# Continuity of partial derivatives in a ball implies differentiability

1. Dec 19, 2009

### Mathmos6

Hi all, I'm looking at the following problem:
Suppose that f:$\mathbb{R}^2\to\mathbb{R}$ is such that $\frac{\partial{f}}{\partial{x}}$ is continuous in some open ball around (a,b) and $\frac{\partial{f}}{\partial{y}}$ exists at (a,b): show f is differentiable at (a,b).

Now I know that if both partial derivatives are continuous in a ball around (a,b) it is differentiable, but I don't know how to deal with the case where we only have continuity of one partial derivative and existence of the other - could anyone help me with this?

I've thought about it for a good hour or so now and don't seem to be getting anywhere (I know I'm meant to post what I've got so far but I literally have nothing!) so the more help you can give me the better!

Many thanks :)

Last edited: Dec 19, 2009
2. Dec 20, 2009

### HallsofIvy

Use the definition of differentiable.

A function of two variables, x and y, is "differentiable" at (a, b) if and only if there exist a function $\epsilon(x,y)$ such that $f(a+ x, b+ y)= f(a, b)+ x (\partial f/\partial x(a,b))$$+ y(\partial f/\partial y(a,b))+ \epsilon(x,y)$ and
$$\lim_{(x,y)\to (a,b)}\frac{\epsilon(x,y)}{\sqrt{x^2+ y^2}}= 0$$

Since both $\partial f/\partial x(a,b)$ and $\partial f/\partial y(a,b)$ exist, we can define $\epsilon(x,y)$ as $f(a+ x, b+ y)- f(a, b)$$+ x (\partial f/\partial x(a,b))+ y(\partial f/\partial y(a,b))$.

Try to use the fact that $\partial f/\partial x$ is continous to show that goes to 0 as (x,y) goes to (0, 0).

3. Dec 20, 2009

### Mathmos6

Got it, thanks very much! :)

4. Dec 20, 2009

### HallsofIvy

I'm impressed! Even with my hint, that was not a simple problem.