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.