, Remember back before, when we talked
about the product rule? You know, it goes
like the derivative of f times g is the
derivative of f times g plus f times the
derivative of g. It's a little bit
mysterious considering that the product
rule has a plus in it. But we proved this
previously, just by going back to the
definition of derivative in terms of
limit. And, calculating the necessary
limit to show that, this product rule was
in fact valid. We've already seen a proof
for the products rule. Originally we
justified the product rule by going back
to the limit definition of derivative and
manipulating that limit. But maybe that
proof didn't speak to you, so now there's
another trick that we can use. We can use
logarithms to replace the product with a
sum. Let's see how. So let's suppose that
f of x is bigger than 0, and g of x is
bigger than 0, say for all x. I just want
to do this for positive functions. Okay.
Now I'm going to use logs, so let's take
the log of f of x times g of x. And what
do I know about logs? Logs turn products
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
of x. I'm going to differentiate both
sides of this equation. So the derivative
log is 1 over, and by the chain rule,
that's 1 over the inside function times
the derivative of the inside function,
which in this case is the derivative of f
of x times g of x. That's what I'd like to
compute. What's the driv of the other
side? Well the derivative of the log is 1
over, so 1 over the inside function times
the derivative of the inside function,
plus log of g of x is 1 over the inside
function times the derivative of the
inside function. Now if I multiply both
sides by f of x times g of x, what
happens? Well if I multiply this side by f
of x times g of x, I've then isolated the
derivative of the product. So this is just
the derivative of f of x times g of x. If
I multiply this side by f of x times g of
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
f prime of x plus, and if I multiply this
term by f of x times g of x, g of x times
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
x. And look, this is the product rule. The
derivative of the product is the, in this
case, g of x times f prime of x plus f of
x times g prime of x. So, I mean, the
order's a little bit different, but it is
the product rule. So we've justified the
product rule another way using logarithms,
but that raises a question, what's the
point of having multiple proofs of a
single mathematical fact? It's not as if
having 2 different proofs of the product
rule makes the product rule any more true.
What this argument has is in its favor is
that it's showing off a nice trick that
you can do with logarithms. There's a
theme that products and quotients are much
more complicated than sums and differences
Armed with logarithms, we can convert
difficult products and quotients into much
easier sums and differences, and that's a
huge win for us.