Differential operators - the rules

1. Apr 21, 2010

randybryan

I always get slightly confused with the rules of differentials.

now $$\frac{d^{2}y}{dx^{2}}$$ is the scond derivative of the function y(x

but rooting this does NOT give the first derivative dy/dx

However, with the operator $$\frac{d^{2}}{dx^{2}}$$, it seems that you can root this and it DOES give the first derivative.

Can someone please explain this to me? I may be wrong, but this seems to be the case in my quantum mechanic notes

2. Apr 22, 2010

tiny-tim

Hi randybryan!

Yes, d2y/dx2 = (d/dx)(d/dx)y, but ≠ [(d/dx)y]2.

It's for the same reason that sin(sin(y)) ≠ [sin(y)]2

3. Apr 22, 2010

HallsofIvy

I suspect that your "quantum mechanic notes" (don't hold mathematicians responsible for what a physicist says!) are using a special notation in which "square root" has some kind of operational definition. That is, "$f^2(x)$" does not mean f(x) times itself but f(f(x)) and $\sqrt{f}$ is the inverse of that.

Last edited by a moderator: Apr 23, 2010
4. Apr 22, 2010

randybryan

Ahh the ongoing feud that is Maths nomenclature vs Physics nomenclature.