Differential and square of differential

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    Differential Square
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Discussion Overview

The discussion centers on the conditions under which the equality \((\frac{dy}{dx})^2 = \frac{d^2y}{dx^2}\) holds. Participants explore this mathematical relationship, questioning its validity across different functions and contexts.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the equality and seeks clarification on the conditions under which it might hold.
  • Another participant asserts that the equality does not generally hold, providing a counterexample with the function \(y = x^2\) where \((dy/dx)^2\) results in \(4x^2\) while \(\frac{d^2y}{dx^2}\) equals \(2\).
  • It is noted that the equality \((dy/dx)^2 = \frac{d^2y}{dx^2}\) holds only for the specific family of functions \(y = -\ln(ax+b)\).
  • A participant mentions that the correct expression is \(\left( \frac{d}{dx} \right)^2y = \frac{d^2y}{(dx)^2\), which is often abbreviated as \(\frac{d^2y}{dx^2}\).
  • Further elaboration is provided on the implications of the equality being treated as a differential equation, leading to a specific solution form for \(y(x)\).
  • Another participant suggests that the original poster may have confused differentiation and integration in their understanding of the equality.

Areas of Agreement / Disagreement

Participants generally agree that the equality is not universally true and that it holds under specific conditions. Multiple competing views regarding the validity of the equality and the types of functions for which it may apply remain present in the discussion.

Contextual Notes

Participants express uncertainty regarding the general applicability of the equality and highlight the need for careful consideration of the functions involved. The discussion reveals a dependence on specific definitions and assumptions related to differential equations.

rsaad
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Hi
I often see the following in books but I do not understand how they are equal. So can someone please tell me for what conditions does the following equality hold?
([itex]\frac{dy}{dx}[/itex]) 2 = (d2 y)/(dx2)
 
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No, generaly this equality doesn't hold.
For example :
y=x²
dy/dx = 2x
d²y/dx² = 2
(dy/dx)² = (2x)² = 4x²
2 is not equal to 4x²

The equality (dy/dx)² = d²y/dx² holds only for one family of functions : y = - ln(ax+b).
It doesn't hold for any other function.
 
The equality that does hold is [itex]\displaystyle \left( \frac{d}{dx} \right)^2y = \frac{d^2y}{(dx)^2}[/itex], which is usually abbreviated (somewhat abuse of notation) as [itex]\displaystyle \frac{d^2y}{dx^2}[/itex].
 
rsaad said:
Hi
I often see the following in books but I do not understand how they are equal. So can someone please tell me for what conditions does the following equality hold?
([itex]\frac{dy}{dx}[/itex]) 2 = (d2 y)/(dx2)
I can't help but wonder where you "often see" that? As others said, it is certainly NOT "generally" true. [itex]d^2y/dx^2= (dy/dx)^2[/itex] is a "differential equation". If we let u= dy/dx, we have the "first order differential equation" [itex]du/dx= u^2[/itex] which can be written as [itex]u^{-2}du= dx[/itex] and, integrating, [itex]-u^{-1}= x+ C[/itex] so that [itex]u= dy/dx= -\frac{1}{x+ C}[/itex]. Integrating that, [itex]y(x)= \frac{1}{(x+ C)^2}+ C_1[/itex]. Only such functions satisfy your equation.
 
HallsofIvy said:
[itex]u= dy/dx= -\frac{1}{x+ C}[/itex]. Integrating that, [itex]y(x)= \frac{1}{(x+ C)^2}+ C_1[/itex]. Only such functions satisfy your equation.
Integrating -1/(x+C) leads to -ln(x+C)+c = -ln(ax+b)
 
JJacquelin said:
Integrating -1/(x+C) leads to -ln(x+C)+c = -ln(ax+b)

He probably got confused and differentiated instead. Anyways, the identity given in the OP is not generally true, actually, it is almost surely false, given the family of functions that can be defined on R2.
 
Yeah, I confuse very easily! Thanks.
 
HallsofIvy said:
Yeah, I confuse very easily! Thanks.
You do get sloppy very often, Halls. Shape up! :smile:
 

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