Proving an equation of f and f^(-1) derivatives

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Homework Help Overview

The discussion revolves around proving a specific equation involving the second derivatives of functions and their inverses. The equation presented is related to the derivatives of a function \( f \) and its inverse \( f^{-1} \). The context is rooted in calculus, particularly in the manipulation of derivatives and the application of the chain rule.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the derivatives of a function and its inverse, questioning how to manipulate the given equation. There is an attempt to substitute expressions for the derivatives, and some participants suggest using the chain rule to facilitate the proof.

Discussion Status

Some participants have offered hints and suggestions regarding the use of the chain rule and the differentiation of inverse functions. However, there is still uncertainty about how to proceed with the proof, and multiple interpretations of the steps involved are being explored.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also a focus on ensuring clarity in communication and understanding of the mathematical concepts involved.

karkas
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Homework Statement



In one of my class's tests I've come across the following equation:

[tex]\frac{d^2 y}{dx^2} \: + \: \left(\frac{dy}{dx}\right)^3 \frac{d^2 x}{dy^2} \: =0[/tex]



Homework Equations



Considering that [tex]\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}[/tex] how does one prove this statement?

The Attempt at a Solution



I've tried substituting like this:

[tex]\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}= \frac{d}{dx}\left(\frac{1}{\frac{dx}{dy}}\right)= \frac{ d \left( \frac{1}{u} \right)}{du}\frac{du}{dx}=-\left(\frac{1}{\frac{dx}{dy}}\right)^2 \frac{d\left(\frac{1}{\frac{dx}{dy}}\right)}{dx}[/tex]

but i don't see how this can continue to finalize the proof...
 
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hi karkas! :smile:

use the chain rule to replace that d/dx at the end by a d/dy :wink:
 
How would that give me the needed second derivative of the inverse function? I can't see it :(
 
hint: what is d/dy of 1/(dx/dy) ? :wink:
 
Its (-d^2x/dy^2)/(dx/dy)^2 isn't it?
 
then doesn't that solve the question? :smile:
 
tiny-tim said:
then doesn't that solve the question? :smile:

oh man I should really learn to improve my writing...
 

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