differential and square of differential


by rsaad
Tags: differential, square
rsaad
rsaad is offline
#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
JJacquelin is offline
#2
Nov22-12, 09:11 AM
P: 744
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
Millennial is offline
#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
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#4
Nov22-12, 11:20 AM
Math
Emeritus
Sci Advisor
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PF Gold
P: 38,879

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
JJacquelin is offline
#5
Nov22-12, 12:42 PM
P: 744
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
Millennial is offline
#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
HallsofIvy is offline
#7
Nov23-12, 08:13 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879
Yeah, I confuse very easily! Thanks.
oay
oay is offline
#8
Nov23-12, 01:00 PM
P: 220
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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