Differential and square of differential

[rsaad]

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?
(\frac{dy}{dx}) ² = (d² y)/(dx²)

 

[JJacquelin]

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.

 

[Millennial]

The equality that does hold is \displaystyle \left( \frac{d}{dx} \right)^2y = \frac{d^2y}{(dx)^2}, which is usually abbreviated (somewhat abuse of notation) as \displaystyle \frac{d^2y}{dx^2}.

 

[HallsofIvy]

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?
(\frac{dy}{dx}) ² = (d² y)/(dx²)

I can't help but wonder where you "often see" that? As others said, it is certainly NOT "generally" true. d^2y/dx^2= (dy/dx)^2 is a "differential equation". If we let u= dy/dx, we have the "first order differential equation" du/dx= u^2 which can be written as u^{-2}du= dx and, integrating, -u^{-1}= x+ C so that u= dy/dx= -\frac{1}{x+ C}. Integrating that, y(x)= \frac{1}{(x+ C)^2}+ C_1. Only such functions satisfy your equation.

 

[JJacquelin]

u= dy/dx= -\frac{1}{x+ C}. Integrating that, y(x)= \frac{1}{(x+ C)^2}+ C_1. Only such functions satisfy your equation.

Integrating -1/(x+C) leads to -ln(x+C)+c = -ln(ax+b)

 

[Millennial]

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]

Yeah, I confuse very easily! Thanks.

 

[skiller]

Yeah, I confuse very easily! Thanks.

You do get sloppy very often, Halls. Shape up!