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[MUSIC] We can drive the product rule by 
just going back, to the definition of 

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derivative. 
So what is the definition of derivative 

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say? It tell us that the derivative of 
the product of f of x and g of x is a 

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limit. 
It's the limit as h approaches zero. 

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Of the function at x+h, which, in this 
case, is the product of f and g, both 

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evaluated at x+h, because I'm thinking of 
this as the function, so I'm plugging in 

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x+h, and I subtract the function 
evaluated at x, which is just f(x)*g(x), 

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and then I divide that by h. 
So, it's this limit of this difference 

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quotient, that gives me the derivative of 
the product. 

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How can I evaluate that limit? Here's the 
trick, I'm going to add a disguised 

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version of zero to this limit. 
Instead of just calculating the limit of 

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f(x+h)g(x+h)-f(x)g(x), I'm going to 
subtract and add the same thing. 

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So here, I've got f(x+h)*g(x+h), just 
like up here. 

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Now I'm going to just subtract 
f(x+h)*g(x+h), and then add it back in, 

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plus f(x+h)*g(x). 
This is just zero, 

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I haven't done anything. 
And I'm going to subtract f(x)*g(x) right 

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here and I'm still dividing by h. 
So these are the same limits, 

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I haven't really done anything, but I've 
actually done everything I need. 

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By introducing these extra factors, I've 
now got a common factor of f(x+h) here 

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and a common factor of g(x) here. 
So, I can collect those out and I'll get 

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some good things happening as a result. 
Let's see exactly how this happens. 

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So this is the limit as h goes to zero, 
I'm going to pull out that common factor 

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of f(x+h) [SOUND]. 
And I'm going to multiply by what's left 

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over g(x+h)-g(x) [SOUND] and I can put it 
over h. 

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So that's these two terms. 
Now, what's left over here? I've got a 

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common factor of g(x). 
And what's left over? f(x+h)-f(x) I'll 

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divide this by h, 
and then the factor I pull out is g(x). 

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So this limit is the same as this limit. 
Now this is a limit of a sum. 

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So that's a sum of the limits provided 
the limits exist and we'll see that they 

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do. 
So this is the limit as h goes to zero of 

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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). 

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Now what do I have here I've got limits 
of products which are the products of 

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limits providing the limits exist, and 
they do and we'll see, so let's rewrite 

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these limits of products as products of 
limits. 

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This is the limit as h goes to zero of 
f(x+h) times the limit as h goes to zero 

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of g(x+h)-g(x)/h. 
You might begin to see what's happening 

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here, 
plus the limit as h goes to zero of 

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f(x+h)-f(x)/h times the limit as h goes 
to zero of g(x). 

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Okay, 
now we've got to check that all these 

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limits exist, in order to justify 
replacing limits [INAUDIBLE] limits. 

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But these limits do exist, 
let's see why? 

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This first limit, the limit of f(x+h) as 
h goes to zero, it's actually the hardest 

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one I think, of all these to see. 
Remember back, we showed that 

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differentiable functions are continuous. 
This is really calculating the limit of f 

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of something, as the something approaches 
x. 

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And that's really what this limit is, and 
because f is continuous, because f is 

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differentiable, this limit is actually 
just f(x). 

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But I think seeing that step is probably 
the hardest in this whole argument. 

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What's this thing here? Well, this is the 
limit of the thing that calculates the 

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derivative of g, and g is differentiable 
by assumption. 

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So, this is the derivative of g at x 
plus, what's this limit? This is the 

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limit that calculates the derivative of 
f, and f is differentiable by assumption, 

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so that's f (x)'. 
This is the limit of g(x), as h goes to 

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zero. 
This is the limit of a constant. 

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Wiggling h doesn't affect this at all, so 
that's just g(x). 

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And look at what we've calculated here. 
The limit that calculates the derivative 

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of the product is f(x)*g'(x)+f'(x)*g, 
that is the product rule. 

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What have we really shown here? Well, 
here is one way to write down the product 

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rule very precisely. 
Confusingly, I'm going to define a new 

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function that I'm calling h. 
So h is just the product of f and g now, 

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h(x)=f(x)*g(x). 
If f and g are differentiable at some 

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point, a, then I know the derivative of 
their product. 

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The derivative of their product Is the 
derivative of f times the value of g plus 

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the value of f times the derivative of g. 
This is a precise statement of the 

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product rule, and you can really see, for 
instance, where this differentiability 

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condition was necessary. 
In our proof, at some point in the proof 

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here, I wanted to go from a limit of a 
product to the product of limits. 

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But in order to do that, I need to know 
that this limit exists. 

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And that limit is exactly calculating the 
derivative of G. 

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So you can really see where these 
conditions are playing a crucial role in 

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the proof of the product rule. 
[MUSIC]