This is X squared times

by Y cubed. So it's a product.

It's a U times by AV, so let me
just write a little U over the

top and a little V there.

You is X squared and
V is Y cubed.

So this is plus.

Let's remember how we
differentiate a product. We take

you. And we multiply it by
the derivative of V. That's the

derivative of Y cubed with
respect to X.

Plus the an we multiply that
by the derivative of U, which

in this case is the derivative
of X squared.

Minus now we can do this one.
The derivative of Cos X with

respect to X. The derivative of
causes minus sign, so a minus

and minus makes a plus sign. X
equals the derivative of and

again my chain rule tells me
that this is the derivative of

two Y with respect to Y times

by. Divide by The X.

So this has been much more
complicated, but notice how it

follows the standard rules that
we've already got for

differentiation. So now the
derivative of sine wired with

respect to Y is just cause why.

DY by X.

Plus X
squared

Times by now this will be the
derivative of Y cubed with

respect to Y.

Times by DY. By The
X.

Plus this one Y cubed times by
now the derivative of X squared

with respect to X is just 2X.

Plus sign X.

We've already done that one
equals and hear the derivative

of two Y with respect to Y.
These two times DY by The X.

Almost done now we still
got a little bit of

differentiation in here to do
so. Let's do that cause

YDY by DX plus I
think I can fairly safely

remove these brackets now. X
squared times 3 Y squared

DY by X +2 XY
cubed plus sign.

X equals 2 DY
by The X.

Where at the same stage as we
were last time we've got the

differentiation Dom and it's
this thing. The why by DX that

we want. So what we gotta do is
get all those terms that

involved why by DX on one side
of the equation and the other

terms on the other side. Now
these are the two terms that

don't have a divided by DX in
them, so I'm going to keep them

at this side. So it's two XY
cubed plus sign X.

Two, XY
cubed plus

sign X

equals. Let's just go back and
see what we've got. We've got a

2 divided by DX here.

2.
DY by The X.

And we're going to bring these
two terms over to this side by

taking them away from both
sides. So we're going to take

away 'cause why do I buy the X?

On both sides, minus cause YDY
by the X, and then we're going

to take away the other term.
This is the X squared times by

three Y squared divided by DX,
right that a little bit

differently. When I do it, so we
have minus three X squared Y

squared divided by X.

Again, we see we've got a set of
terms here, each with divided by

DX as a common factor.

Two, XY cubed plus sign
X is equal to.

Let's take out this common
factor of DY by The X.

Well, it's multiplying two, so
we've got a two there. It's

multiplying minus cause Y, so
we've gotta minus cause why

there? And it's multiplying
minus three X squared Y squared

minus three X squared Y squared.
And now finally we can get

divided by DX on its own,
because we can divide throughout

divide both sides of the
equation by what's in this

bracket. So we have two XY cubed

plus sign. X all over 2
minus cause Y minus three

X squared Y squared.

And there's our divide by
the eggs.

Now let's just look back at this

one. Look at all this
complicated differentiation that

we had here.

Now I went through it slowly and
carefully, but when we're doing

calculations on our own, we
might make slips. It would be

helpful if we could automate
some of this process so it

instead of writing down the
derivative of sine wave with

respect to X is and going
through the chain rule.

We went automatically to all we
need to do is differentiate sign

wired with respect to Y. That's
cause Y an multiplied by divided

by DX. So we automate we would
miss out at least these two

lines and go direct from there
to there. Similarly, the

derivative of two Y with respect
to X would be the derivative of

two Y with respect to Y two
times divided by DX. And so we

go direct from there.

To there, so let's have a look
at that in another example.

So we'll take
Y squared plus

X cubed minus
XY plus cause

Y. Equals note this time
we're going to go.

Direct to the differentiation,
we're not going to go by the

chain rule. We're going to use
it, of course, but we aren't

going to write it down, so we
want to differentiate this with

respect to X. So the first term
is Y squared, so we know that to

differentiate Y squared with
respect to X, we differentiate

this with respect to why that's

too why. And we multiply by
DY by DX.

