Differentiating \frac{dx}{dy} = (\frac{dy}{dx})^{-1} with Respect to x

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Homework Help Overview

The discussion revolves around differentiating the relationship between \(\frac{dx}{dy}\) and \(\frac{dy}{dx}\), specifically focusing on deriving the second derivative \(\frac{d^2x}{dy^2}\) in terms of \(\frac{d^2y}{dx^2}\). The subject area is calculus, particularly concerning derivatives and their relationships.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to differentiate the given equation with respect to \(x\) but expresses uncertainty about how to proceed. Some participants suggest starting with the definition of \(\frac{d^2x}{dy^2}\) and differentiating \(\frac{dx}{dy}\) with respect to \(y\).

Discussion Status

The discussion is ongoing, with participants providing guidance and clarifications on the differentiation process. There is a recognition of the need for supportive dialogue rather than complete solutions, emphasizing the learning aspect of the forum.

Contextual Notes

There is a note regarding the forum's guidelines, highlighting that complete solutions should not be provided, and participants are encouraged to help the original poster develop their understanding instead.

John O' Meara
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Given that [tex]\frac{dx}{dy} = (\frac{dy}{dx})^{-1}\\[/tex], differentiate throughout with respect to x and show that [tex]\frac{d^2x}{dy^2} = \frac{- \frac{d^2y}{dx^2}}{(\frac{dy}{dx})^3}\\[/tex].

An attempt: [tex]\frac{d^2 x}{dx dy} = \frac{d (\frac{dy}{dx})^{-1}}{dx} \\[/tex].

I need help to get me started. Thanks for the help.
 
Last edited:
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Sorry for the second post. It wasn't meant to happen.
 
It is better to start with [tex]\frac{d^2x}{dy^2} = \frac{d}{dy}\big(\frac {dx}{dy})[/tex]
 
Last edited:
I answered your question in your other post.
 
Kummer said:
I answered your question in your other post.

That's the second time you done this today (at least). The idea of the Homework Forum is NOT to post complete solutions. The idea is to HELP the poster to solve the problem themselves by giving them guidance. Not to do the work for them. With alll due respect, you've been around for a while, hasn't anyone pointed this out to you before?
 

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