[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]