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.