# Partial derivatives (question I am grading).

1. Mar 30, 2012

### MathematicalPhysicist

We have a function f:R^2->R and it has partial derivative of 2nd order.
Show that $f_{xy}=0 \forall (x,y)\in \mathbb{R}^2 \Leftrightarrow f(x,y)=g(x)+h(y)$

The <= is self explanatory, the => I am not sure I got the right reasoning.

I mean we know that from the above we have: $f_x=F(x)$ (it's a question before this one), but now besides taking an integral I don't see how to show the consequent.

Any thoughts how to show this without invoking integration?

2. Mar 30, 2012

### hunt_mat

My first thought is the mean value theorem which states that if $f'(x)=0$ then $f\equiv$constant., apply this to $x$ when $y$ yields $\partial_{y}f\equiv$constant, but this constant will be dependent on $y$ and therefor is a function of $y$.

Get the general idea now?

3. Mar 30, 2012

### MathematicalPhysicist

That's basically what I have written, so I know that f_x =F(x) and f_y=G(y). But this is where I am not sure how to procceed.

I mean I know that: if F(x)=h'(x), then G(x,y)= f(x,y)-h(x) then G_x=0 and then G=g(y).

The problem is how do I know that F(x)=h'(x) I am assuming that F is of this form, aren't I?

4. Mar 30, 2012

### hunt_mat

I think that as you effectively have the equation $f'(x)=g(x)$ then you want to show that there is a function $F(x)$ with the property $F'(x)=f(x)$, and I think that this follows from a version of the fundamental theorem of calculus. By extension $G(x)=F(x)+C$ will also satisfy this equation. I think you can say this because $f\in C^{1}$ as you have $f'(x)$.