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, Remember back before, when we talked
about the product rule? You know, it goes

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like the derivative of f times g is the
derivative of f times g plus f times the

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derivative of g. It's a little bit
mysterious considering that the product

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rule has a plus in it. But we proved this
previously, just by going back to the

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definition of derivative in terms of
limit. And, calculating the necessary

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limit to show that, this product rule was
in fact valid. We've already seen a proof

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for the products rule. Originally we
justified the product rule by going back

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to the limit definition of derivative and
manipulating that limit. But maybe that

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proof didn't speak to you, so now there's
another trick that we can use. We can use

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logarithms to replace the product with a
sum. Let's see how. So let's suppose that

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f of x is bigger than 0, and g of x is
bigger than 0, say for all x. I just want

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to do this for positive functions. Okay.
Now I'm going to use logs, so let's take

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the log of f of x times g of x. And what
do I know about logs? Logs turn products

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into sums. So the log of f of x times g of
x is the log of f of x plus the log of g

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of x. I'm going to differentiate both
sides of this equation. So the derivative

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log is 1 over, and by the chain rule,
that's 1 over the inside function times

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the derivative of the inside function,
which in this case is the derivative of f

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of x times g of x. That's what I'd like to
compute. What's the driv of the other

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side? Well the derivative of the log is 1
over, so 1 over the inside function times

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the derivative of the inside function,
plus log of g of x is 1 over the inside

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function times the derivative of the
inside function. Now if I multiply both

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sides by f of x times g of x, what
happens? Well if I multiply this side by f

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of x times g of x, I've then isolated the
derivative of the product. So this is just

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the derivative of f of x times g of x. If
I multiply this side by f of x times g of

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x, f of x times 1 over f of x is just 1,
but I'm left with a factor of g of x times

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f prime of x plus, and if I multiply this
term by f of x times g of x, g of x times

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1 over g of x is just one, but I'm left
with an f of x, so f of x times g prime of

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x. And look, this is the product rule. The
derivative of the product is the, in this

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case, g of x times f prime of x plus f of
x times g prime of x. So, I mean, the

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order's a little bit different, but it is
the product rule. So we've justified the

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product rule another way using logarithms,
but that raises a question, what's the

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point of having multiple proofs of a
single mathematical fact? It's not as if

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having 2 different proofs of the product
rule makes the product rule any more true.

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What this argument has is in its favor is
that it's showing off a nice trick that

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you can do with logarithms. There's a
theme that products and quotients are much

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more complicated than sums and differences
Armed with logarithms, we can convert

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difficult products and quotients into much
easier sums and differences, and that's a

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huge win for us.