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Differential and square of differential

by rsaad
Tags: differential, square
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rsaad
#1
Nov22-12, 08:54 AM
P: 77
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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JJacquelin
#2
Nov22-12, 09:11 AM
P: 759
No, generaly this equality doesn't hold.
For example :
y=x
dy/dx = 2x
dy/dx = 2
(dy/dx) = (2x) = 4x
2 is not equal to 4x

The equality (dy/dx) = dy/dx holds only for one family of functions : y = - ln(ax+b).
It doesn't hold for any other function.
Millennial
#3
Nov22-12, 09:28 AM
P: 295
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].

HallsofIvy
#4
Nov22-12, 11:20 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,502
Differential and square of differential

Quote Quote by rsaad View Post
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.
JJacquelin
#5
Nov22-12, 12:42 PM
P: 759
Quote Quote by HallsofIvy View Post
[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)
Millennial
#6
Nov22-12, 03:07 PM
P: 295
Quote Quote by JJacquelin View Post
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.
HallsofIvy
#7
Nov23-12, 08:13 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,502
Yeah, I confuse very easily! Thanks.
skiller
#8
Nov23-12, 01:00 PM
P: 234
Quote Quote by HallsofIvy View Post
Yeah, I confuse very easily! Thanks.
You do get sloppy very often, Halls. Shape up!


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