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Studying for a test in AP Calculus AB? Prepare with this study guide on Product, quotient, & chain rules.

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# Chain rule proof

Video transcript

- What I hope to do in this video is a proof of the famous and useful and somewhat elegant and
sometimes infamous chain rule. And, if you've been
following some of the videos on "differentiability implies continuity", and what happens to a continuous function as our change in x, if x is
our independent variable, as that approaches zero, how the change in our function approaches zero, then this proof is actually
surprisingly straightforward, so let's just get to it, and this is just one of many proofs of the chain rule. So the chain rule tells us that if y is a function of u, which is a function of x, and we want to figure out
the derivative of this, so we want to differentiate
this with respect to x, so we're gonna differentiate
this with respect to x, we could write this as the derivative of y with respect to x, which is going to be
equal to the derivative of y with respect to u, times the derivative
of u with respect to x. This is what the chain rule tells us. But how do we actually
go about proving it? Well we just have to remind ourselves that the derivative of
y with respect to x... the derivative of y with respect to x, is equal to the limit as
delta x approaches zero of change in y over change in x. Now we can do a little bit of
algebraic manipulation here to introduce a change
in u, so let's do that. So this is going to be the same thing as the limit as delta x approaches zero, and I'm gonna rewrite
this part right over here. I'm gonna essentially divide and multiply by a change in u. So I could rewrite this as delta y over delta u times delta u, whoops... times delta u over delta x. Change in y over change in u, times change in u over change in x. And you can see, these are
just going to be numbers here, so our change in u, this
would cancel with that, and you'd be left with
change in y over change x, which is exactly what we had here. So nothing earth-shattering just yet. But what's this going to be equal to? What's this going to be equal to? Well the limit of the product is the same thing as the
product of the limit, so this is going to be the same thing as the limit as delta x approaches zero of,
and I'll color-coat it, of this stuff, of delta y over delta u, times-- maybe I'll put parentheses around it, times the limit... the limit as delta x approaches zero, delta x approaches zero, of this business. So let me put some parentheses around it. Delta u over delta x. So what does this simplify to? Well this right over here,
this is the definition, and if we're assuming, in
order for this to even be true, we have to assume that u and y are differentiable at x. So we assume, in order
for this to be true, we're assuming... we're assuming y comma
u are differentiable... are differentiable at x. And remember also, if
they're differentiable at x, that means they're continuous at x. But if u is differentiable at x, then this limit exists, and
this is the derivative of... this is u prime of x, or du/dx, so this right over here... we can rewrite as du/dx, I think you see where this is going. Now this right over here, just looking at it the way
it's written out right here, we can't quite yet call this dy/du, because this is the limit
as delta x approaches zero, not the limit as delta u approaches zero. But we just have to remind ourselves the results from, probably,
the previous video depending on how you're watching it, which is, if we have a function u that is continuous at a point, that, as delta x approaches zero, delta u approaches zero. So we can actually rewrite this... we can rewrite this right over here, instead of saying delta x approaches zero, that's just going to have the effect, because u is differentiable at x, which means it's continuous at x, that means that delta u
is going to approach zero. As our change in x gets smaller
and smaller and smaller, our change in u is going to get smaller and smaller and smaller. So we can rewrite this, as our change in u approaches zero, and when we rewrite it like that, well then this is just dy/du. This is just dy, the derivative
of y, with respect to u. So just like that, if we assume y and u are differentiable at x, or you could say that
y is a function of u, which is a function of x, we've just shown, in
fairly simple algebra here, and using some assumptions about differentiability and continuity, that it is indeed the case that the derivative of y with respect to x is equal to the derivative
of y with respect to u times the derivative
of u with respect to x. Hopefully you find that convincing.