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, The Chain rule is so important, it's
worth thinking through a proof of its

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validity. You might be tempted to think
that you can get away with just

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cancelling. What I mean is you might be
tempted to think that something like this

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works. let's say you wanted to
differentiate g of f of x. You might set f

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of x equal to y. And then, you might
think, well, what you're really trying to

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calculate is the derivative of g of y and
okay. And then, you might say, well, the

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derivative of g of y, that will be the
derivative of g with respect to y times

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the derivative of y with respect to x, and
that's really the Chain rule. I mean, this

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first thing is the derivative of g and
this other thing is the derivative of f

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just like you'd expect, and then you're
trying to say that you can just cancel,

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alright? You're not allowed to just
cancel. I mean, the upshot here is just

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that dy/dx is not a fraction, alright? You
can't justify this equality by just

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canceling because these objects, the way
you're supposedly doing the canceling,

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they're not fractions. We need a more
delicate argument than that. One way to go

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is to give a slightly different definition
of derivative. Well, here's a slightly

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different way of packaging up the
derivative. The function f is

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differentiable at a point a, provided
there's some number, which I"m

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suggestively calling f prime of a, if the
derivative of f at a, so that the limit of

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this error function is equal to 0. And
what's this error function? Well, it's

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measuring how far my approximation is that
I'd get using the derivative from the

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actual functions output if I plug in an
input value near a. In some ways, that's

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actually a nicer definition of derivative,
since it really conveys that the

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derivative provides a way to approximate
output values of functions. In any case,

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now, let's take this new definition of
derivative and try to prove the Chain

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rule. Try to approximate g of f of x plus
h. Try to discover the Chain rules. So, I

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want to be able to express this in terms
of the derivative s of g and f, at least,

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approximately, and then control the error.
Well, I can do that for f because I'm

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assuming that f is deferential about the
point x. So, this is g of, instead of f of

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x plus h, f of x plus the derivative of f
at x times h plus an error term, we should

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be calling error of f of h times h. I'm
going to play the same game with g. This

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is g of f of x plus a small quantity. And
if I assume that g is differentiable at

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the point f of x, then this is g of f of x
plus the derivative of g at f of x times

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how much I wiggle by, which, in this case,
is f prime of x h plus that error term

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plus an error term for g, which is the
error term for g. And I have to put in how

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much I wiggled by, which, in this case, is
f prime of xh plus the error term for f at

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h times h times that same quantity, f
prime of xh plus the error term for f at h

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times h. Alright, so that's exactly equal
to g of f of x plus h and I'm including

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all of the error terms. Now, we can expand
out a bit. So, this is g of f of x plus,

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you can multiply these two terms together,
g prime of f of x times f prime of xh.

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That's looking really good, because that's
what the Chain rule is, right? It's

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supposed to give me this as the derivative
of g composed with f. Plus, I've got a ton

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of error terms now. All those error terms
have an h, so I'm going to collect all the

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h's at the end. The first error term is g
prime of f of x, this term here, times the

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error of f at h. The next ones, plus the
error of g, at this complicated quantity,

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I was going to abbreviate hyphen, times f
prime of x times h, I'm collecting all of

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these h's the end, plus the error term for
g, at that complicated quantity, times the

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error for f at h, and all of that is times
h. Alright. Now, this is almost giving me

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the derivative of the composite function
provided that I can control the size of

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this error term, right? What I need to
show now is that the limit as h approches

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0 of this error term is really 0, and the
error term, right, it's the part before th

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e times h, and it's g prime of f of x
times the error term for f at h plus the

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error term for g times f prime of x, plus
the error term for g times the error term

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for f at h. Now, why do I know that, that
limit is equal to 0? Well, I can do it in

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pieces, right? It's the limit of the sum,
so it's the sum of the limits. and I know

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that this first term is 0 because it's got
an error f h term in it, and because f is

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differentiable, the error term goes to 0.
I likewise know the same for this,

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alright? This is, it also got an error f
of h term in it. The most mysterious term

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is this. But if you think a little bit
more about it, the error of g at this

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hyphen thing, which I'm abbreviating this
whole thing here, also goes to 0 as h goes

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to 0. And that's another thing to know
that the limit as h goes to 0 of this

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quantity is 0 which is then enough to say
that g of f of x plus h equals this

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quantity, actually implies that this is
the derivative. So, here's what we've

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actually shown. Suppose that f is
differentiable at a point a and g is

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differentiable at the point f of a, then
the composite function, g composed with f,

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is differentiable at a, with the
derivative of g of f at the point a, equal

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to the derivative of g at f of a times the
derivative of f at a.