Understanding Variable Changes in the Chain Rule: A Guide for Mathematicians

In summary, the conversation discusses the use of variable changes in the proofs of Proposition 1.3.4 and Proposition 1.3.6, and whether partial variable changes are acceptable in the first proposition. The conversation also touches on the use of Leibniz notation and the confusion it can cause, with one person suggesting to use a different notation to make the functions involved more obvious.
  • #1
Buri
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I have a question about the variable changes in the proofs of Proposition 1.3.4 and Proposition 1.3.6. In the first one, it seems like the author does the variable change but once he applies the chain rule he doesn't do it completely. While in the second it seems like it does the variable changes like usual in the chain rule. Are these 'partial' variable changes in the first one okay? I don't see what's the point of the variable change if you're not going to apply it completely - though I do see the purpose of not doing it completely in the first proposition. And another question, when applying the chain rule in Liebniz notation do you HAVE to make the variable change?

I'm not very comfortable with Liebniz notation since I've never used it before really. I'm more used to the f'(x) notation or Df notation.

Any clarification would be appreciated!
 

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  • #2
I'm not a big fan of the notation, but they both look like straightforward applications of the chain rule. What is it about the first one that doesn't look "complete" to you?

What the author is doing in the first one is to start with

[tex]t=\phi\circ\phi^{-1}(t)[/tex]

and apply d/dt to both sides, which combined with the definitions [itex]\psi=\phi^{-1}[/itex] and [itex]\bar t=\psi(t)[/itex] yields

[tex]1=\phi'(\phi^{-1}(t))(\phi^{-1})'(t)=\phi'(\psi(t))\psi'(t)=\phi'(\bar t)\psi'(t)=\frac{d\phi}{d\bar t}\frac{d\psi}{dt}[/tex]

Edit: I edited the math above, because the [itex]\phi^{-1}[/itex] was missing a ' symbol.
 
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  • #3
Fredrik said:
I'm not a big fan of the notation, but they both look like straightforward applications of the chain rule. What is it about the first one that doesn't look "complete" to you?

Well there shouldn't be a d(psi) in the first one. Shouldn't it be instead a dt (tilda)?
 
  • #4
I'm not sure if it really matters though. I just find that the noation is sometimes used like literal fractions too - I haven't studied differential forms yet, so maybe they actually can be treated like separate entitities, but I've always heard you can't. So..
 
  • #5
Buri said:
Well there shouldn't be a d(psi) in the first one. Shouldn't it be instead a dt (tilda)?
That's what I don't like about the notation. :smile: See if you understand the calculation I added to my previous post. (I think I wrote it before you posted your reply, but I lost contact with the site for a few minutes, and saved the changes when I was able to reconnect).
 
  • #6
Fredrik said:
I'm not a big fan of the notation, but they both look like straightforward applications of the chain rule. What is it about the first one that doesn't look "complete" to you?

What the author is doing in the first one is to start with

[tex]t=\phi\circ\phi^{-1}(t)[/tex]

and apply d/dt to both sides, which combined with the definitions [itex]\psi=\phi^{-1}[/itex] and [itex]\bar t=\psi(t)[/itex] yields

[tex]1=\phi'(\phi^{-1}(t))\phi^{-1}(t)=\phi'(\psi(t))\psi'(t)=\phi'(\bar t)\psi'(t)=\frac{d\phi}{d\bar t}\frac{d\psi}{dt}[/tex]

I do understand what is going on. But my question is more as to if the partial variable changes like when its supposed to be d t (tilda) instead of d (psi) are those correct? Because just to make notation easier, what he's doing is this:

df/du dg/dx when normally it should be df/du du/dx. See what I imean?
 
  • #7
Fredrik said:
That's what I don't like about the notation. :smile: See if you understand the calculation I added to my previous post. (I think I wrote it before you posted your reply, but I lost contact with the site for a few minutes, and saved the changes when I was able to reconnect).

I do. And I guess what you don't like about the noation is exactly what I find confusing about it. This course is in Differential Geometry, but its geared towards physics students actually and so they use different notation and such as to what math major classes use and so I just find this rather confusing, eventhough I do understand its just the chain rule.
 
  • #8
I often find the d(something)/d(something) notation very confusing myself, but when I do, I just ignore the book and do the calculation myself in a notation that makes it obvious what functions I'm dealing with. When I use the chain rule, it looks something like this:

[tex](f\circ g)_{,i}(x)=f_{,j}(g(x))g^j_{,i}(x)[/tex]

(I'm using the summation convention, and ",i" for the ith partial derivative).
 
  • #9
Fredrik said:
I often find the d(something)/d(something) notation very confusing myself, but when I do, I just ignore the book and do the calculation myself in a notation that makes it obvious what functions I'm dealing with. When I use the chain rule, it looks something like this:

[tex](f\circ g)_{,i}(x)=f_{,j}(g(x))g^j_{,i}(x)[/tex]

(I'm using the summation convention, and ",i" for the ith partial derivative).

Thanks for the idea. I'll do the same from now on. I've spent so much time trying to unravel calculations with Leibniz notation, but every time I figure something out something else crops up which doesn't follow the rules, so there isn't much of a point I guess in trying to figuring out all the nuances of the notation.

Thanks for your help! :smile:
 

1. What is the chain rule?

The chain rule is a rule in calculus that allows us to find the derivative of a composite function. In other words, it helps us to find the rate of change of one function with respect to another function.

2. How do you apply the chain rule?

To apply the chain rule, you need to first identify the inner and outer functions in the composite function. Then, you can use the formula (f o g)'(x) = f'(g(x)) * g'(x) to find the derivative, where f is the outer function and g is the inner function.

3. What are some common mistakes when using the chain rule?

One common mistake is not properly identifying the inner and outer functions. Another mistake is forgetting to multiply the derivative of the inner function by the derivative of the outer function. It is also important to be careful with chain rule when dealing with trigonometric functions, as they require special rules.

4. What is a substitution?

A substitution is a method in calculus used to simplify integrals by replacing a variable with a new one. This allows us to solve integrals that may not be easily solvable with the standard techniques.

5. When should you use substitution?

Substitution is most commonly used when dealing with integrals that involve variables raised to a power, such as x^2 or x^3. It is also useful for trigonometric integrals and rational functions. If the integrand is a composition of two functions, the chain rule can also be used in conjunction with substitution.

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