Chain rule and change of variables again

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Discussion Overview

The discussion revolves around the application of the chain rule in calculus, specifically focusing on the transformation of derivatives when considering x as a function of y instead of the conventional approach of y as a function of x. Participants explore the implications of this change on first and second derivatives.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about the second derivative transformation, particularly the presence of a cubic term in the denominator of the equation d2y/dx2 = -d2x/dy2 / (dx/dy)3.
  • Another participant attempts to clarify the transformation by rewriting the first derivative and applying the chain rule to derive the second derivative, suggesting that this is a starting point for understanding.
  • A question is raised regarding the differentiation of dx/dy with respect to x, leading to a discussion about the roles of x and y as dependent and independent variables.
  • One participant asserts that if x is a function of y, then x should be considered the dependent variable, challenging the notion that it is independent in this context.
  • Further clarification is provided that differentiating d/dx(dx/dy) can be expressed using the chain rule, indicating a deeper layer of differentiation is involved.

Areas of Agreement / Disagreement

Participants exhibit some agreement on the application of the chain rule but remain divided on the interpretation of variables as dependent or independent, leading to unresolved questions about the differentiation process.

Contextual Notes

The discussion includes assumptions about the relationships between variables and the application of differentiation rules, which may not be fully articulated or agreed upon by all participants.

jonjacson
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We start with:

d2y/dx2

And we want to consider x as function of y instead of y as function of x.

I understand this equality:

dy/dx = 1/ (dx/dy)

But for the second order this equality is provided:

d2y/dx2 =- d2x/dy2 / (dx/dy)3

Does anybody understand where is it coming from? The cubic term in the denominator looks quite strange, I don't know how to understand that equation.

ANy help is welcome.
 
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jonjacson said:
We start with:

d2y/dx2

And we want to consider x as function of y instead of y as function of x.

I understand this equality:

dy/dx = 1/ (dx/dy)

But for the second order this equality is provided:

d2y/dx2 =- d2x/dy2 / (dx/dy)3

Does anybody understand where is it coming from? The cubic term in the denominator looks quite strange, I don't know how to understand that equation.

ANy help is welcome.
Write your second equation above as ##dy/dx = (dx/dy)^{-1}##
Now take the derivative with respect to x of both sides:
##d^2y/dx^2 = (-1) (dx/dy)^{-2} \cdot d/dx(dx/dy)##
On the right side above, I'm using the chain rule.
Is that enough of a start?
 
Mark44 said:
Write your second equation above as ##dy/dx = (dx/dy)^{-1}##
Now take the derivative with respect to x of both sides:
##d^2y/dx^2 = (-1) (dx/dy)^{-2} \cdot d/dx(dx/dy)##
On the right side above, I'm using the chain rule.
Is that enough of a start?

It looks like a great start but I have a question.

If we have x as function of y, dx/dy will be a function of y too, WHat is the meaning of deriving this respect to x?
X is the independent variable now, Does it make sense to differenciate a function respect itself?

Maybe I am confused and I should go to bed.
 
jonjacson said:
It looks like a great start but I have a question.

If we have x as function of y, dx/dy will be a function of y too, WHat is the meaning of deriving this respect to x?
You can differentiate it with respect to x (not derive it).
jonjacson said:
X is the independent variable now
No, if x is a function of y, x is the dependent variable, and y is the independent variable.
jonjacson said:
, Does it make sense to differenciate a function respect itself?

Maybe I am confused and I should go to bed.
In the last part of what I wrote I have ##d/dx(dx/dy)##. That's the same as ##\frac d {dy} \left(\frac {dx}{dy} \right)\cdot \frac{dy}{dx}##, with the chain rule being applied once more.
 

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