Are Partial Derivatives Commutative for Functions of Multiple Variables?

[Kyle.Nemeth]

Homework Statement

I would just like to know if this statement is true.

Homework Equations

$$\frac {\partial^2 f}{\partial x^2} \frac{\partial g}{\partial x}=\frac{\partial g}{\partial x} \frac {\partial^2 f}{\partial x^2}$$

The Attempt at a Solution

I've thought about this a bit and I haven't come to a conclusion. Thanks for the help!

 

[Mr-R]

Well, it depends on $f$ and $g$ and not so on the partial derivative. If $f$ and $g$ are "normal" functions like $f(x)=x^2$ for example, then the statement is true. On the other hand, if they represent matrices then generally they wouldn't commute, ie. $f\cdot g\neq g\cdot f$ because $g$ and $f$ do not commute generally.

 

[Ray Vickson]

Homework Statement

I would just like to know if this statement is true.

Homework Equations

$$\frac {\partial^2 f}{\partial x^2} \frac{\partial g}{\partial x}=\frac{\partial g}{\partial x} \frac {\partial^2 f}{\partial x^2}$$

The Attempt at a Solution

I've thought about this a bit and I haven't come to a conclusion. Thanks for the help!

If you set $A = \partial g/\partial x$ and $B = \partial^2 f/\partial x^2$, you have written $A B = B A$, which is true for any two real numbers.

However, if what you really meant was to have
$$\frac{\partial}{\partial x} \left( g \frac{\partial^2 f}{\partial x^2} \right)$$
on one side and
$$\frac{\partial^2} {\partial x^2} \left( f \frac{\partial g}{\partial x} \right)$$
on the other, then that is a much different question.

Which did you mean?

 

[Kyle.Nemeth]

I intended for the original question you had answered about $$AB=BA$$ for any real number. I was assuming that the second derivative had acted on f and the first derivative had acted on g.