0:00

[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).

3:03

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]