So we've just gone over the product rule and we did the chain rule as well.
We did a proof for chain rule so it would be unfair not to do a proof for the product rule as well.
So let's have a look at how we could algebraically prove the product rule.
So if we now move on here, so we'll do the proof for our product rule.
This is quite a standard proof so you'll be able find this in textbooks or internet websites
where they follow this standard proof to prove the product rule and the definition of it.
So if we have a function dy/dx we're going to start with just reminding you of differentiation from first principles.
Remember that as the limit of delta x tends to zero, we have f of x plus delta x minus f of x over delta x.
So these are the things that we've done and we've applied a few times
through this course so just look back if you're struggling with this definition.
Okay, so now we're looking at the product of two functions.
So we now have f of x multiplied by g of x and we're now trying to prove or to find the derivative for this.
So let's just say we're now trying to differentiate f of x multiplied by g of x.
We can say this is the same as we want the limit of delta x tending to zero
rather than using this part here we can say that we're going to use f of x plus delta x,
g of x plus delta x minus f of x, g of x. But unfortunately it's not as straightforward
because we're dealing with two functions. We also add this extra little bit to this,
now I'll show you what it is and then I'll explain it to you shortly so we add f of x plus delta x, g of x
and also while we subtract that and then we also add f of x plus delta x, g of x then we have a bracket.
And remember that this is all being divided by the whole thing is being divided by delta x.
Okay, now up to here is fine so this is just differentiation from first principles.
Observe this little term here, they are exactly the same.
We have a minus and a plus so essentially I am adding zero to it
because these two terms technically should cancel each other or because you have a minus term
and you have a plus term so you could just cancel them out.
So technically if you really think about it I've just added nothing to it.
So we're still sticking to our actual definitions of first principles.
I've added a little bit extra here and you'll see why that is shortly.
It just makes the proof a lot neater and easier to derive.
Okay, we've got four terms here at the top as the numerator.
We are now going to put the like terms together and try and factorize.
So if we now take the limit of delta x times to zero, I'm just going to write together f of x plus delta x
which is my first term. G of x plus delta x so that's this term here I've just written that.
And then I bring in this term here so I've got minus f of x plus delta x, g of x.
And that's only so I can have this f of x plus delta x terms together.
The other two that are left, I have f of x plus delta x multiplied by g of x
which is this term and then I also have minus f of x, g of x.
So I haven't changed anything, I just rewritten it. And all of this is being divided by delta x.
So we're now going to factorize these two terms, so we put brackets around this term here
and brackets around this term here. So we're going to factorize it,
by that I mean look for common factors so you see that it has a common factor here and here,
so we will take that out so we can rewrite this as the limit of delta x tends to zero.
We can take out f of x plus delta x as a common factor leaving us with g of x plus delta x minus g of x all over delta x.
And we do exactly the same thing for the second part of the function here.
So here we have a common factor of g of x, g of x so if we take g of x out
and we are now left with f of x plus delta x minus f of x over delta x.
Okay, if I now use the fact that the limit of products is the product of limits,
I can apply this limit to this by itself, and to this term, and to this term, and to this term.
So imagine now rather than me having to write it all again that I am now applying that the limit of delta x
times to zero for each individual term. So let's see if we can work that out now,
so if I apply the delta x tends to zero to f of x plus delta x, I am just left with f of x.
Look closely at this bracket here, so we have g of x plus delta x minus g of x over delta x.
Now this is almost the same as this function here, so if I just highlight this for a second,
but with g's in it so rather than an f term we have a g term.
So essentially, this is just the differential of g's so we can write this as g differentiated.
When you apply delta x tends to zero to g of x, nothing changes
because there's no delta x there so this just stays as g of x.
And lastly, if you look at this function here this is exactly the same as the definition at the start
so we're looking at this function once again so lots of arrows pointing towards that definition.
So this function again, so this is just a differential of f.
And there you have it, the definition or the derivative of the product rule.
It's f of x left as is, let's just tidy it up here so we've got f of x, g dash of x, plus g of x, f dash of x,
so just to make some space remember what we said that this is the same as uv dash plus vu dash
which is the product rule here. So what we've done here is prove the product rule using algebra
and if you ever need to prove it you can obviously go back or refer back to this proof
and even if you're not proving the product rule or the chain rule,
these are just skills that will always help you derive math formulas algebraically.
If you're ever faced by derivative of functions that you were unable to find,
you can always go back to basics and you can always use these definitions,
you can always use these skills, skills of logic, skills of derivation, and skills of algebra
to come to your conclusions or come up with new rules of differentiation if you can.