So we've spoken of two ways of dealing with the function of a function.
We obviously have the full definition of the chain rule
and also just by observation, what we can do to just differentiate faster.
Now we said we're going to do a proof so we'll do a proof now
and the proofs for chain rule are usually fairly complicated
but there's a slightly simpler one that we'll talk about now
which hopefully convinces you that this is true or you can just go back to our mathematical calculation
and use that as a proof but either way, I'll show you both
and then you can pick and choose whichever one you prefer.
So if we move here to the chain rule proof,
like I said before there are plenty of proofs out there.
Some of them are numerically and algebraically fairly complicated
but thanks to some clever people who've made things a little bit easier,
I found a fairly simpler proof that we can go through that hopefully convince us of the definition.
So, if we are to differentiate y as a function of something else,
so we're wanting to differentiate this, so let's just say we want to d/dx, this function
and that should then give us the gradient dy/dx. So that's what we're trying to prove here.
Now, recall back to differentiation from first principles,
where we said that dy/dx as the limit of delta x tends to zero is delta y over delta x,
so that's the main definition. I'm not writing the whole function out,
I'm just gonna write this for now. We now have delta y over delta x
which we can re-write as the limit, as delta x tends to zero of delta y over delta u,
multiplied it by delta u by delta x, so remember that's the chain rule
if you're using two functions, we can re-write this in this manner.
So we're applying that the limit of delta x tends to zero to those two functions.
Okay now, here's a little property of limits which sets that the product of limits
is the limit of the product, that means that I can break this down into two sections,
so I can re-write this as the limit, delta x tends to zero of delta y by delta u
and this is just the property that I'm using of limits and I can re-write this function
as limit delta x tends to zero of delta u of delta x.
So all I've done is I've split this here and this here and I applied the concept of the limit
to both of them separately, they're still multiplying.
Okay, this one is fairly simple because if you look at it, this is just delta u by delta x
so change in u over change in x as delta x tends to zero.
If you look back at differentiation from first principles, this part here will just give you, du/dx.
So if we were to define du/dx, we'd used this definition.
However, this little bit is slightly more problematic because we have delta y over delta u
but our limit now is saying as delta x tends to zero.
Now, this is where the proof gets a bit complicated
but it's fairly easy just to change your limit to delta x to delta u.
Okay, so all we're doing is we're making this little change here and we're changing our limit.
So if we say, not delta x, we'll say delta u tends to zero and if we do that,
you'll see that we come up with the derivative or the definition of the derivative
for dy/du and again we have just decided that dy/dx is dy/du multiplied by du/dx
if I just write that here what we've just derived so dy/dx is dy/du
multiplied by du/dx and you will remember that this is the chain rule
that we have previously derived and we have now proved using limits or the property of limits.