Sometimes we given functions
that are actually products of

functions. That means two
functions multiplied together.

So an example would be Y equals

X squared. Times by the cosine
of 3 X so there we've got a

product X squared is one
function multiplied by the

cosine of 3 XA second function.

Usually we write this
as UV.

And there is a formula which we
can use to be able to

differentiate this DY by the X

is UDVIDX. Plus

V times by du by

DX. So every time we've
got a product, we can use

this formula.

We've identified you with X

squared. We've identified
V with calls 3X, So what

we need to do now is
write down those

derivatives. So do you
buy the X?

You, we've identified
as X squared. So do

you buy the X is 2 X.

DV by DX, we identified V as
costs 3X, so we need to write

down the derivative of DV by DX,
and we know that that will be

minus three sign 3X.

Now we can put these two.

Together with these two
into this formula. So

let's do that. Why by
DX is equal to.

You

X squared.

Times DV by the X so
it's times by minus three

sign 3X.

Plus the.

That's cause 3X.

Times by du

by DX2X. Now

this. Looks ugly and really we
need to tidy it up a little bit.

We want to write these terms in
a nice order, but at the same

time we want to try and identify
any common factors that there

are. The reason is that what you
might want to do is solve an

equation like this by putting it
equal to 0 to find solutions for

Maxima and minima. So it's
important to be able to spot the

common factors and take them
out. And if we look in this

term, we've gotten X squared,
and in this term we've gotten X,

so we actually got a common
factor of X, and we can take

that out and put it at the front
of the bracket so X and then a

bracket. Let's think what we've
got left here. We've got minus

three and an X, so we have minus
three X sign 3X.

Plus now we took out the X, so
we've got the two left to

multiply by the cost 3X, and so
we have two cars 3X and we just

close the bracket there.

We've got that finished and it's
in a nice factorize form so that

if we wanted to do something
more with it, as I said, find a

maximum or minimum we could go
on and do that.

Let's have a look
at another example.

This time, let's have a
look at one.

Where we've got all just
functions of X know trig

functions in just nice functions
of X except will make this one

have a fractional index will
make it be to the power 1/2. In

other words, it's a square root.

So why is a U times of
E? So let's try and get into

the habit of identifying these
functions very specifically.

So we've identified the
two bits that would be

multiplied together to
make the product U&V.

Now let's do that derivatives.
Do you buy the axis

three X squared?

Member multiplied by the index
and take one off it so that

gives us a three X squared.

TV by The X.

Now this one is going to be a

little bit different. We've got

a half there. We've gotta minus
X in there.

Making use of our idea of
function of a function, we

differentiate this inside, so
that's the derivative of minus.

X is minus one.

And then we must differentiate
this so we get a half.

Times 4 minus X and we
take one away from 1/2.

That gives us minus 1/2.

Let's quote our formula DY by
the X is equal to.

And we know that it's
UDV by the X plus

VU by The X.

And we can now plug in
the bits that we need,

so we you is X cubed.

Times by this minus 1/2
of 4 minus X to

the minus 1/2.

Plus VDU by DX.

That's 4 minus
X to the half.

Times by three X
squared.

Now again, this doesn't look a
nice lump of algebra really. We

might have to do something with
it later on, and if we did have

to do something with it later
on, we need it to look a lot

better. We need it to look a lot

nicer. So I'm going to go
into a new sheet and I'm

going to rewrite this at the
top of the page, and then

we're going to have a look at
how we might simplify it.

So the why by DX?

Equal 2X cubed times
minus half 4 minus

X to the minus

1/2. Plus now we have
V which was 4 minus X to the

half times EU by DX, which was
three X squared and we want to

put all this together.

It's this term which doesn't
look awfully good. Let's

remember that. To the power
minus 1/2 means that it's

in the denominator, so we
have minus X cubed over

2 * 4 minus X
to the half plus.

Haven't done anything with
this side yet, so we can

leave it as it is what we
want to do is put everything

over this denominator. 2 * 4
minus X to the half. We treat

this as though it's over a
denominator of one, so I'll

common denominator.

Will be 2, four minus X
to the half.

This stays just the same no
problem. We haven't done

anything with the denominator,
so we don't do anything with the

top. Plus this we have
multiplied a one here.

By this so we must multiply
everything here. So two times

the three is going to give us
six X squared, and then we have

4 minus X to the power half
multiplied by 4 minus X to the

power half. So that just gives
us 4 minus X, 'cause if we have

the half and half together,
that's one, and to the power one

is just traditionally written is
just 4 minus X.

Equals the denominator can stay
the same. We've now got it all

over this common denominator.

And we can look at what we've

got on the top. Again, we're
looking for the idea of a

common factor. Can we take out
a common factor? And yes, we

can. There's an X squared
there, and there's an X

squared inside that X cubed X
cubed is X squared times by X,

so we can pull out that X
squared.

Then what are we left with? If I
take out the X squared out of

this bracket? I've got 6 times
by 4 which is 24 and I've got 6

times by minus X which is minus

6. Thanks, but I mustn't forget
if I take an X squared out of

here. I've also got another
minus X left. So altogether

that's minus 7X.

And again, that's in a nice tidy
form, so that if we need too

next time, we can actually do
something else with it. We could

put this equal to 0 and take the
top and solve it.

We just look at one more

example. Let's take Y
equals 1 minus X cubed

all times by E to
the 2X.

First of all, let's identify our
U and our V, so will put that

you is going to be equal to 1
minus X cubed.

The V is going to be equal to
E to the 2X.

And now we differentiate you.

So we have DU by the X
is equal to the

derivative of one is 0
because one is a constant

and the derivative of
minus X cubed is minus

three X squared.

We want the derivative of V.

V is E to the 2X and remember
that in order to differentiate

the exponential function, we
differentiate the power.

And we multiply the derivative
of that power by E to the 2X

in this case, so that DV by the
X is equal to the derivative of

the power. The derivative of 2X,
which is just two times by E to

the 2X. And let's remember
our formula that if Y is

equal to U times by V
then D why by DX is

equal to you?

TV by the
X plus VD

you by X.

So we've got U. We've got DV
by DX. We've got V and we've

got du by DX here, so let's
make those substitutions DY

by ZX is equal to.

U to begin with, that's one
minus X cubed.

Times by and we want the
derivative of VDV by DX, so

that's here 2 times E to the

power 2X. Plus

and now we want V. That's E to

the 2X. And now we want the
ubaidi X, and that's minus three

X squared. That's times by minus
three X squared. And as we've

done before, let's look for any
common factors that there might

be. And here we've got a common
factor of E to the 2X in each

term, so let's take that out E
to the 2X, which will leave us

with. Well, here we've got 2
times by one minus X cubed, so

let's do that multiplication. So
we've got 2.

2 times by 1 -
2 times by X cubed.

And then from this one we've got
minus three X squared. So we put

that minus three X squared and
its usual. When you've got an

expression like this, a
polynomial in terms of X to

write it so that we've got the
powers of X in some kind of

order. And in this case I'll
write them in ascending order

from the smallest powers of X up
to the largest, so that would be

2 - 3 X squared.

Minus two X cubed and being
factorized. That derivative is

now in a position where we can
use it for other things, perhaps

to solve the why by DX equals 0.