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The Chain Rule

 
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Jul30-12, 08:46 AM   #1
 

The Chain Rule


I have read a few sources regarding the chain rule, and a pervasive explanation that most of the sources share is this, which is way to sort of make sense of it:

"Regard du/dx as the rate of change of u with respect to x, dy/du as the rate of change of y with respect to u, and dy/dx as the rate of change of y with respect to x. If u changes twice as fast as x and y changes three times as fast as u, then it seems reasonable that y changes six times as fast as x, and so we expect that dy/dx = dy/du * du/dx."

I don't understand why it is a simple operation of multiplication to find how fast y changes compared to x. Maybe I am missing something. I'd like to mention, though, that I do understand the chain rule; but when I read this description of it, I just don't seem to understand.
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Jul30-12, 09:44 AM   #2
 
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The simplest way of looking at it (though not as rigorous as the proof of the chain rule) is that, while dy/dx is not a fraction, it is the limit of a fraction- go back before the limit, to [itex]\Delta y/\Delta x[/itex], and with x a function of u, [tex]\Delta y/\Delta x= (\Delta y/\Delta u)(\Delta u/\Delta x)[/tex]. You can "cancel" in that fraction and take the limit.
Aug11-12, 08:06 AM   #3
 
Quote by HallsofIvy View Post
The simplest way of looking at it (though not as rigorous as the proof of the chain rule) is that, while dy/dx is not a fraction, it is the limit of a fraction- go back before the limit, to [itex]\Delta y/\Delta x[/itex], and with x a function of u, [tex]\Delta y/\Delta x= (\Delta y/\Delta u)(\Delta u/\Delta x)[/tex]. You can "cancel" in that fraction and take the limit.
In rigorous mathematics, I heard you can't treat dy/dx like a fraction. But I don't think there's anything wrong by doing that, since dy/dx is like (as HallsoIvy said above) Δy/Δx, so isn't there sort of like a value in Δx and Δy? So why can't you treat it like a fraction?
Aug11-12, 10:41 PM   #4
 

The Chain Rule


Quote by Byron Chen View Post
In rigorous mathematics, I heard you can't treat dy/dx like a fraction. But I don't think there's anything wrong by doing that, since dy/dx is like (as HallsoIvy said above) Δy/Δx, so isn't there sort of like a value in Δx and Δy? So why can't you treat it like a fraction?
The simple reason is that you are not dealing with a number so to speak: dy, dx and all the others are not numbers but the results of limits.

The proofs for calculus show when they can be treated like fractions, but remember that in general, they are the result of limiting processes.

One thing you can do however, is to get an understanding by using fixed delta values instead of the infinitesimal ones and see how they affect calculations.

I stress that these are not the same things and only to be used as a naive guide to what's happening, but never the less they can be useful.

Consider the following: the total differential. Lets say you have a function f of two variables x and y that are independent.

Then you can obtain the total differential df = df/dx*dx + df/dy*dy.

If you change the infinitesimals to deltas, then you can see that what this is doing is it's taking the rate of change with respect to each variable and then incrementing it in the right way.

Lets simplify it and let triangle_x = triangle_y = 1 (in place of dx and dy). Also triangle_f is the distance between successive measurements.

So for the change in f with respect to deltas of 1, we get the change in the x and y direction.

If you add these together as vectors you get the result of a right angled triangle where the change in f is the hypotenuse of the triangle and the other sides are the rates of change with a step-size of 1 (not infinitesimal).

But because we are dealing with infinitesimals and limits as opposed to constant numbers we can't just simply "cancel" terms because again they are not numbers.
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