# Trying to prove something

1. Apr 11, 2010

### Jamin2112

I'm trying to the follow the work shown in the textbook, but getting a little confused.

Theorem I. Suppose the function f(x,y) is defined in some neighborhood of the of the point (a,b). Suppose one of the partial derivatives, say, ∂f/∂x, exists at each point of the neighborhood and is continuous at (a,b), while the other partial derivative is defined at least at the point (a,b). Then f is differentiable at (a,b).

The first thing the book does is write down

f(a+h, b+k) - f(a,b) = f(a+h, b+k) - f(a, b+k) + f(a, b+k) - f(a,b).

Okay. So far, so good.

Now, the part I put in red looks like the [f(a,b+k)-f(a,b)]/k, the partial derivative of f with respect to y at (a,b) when k approaches zero. But since k is just k, we can write this as

f(a+h, b+k) - f(a, b+k) = k ∂f/∂y|(a,b) + Ω1k, where Ω1 approaches zero as k approaches 0.

The part I put in blue, I suppose, could similarly be rewritten as h ∂f/∂x|(a,b+k) + hΩ2. Instead the book uses the mean value theorem as such:

f(a+h, b+k) - f(a, b+k) = h ∂f/∂x|(a+øh, b+k), where 0<ø<1.

Because ∂f/∂x is assumed to be continuous at (a,b), we can write

∂f/∂x|(a+øh, b+k) = ∂f/∂x|(a,b) + Ω2, where Ω2 --> 0 as k --> 0 and h --> 0.

The part in green is what I do not understand. Please explain.

2. Apr 11, 2010

### snipez90

That's just what it means for the partial derivative to be continuous at (a,b). Since ∂f/∂x|(a+øh, b+k) -> ∂f/∂x|(a,b) as h -> 0 and k -> 0, we can for example define Ω2 to be the difference ∂f/∂x|(a+øh, b+k) - ∂f/∂x|(a,b) and see that Ω2 -> 0 when h -> 0 and k -> 0.