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Sometimes we given functions
that are actually products of

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functions. That means two
functions multiplied together.

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So an example would be Y equals

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

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product X squared is one
function multiplied by the

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cosine of 3 XA second function.

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Usually we write this
as UV.

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

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differentiate this DY by the X

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is UDVIDX. Plus

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V times by du by

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

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this formula.

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We've identified you with X

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

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we need to do now is
write down those

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derivatives. So do you
buy the X?

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

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you buy the X is 2 X.

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

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

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minus three sign 3X.

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Now we can put these two.

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Together with these two
into this formula. So

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

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You

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X squared.

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

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sign 3X.

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Plus the.

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That's cause 3X.

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Times by du

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by DX2X. Now

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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close the bracket there.

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

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

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maximum or minimum we could go
on and do that.

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Let's have a look
at another example.

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

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

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

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

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other words, it's a square root.

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

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the habit of identifying these
functions very specifically.

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So we've identified the
two bits that would be

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multiplied together to
make the product U&V.

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

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three X squared?

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

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gives us a three X squared.

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TV by The X.

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Now this one is going to be a

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little bit different. We've got

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

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

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

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X is minus one.

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

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

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That gives us minus 1/2.

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

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

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VU by The X.

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

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so we you is X cubed.

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

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the minus 1/2.

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Plus VDU by DX.

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That's 4 minus
X to the half.

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Times by three X
squared.

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

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

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

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better. We need it to look a lot

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

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

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

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So the why by DX?

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Equal 2X cubed times
minus half 4 minus

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X to the minus

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

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

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put all this together.

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

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

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in the denominator, so we
have minus X cubed over

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2 * 4 minus X
to the half plus.

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

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

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

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

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common denominator.

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Will be 2, four minus X
to the half.

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

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

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top. Plus this we have
multiplied a one here.

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

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

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

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

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

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is just traditionally written is
just 4 minus X.

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

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over this common denominator.

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And we can look at what we've

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

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

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

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

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so we can pull out that X
squared.

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

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

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times by minus X which is minus

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

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

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that's minus 7X.

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

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

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

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We just look at one more

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

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all times by E to
the 2X.

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

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

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

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And now we differentiate you.

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So we have DU by the X
is equal to the

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derivative of one is 0
because one is a constant

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and the derivative of
minus X cubed is minus

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three X squared.

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We want the derivative of V.

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

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the exponential function, we
differentiate the power.

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

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

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

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

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

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equal to you?

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TV by the
X plus VD

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you by X.

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

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

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by ZX is equal to.

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

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

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that's here 2 times E to the

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power 2X. Plus

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and now we want V. That's E to

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

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

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

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

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

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

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

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2 times by 1 -
2 times by X cubed.

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

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

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expression like this, a
polynomial in terms of X to

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

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

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

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2 - 3 X squared.

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Minus two X cubed and being
factorized. That derivative is

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

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to solve the why by DX equals 0.