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Thread: Continuity and Differentiability

  1. #1
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    Question 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).

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  2. #2
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    Let us start with the first one.
    Quote Originally Posted by catcat103 View Post
    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 $\displaystyle f$ 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 $\displaystyle \partial_x f(0,0)$. By definition it is $\displaystyle \lim_{h\to 0} \tfrac{f(h,0)-f(0,0)}{h} = \lim_{h\to 0}\tfrac{0}{h} = 0$. Thus, $\displaystyle \partial_x f(0,0) = 0$. Similarly $\displaystyle \partial_y f(0,0)=0$. And so the partials exist.
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  3. #3
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    Also, what is the meaning of continuously differentiable?
    - differentiable
    - the derivative is continuous

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