Continuity of partial derivatives in a ball implies differentiability

[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 :)

 

[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 continuous to show that goes to 0 as (x,y) goes to (0, 0).

 

[Mathmos6]

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 continuous to show that goes to 0 as (x,y) goes to (0, 0).

Got it, thanks very much! :)

 

[HallsofIvy]

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