Let's look at the product rule now, as the title suggests we are now looking at the products of two functions.
A product means two things that are multiplying together.
So we'll be looking at two functions that are multiplying together
and we'll be applying the product rule to it in order to differentiate.
Let's look at a numerical example and define the rule firstly.
Here we have our first question and it's important to observe
what makes this different to the kind of questions that we've been looking at.
We have y equals to 5x squared and then the square root of 6x plus 7.
Now if you start to look at this closely you will see that this is in fact two functions multiplying together.
We have 5x squared as one function and then we have the square root of 6x plus 7 as another function.
They're not adding or subtracting, they are in fact multiplying.
In order to differentiate this we are going to have to use the product rule.
I'll define the product rule first or give you the formal definition of it
and then we'll start to do this numerically, we'll use this example to work through it
and then we will also look at an algebraic proof to prove that the product rule is true.
So if you have any function that is a multiple of two functions,
so say in our example it's 5x squared and square root of 6x plus 7.
Let's split them and call them two different functions so one of them we can call, call f of x and the other one we'll call g of x.
The formal definition of the product rule states that you leave f of x as it is so you don't touch it,
you don't differentiate it, and you multiply it with the differential of g of x.
So here, we're calling it g dash of x, so that's g of x differentiated.
And then you add g of x, so without changing g of x you leave it as it is
and you multiply it with the differential of f of x. It's fairly simple if you think about it,
you leave one function as it is times it with the differential of the other.
You leave the other function as it is and times it with the differential of the first.
We'll give it some formal terminology shortly so you can learn it as a rule and then apply it to product.
You could to make your writing easier to make it easier to write when you're doing an exam questions.
Let u equal to f of x and v equal to g of x. So this is no different to what we've already said,
we're just using different letters to make it a little bit easier.
And then, you can say that the differential of u and v is the same as vdudx, so v left as it is and u differentiated
plus u left as it is and v differentiated. In a lot of newer textbooks and on the internet and websites
you will see that the second notation is used just because it's faster
and it's a bit more efficient but it means exactly the same thing as we said earlier.
So, let's try and work this out now using the product rule.
I have a question here, so same example as we spoke of earlier.
We have 5x squared square root of 6x plus 7. Now the first thing that we don't like about this question
is the square root so let's just deal with that first, we can rewrite this as 5x squared
and then I have 6x plus 7 to the power of a half. Now you can see that these are two products,
so I have this function and then I have this function timesing with it.
It's often easier to just multiply them through so you can times them through
but not in this case because you have a power of a half and remember we don't want to go
into binomial theorem and make this more complicated than it needs to be.
We'll just treat this as two separate functions. Let's apply the product rule now,
so the product rule said that you could split them into two functions, u and v, or f of x and g of x,
anything that you want. I'll say that my u function is this function here so that's 5x squared
and I'll say that my u function is 6x plus 7 to the power of a half.
We now differentiate each one of them separately, so we can say that du/dx
which we can rewrite as u dash whatever notation you prefer to write.
So if I differentiate this type gives me 10x and then here I can say dv/dx
which we can also write as v dash which just means v differentiated.
And now when I differentiate this function, be a little bit careful
because this isn't just a straightforward function, this is in fact a function of a function.
So you can see that I have a function here and then I have a function here.
So when I differentiate v, I am actually using the chain rule to differentiate it.
So remember the fast way of doing the chain rule, you differentiate the outside
so bring the power to the front so that gives me a half.
Whatever is on the inside can stay as is and you decrease the power by 1.
So I will have a half minus 1 which will give us minus a half.
The next step, you then multiply with the differential of the inside which gives you just 6.
We got to tidy this up a little bit so that gives me a half, I can bring the 6 next to it here,
I've got 6x plus 7 to the minus a half. This can cancel out, so 6 divided by 2 gives you 3 so I have 3,
6x plus 7 to the minus a half. So it's getting fairly interesting now
because you're not just using the product rule, you're using the chain rule within the product rule
and all of the rules are coming together and we're dealing with a lot more complicated functions.
Anyway, back to the product rule. We said that the product rule as a definition
if you are differentiating u and v, so if you have a product like so
and you are differentiating it so you want to differentiate this whole thing,
we said that the rule is vdudx, plus udvdx. If you want to write it in full,
you can write it as v times du/dx plus u times dv/dx.
It's really your preference how you choose to learn it but you do need to know the rule when you are applying this.
Okay, let's bring in our numbers. So if we start with v, remember this is my v function here,
so if I use my v function I've got 6x plus 7 to the half multiply this with du/dx,
which is here so we'll times this with 10x, that's the first part then. We then plus with u,
so this is my u function which is 5x squared. So it's quite important that you label it all correctly,
you can see where everything is and then you multiply it with dv/dx, which is this function here.
So, if I just put brackets around it so we've got 3, 6x plus 7 to the minus a half and then we close that.
All we have to do now is tidy this up, so if I bring the 10x to the front that's 6x plus 7 to the half.
These two numbers can multiply together so that gives me 15x squared
and then I have 6x plus 7 to the minus a half. If you want to you can take this negative power down
so you could rewrite this as 10x, 6x plus 7 to the half, plus 15x squared over 6x plus 7 to the half
so you can make it positive. And this here is your differential using the product rule
and in between we also used the chain rule. So all we've done is applied this function here,
or all this rule here which is our product rule. Let's just quickly remind you what that is,
so that's the rule that you use when you're dealing with two products.
You have vdudx plus udvdx, or vu dash plus uv dash or you could just think about it.
It's one of the functions left alone, the other function differentiated, and then it's the second function left alone,
and the first function differentiated. And so like we said we like shortcuts,
we will start to use it a little bit faster without the rule in shortly in the next few questions.