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U2S4V06 Deriving the Chain Rule.txt
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#
# File: content-mit-18-01-1x-captions/U2S4V06 Deriving the Chain Rule.txt
#
# Captions for MITx 18.01.1x module [kbYVRmQdX9E]
#
# This file has 77 caption lines.
#
# Do not add or delete any lines. If there is text missing at the end, please add it to the last line.
#
#----------------------------------------
In this video, we're going to justify the chain rule
a little bit more formally.
But you'll see that the key idea is something that we
discussed in the last video.
And it's thinking about derivatives
as ratios of deltas.
So we have our function h as the composition f of g of x.
And as before, we'll draw this using some number lines.
So g is taking an input here and giving an output on this line.
And f takes that as its input.
And it spits out an output on this bottom line.
And so when we have that, then h is this composition of the two.
And I want to assign variables to these number lines.
So the input we're calling x.
The final output let's denote as y.
And then the intermediate variable
is traditionally called u.
So u is really g of x.
And we can write h of x as f of u, if we wish.
Now, our goal is the derivative of h.
So in Leibniz's notation, that's dy dx.
And, by definition, this is delta y
over delta x taken as a limit.
It's as delta x goes to 0.
So let me draw delta x here.
We've got a change in x.
And that causes a change in u down on this line.
So we have a delta u.
We've got then a change in y based on that, once we apply f.
So we're looking for delta y over delta x.
That's the ratio of this change to this one.
But we know that we can write that
as the product of the intermediate ratios,
just like we did in the last video.
So in other words, delta y over delta u times delta
u over delta x.
And we need to take the limits of both of them
as delta x goes to 0.
Now, this delta u over delta x as delta x goes
to zero, that's just du dx.
And for this part, what we know as delta x goes to 0,
delta u is also going to go to 0.
So we can replace this limit with delta u
going to 0 instead of delta x.
And now this term is exactly dy du.
So that's pretty much it.
We've got dy dx on the left equaling dy du times du dx.
It's a very suggestive notation, isn't it?
So it almost looks like the du's cancel.
They don't.
The du by itself isn't actually a thing.
It's got to be part of this entire Leibniz notation.
But this reminds us that this is true, essentially,
because the delta u's do cancel.
So one bit of business that we have to take care of,
we need to say where these derivatives are being taken,
where they're being located.
So if this initial point up here is a,
then dy dx is being taken at x equals a.
And the same thing for du dx, also taken at x equals a.
But the input variable for dy du is u.
So where is u?
Well, u is here.
And that's g of a.
So that's where we're going to be taking its derivative.
If you want to use prime notation, dy dx is h prime.
And that's at a.
dy du is f prime.
That's being taken at g of a.
And, finally, du dx-- that's g prime at a.
So this is our chain rule.
It's fabulous.
We're going to do plenty of examples.
So stay tuned.