Now we want the derivative of X
cubed with respect to X, so

that's 3X. Squared

Minus.

Now I am going to write this
down in full because it's the

derivative of XY with respect to
X. It's a product again, it's X

times by Y.

Plus, the derivative of Cos Y
with respect to X, which we do

as a derivative of cause Y with
respect to Y that's minus Sign

Y. Times DY by DX,
the derivative of 0 is

just zero. So let's get
these two terms together because

they both got the why by DX. So
I have two Y minus Sign Y.

Times Ty by DX
plus three X squared.

Minus. Now this is
a product. It is a U times

by AV. So we

want you. Times the derivative
of Y with respect to X, which

is just the wise by The X.

Plus V, which is Y Times
the derivative of X, which is

just one. Equals
0.

So we see here that we've got
another term now involving the

why by DX it's minus X, so we
can put that in the bracket so

we can have two Y minus sign Y

minus X. DY by the
X plus three X squared

and then minus this term
here minus Y equals 0.

If we take this over to the
other side, In other words, we

take this away from both sides
and add that to both sides. You

have two Y minus sign Y minus X
times DY by X is equal to.

Adding this one to both sides.
Why taking this one away from

both sides, we get that.

Now it's clear how we would
finish this off. We will take

this factor here and divide both
sides by it.

So we get DY by
the X was equal 2.

Just go back. We've got Y
minus three X squared to go

in the numerator on the top.

And this factor to go on the
bottom in the denominator two Y

minus Sign Y.

Minus.

X. And then

we have. Our derivative,
Let's just take one

more example.
Y

cubed
minus

X. Sign

Y. Plus Y squared

over X. Equals
8.

Again, all the axes and Wise
bundled up together if possible.

We want to be able to do this
directly. We want to be able to

differentiate it straight away
without going through the chain

rule. So the derivative of Y
cubed with respect to Y times by

DY by DX. So that's three Y
squared times DY by X minus.

We want the derivative of this.
This is a product so let's see

if we can do it all in one
go. Again you want X Times the

derivative of sine Y with
respect to X. That's X times

cause YDY by X.

And now we want sign Y times the
derivative of X so that sign Y

and the derivative of X is just

one. This one is a bit trickier.
This is a quotient Y squared

over X, so we want plus. Now
let's remember what we do with

the quotient. It's V which is on
the bottom that's X Times the

derivative of what's on the top.
The derivative of Y squared with

respect to X, which is the
derivative of Y squared with

respect to Y2Y times DY by the
X minus Y squared times the

derivative of V, which in this
case is just X.

All over V squared, which is
X squared equals 0.

Now this needs a little bit of

tidying up. We've got a
denominator here that we can

probably multiply out by, so
let's do that.

And do some tidying on the way,
so this will be three X squared

Y squared DY by The X.

Minus X cubed cause
YDY by the X

minus X squared Sign

Y. +2 XYDY by
X minus Y squared equals 0,

so I've multiplied throughout by
this X squared so the X

squared is appeared there,
multiplying that it's appeared

there inside that X cubed,
multiplying that it's appeared

there, multiplying the sign Y,
and it's gone. From here, 'cause

we've multiplied by so.

Multiplying and dividing by, in
effect, leaving this on changed.

Now let's get together all the
terms in DY by DX, so we

have the why by DX times this
term, three X squared Y squared.

This term minus X cubed

cause why? This
term +2 XY.

And here I've got minus X
squared sign Y or I think I want

to add that to the other side.
So that's plus X squared sign Y,

and here I've got minus Y
squared. Again, I think I want

to add that to both sides, so I
get plus Y squared over there.

Now why by X is
equal to X squared sign

Y plus Y squared in
the numerator and dividing by

this expression as the
denominator. Three X squared Y

squared minus X cubed cause
Y, +2 XY.

Notice how much shorter
automating that

Differentiation's made? What's
quite a complicated problem, and

that's something you want to
work at. Trying to automate your

differentiation so you don't
have to go through the rules and

write them down every time.