Continuity and Differentiability

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June 29th 2008, 01:53 AM
catcat103

Continuity and Differentiability

I am confused about the definition of continuity and differentiability of n-variable functions. Also, what is the meaning of continuously differentiable?

Here are some examples that I don't understand fully. Please explain.

1. Let function f be defined on the whole xy plane. f(x,y)= 1 if x=y=/=0 and f(x,y)=0 otherwise. In this case, f is not continuous at (0,0) but both partial derivatives fx and fy exists at (0,0).

2. Let function f be defined as f(x,y)=(x^1/3 + y^1/3)^3. f(x,y) is continuous and has partial derivatives at origin (0,0) but is not differentiable there.

3. Let f be defined by f(x,y)=y^2 + x^3 sin (1/x) for x=/=0, and f(0,y)=y^2. f is differentiable at (0,0), but is not continuously differentiable there because fx(x,y) is not continuous at (0,0).

Thanks. > <
June 29th 2008, 08:29 AM
ThePerfectHacker

Let us start with the first one.

Quote:

Originally Posted by catcat103

1. Let function f be defined on the whole xy plane. f(x,y)= 1 if x=y=/=0 and f(x,y)=0 otherwise. In this case, f is not continuous at (0,0) but both partial derivatives fx and fy exists at (0,0).

In single-variable calculus if $f$ is differenciable at the point then it must be continous there. This example is showing that the existence of partial derivatives does not gaurentte the continuity of the function. Look at $\partial_x f(0,0)$ . By definition it is $\lim_{h\to 0} \tfrac{f(h,0)-f(0,0)}{h} = \lim_{h\to 0}\tfrac{0}{h} = 0$ . Thus, $\partial_x f(0,0) = 0$ . Similarly $\partial_y f(0,0)=0$ . And so the partials exist.
June 29th 2008, 11:12 PM
Moo

Quote:

Also, what is the meaning of continuously differentiable?

- differentiable
- the derivative is continuous

(Wink)