00:01
So we've looked at the product
rule in more than one way,
we've looked at the
product rule numerically,
we've looked at the product
rule and its definition.
00:11
And we've also done an exhaustive
derivation of the product rule.
00:15
So let's do one more example
before we move on to the
next and the final rule.
00:22
Let's have a look at this
questions y=x^3(3x^2-1)^4
First thing that we need to do is we need
to spot what kind of function this is.
00:35
Firstly, you can see that
we have a function x^3
and that is multiplying
with another function.
00:41
So there's definitely
a product rule.
00:43
But the second function, you will also
notice is a function within a function.
00:47
So we're using more than one rule
here to work out the differential.
00:52
Let's just remind ourselves of
what the product rule states.
00:55
So the product rule says
that dy/dx= vdu/dx+udv/dx
or you could use the dash
notation, whichever one you prefer.
01:08
It doesn't matter what
order you put them in either
because they're plussing, so you could put
your udv/dx first, and then vdu/dx next
because they're adding,
it doesn't really make a difference.
01:20
So we're going to
split this into two.
01:21
So we're going to say this is
my u function, which is x^3
and this is my v function,
which is (3x^2-1)^4.
01:31
We differentiate each
one of them separately,
so I have du/dx,
which gives me 3x^2.
01:39
And then I have dv/dx,
which gives me,
remember that this is now the chain rule.
01:45
So firstly, the outside
function, bring the power down 4,
leave everything on
the inside as it is
decreased the power by 1,
and then don't forget to
multiply with the differential
of the inside
function, which is 6x.
02:00
This gives me a final
derivative of 24x(3x^2-1)^3.
02:09
All of that is done.
02:11
So now it's just a matter of
putting this into the product rule.
02:14
So the product rule
is saying dy/dx,
we want vdu/dx, so v is here.
02:20
So I've got (3x^2-1)^4.
02:25
I now need to multiply this with
du/dx, which is this function.
02:29
So it's multiplying with 3x^2,
put a plus in the middle because
that's what the rule says.
02:34
We now take our u
function, which is x^3.
02:38
And we multiply this with dv/dx,
that's this function here.
02:43
So if I just put brackets around,
it's because this has a bit more happening,
I got (3x^2-1)^3.
02:51
And we just tidy this up,
so I've got 3x^2(3x^2-1)^4.
03:01
So we multiply this term
together, which gives me 24x^4,
and then (3x^2-1)^3.
03:11
There's one last
thing you can do here,
you can see here that within
this whole expression,
you have a common
factor of (3x^2-1)^3.
03:19
So you can take that
out of this bracket.
03:23
This is just algebra now.
03:25
So just working on our algebra skills
here, leaving us now.
03:28
So you've taken (3x^2-1)^3 out,
so that leaves you
with 3x^2(3x^2-1).
03:39
And in this term, here,
you've taken the (3x^2-1)^3 out,
leaving you with just 24x^4.
03:49
So this is just an extra
bit of factorizing,
if you would like to do,
we can neaten this up,
that gives you (3x^2-1)^3,
multiply that through,
we end up with 9x^4-3x^2+24x^4.
04:11
You can now see that you can add x
to the 4x to the 4 terms together.
04:19
So got (3x^2-1)^3,
you can add 9x^4 with 24x^4,
giving you (33x^4-3x^2).
04:34
So just an extra little
part of expanding things out
and you can see that it's made
the solution a little bit easier
or a little bit easier to read
than what we had here previously.
04:44
So we've just tidy this up to make
it into a slightly more simpler,
a bit more readable
differential.
04:51
So we've now used the product
rule to find the derivative
for x^3 (3x^2-1)^4
We use the product rule
and the chain rule,
and we did some algebraic simplifying
to get to our final answer.