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?
([itex]\frac{dy}{dx}[/itex]) 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 [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

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

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

Millennial

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.

oay

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