[MUSIC] We can drive the product rule by 
just going back, to the definition of 
derivative. 
So what is the definition of derivative 
say? It tell us that the derivative of 
the product of f of x and g of x is a 
limit. 
It's the limit as h approaches zero. 
Of the function at x+h, which, in this 
case, is the product of f and g, both 
evaluated at x+h, because I'm thinking of 
this as the function, so I'm plugging in 
x+h, and I subtract the function 
evaluated at x, which is just f(x)*g(x), 
and then I divide that by h. 
So, it's this limit of this difference 
quotient, that gives me the derivative of 
the product. 
How can I evaluate that limit? Here's the 
trick, I'm going to add a disguised 
version of zero to this limit. 
Instead of just calculating the limit of 
f(x+h)g(x+h)-f(x)g(x), I'm going to 
subtract and add the same thing. 
So here, I've got f(x+h)*g(x+h), just 
like up here. 
Now I'm going to just subtract 
f(x+h)*g(x+h), and then add it back in, 
plus f(x+h)*g(x). 
This is just zero, 
I haven't done anything. 
And I'm going to subtract f(x)*g(x) right 
here and I'm still dividing by h. 
So these are the same limits, 
I haven't really done anything, but I've 
actually done everything I need. 
By introducing these extra factors, I've 
now got a common factor of f(x+h) here 
and a common factor of g(x) here. 
So, I can collect those out and I'll get 
some good things happening as a result. 
Let's see exactly how this happens. 
So this is the limit as h goes to zero, 
I'm going to pull out that common factor 
of f(x+h) [SOUND]. 
And I'm going to multiply by what's left 
over g(x+h)-g(x) [SOUND] and I can put it 
over h. 
So that's these two terms. 
Now, what's left over here? I've got a 
common factor of g(x). 
And what's left over? f(x+h)-f(x) I'll 
divide this by h, 
and then the factor I pull out is g(x). 
So this limit is the same as this limit. 
Now this is a limit of a sum. 
So that's a sum of the limits provided 
the limits exist and we'll see that they 
do. 
So this is the limit as h goes to zero of 
f(x+h)*g(x+h)-g(x)/h plus the lim as h 
goes to zero of f(x+h)-f(x)/h*g(x). 
Now what do I have here I've got limits 
of products which are the products of 
limits providing the limits exist, and 
they do and we'll see, so let's rewrite 
these limits of products as products of 
limits. 
This is the limit as h goes to zero of 
f(x+h) times the limit as h goes to zero 
of g(x+h)-g(x)/h. 
You might begin to see what's happening 
here, 
plus the limit as h goes to zero of 
f(x+h)-f(x)/h times the limit as h goes 
to zero of g(x). 
Okay, 
now we've got to check that all these 
limits exist, in order to justify 
replacing limits [INAUDIBLE] limits. 
But these limits do exist, 
let's see why? 
This first limit, the limit of f(x+h) as 
h goes to zero, it's actually the hardest 
one I think, of all these to see. 
Remember back, we showed that 
differentiable functions are continuous. 
This is really calculating the limit of f 
of something, as the something approaches 
x. 
And that's really what this limit is, and 
because f is continuous, because f is 
differentiable, this limit is actually 
just f(x). 
But I think seeing that step is probably 
the hardest in this whole argument. 
What's this thing here? Well, this is the 
limit of the thing that calculates the 
derivative of g, and g is differentiable 
by assumption. 
So, this is the derivative of g at x 
plus, what's this limit? This is the 
limit that calculates the derivative of 
f, and f is differentiable by assumption, 
so that's f (x)'. 
This is the limit of g(x), as h goes to 
zero. 
This is the limit of a constant. 
Wiggling h doesn't affect this at all, so 
that's just g(x). 
And look at what we've calculated here. 
The limit that calculates the derivative 
of the product is f(x)*g'(x)+f'(x)*g, 
that is the product rule. 
What have we really shown here? Well, 
here is one way to write down the product 
rule very precisely. 
Confusingly, I'm going to define a new 
function that I'm calling h. 
So h is just the product of f and g now, 
h(x)=f(x)*g(x). 
If f and g are differentiable at some 
point, a, then I know the derivative of 
their product. 
The derivative of their product Is the 
derivative of f times the value of g plus 
the value of f times the derivative of g. 
This is a precise statement of the 
product rule, and you can really see, for 
instance, where this differentiability 
condition was necessary. 
In our proof, at some point in the proof 
here, I wanted to go from a limit of a 
product to the product of limits. 
But in order to do that, I need to know 
that this limit exists. 
And that limit is exactly calculating the 
derivative of G. 
So you can really see where these 
conditions are playing a crucial role in 
the proof of the product rule. 
[MUSIC]