# Continuity of partial derivatives in a ball implies differentiability

• Mathmos6
In summary, the conversation discusses the problem of proving differentiability of a function, f, at a point (a, b) when one partial derivative is continuous and the other exists at that point. The definition of differentiability is used to show that a function with both continuous partial derivatives is differentiable. The conversation then explores how to use the continuity of one partial derivative to prove differentiability at a point. After considering the problem for a while, the individual is able to solve it with the help of a hint.
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:
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).

HallsofIvy said:
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! :)

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

## What is the definition of continuity of partial derivatives in a ball?

The continuity of partial derivatives in a ball refers to the smoothness of a multivariable function within a certain region, or "ball," in space. It means that the function has continuous partial derivatives at every point within the ball.

## How is continuity of partial derivatives in a ball related to differentiability?

If a function has continuous partial derivatives in a ball, then it is also differentiable at every point within that ball. This means that the function has a well-defined derivative at each point, and the gradient of the function is well-defined throughout the ball.

## What is the significance of continuity of partial derivatives in a ball in real-world applications?

In real-world applications, continuity of partial derivatives in a ball is important because it allows us to make accurate predictions and calculations. It ensures that a function is smooth and well-behaved within a certain region, making it easier to model and analyze.

## How can we prove that continuity of partial derivatives in a ball implies differentiability?

We can prove this statement using the mean value theorem for multivariable functions, which states that if a function is continuous and has continuous partial derivatives in a ball, then it is also differentiable within that ball. This can be shown through a rigorous mathematical proof.

## Are there any exceptions to the statement that continuity of partial derivatives in a ball implies differentiability?

Yes, there are certain cases where a function may have continuous partial derivatives in a ball but is not differentiable at a specific point within that ball. This can occur if the function has a "corner" or sharp point at that point, which violates the smoothness required for differentiability. However, in most cases, continuity of partial derivatives in a ball does imply differentiability.

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