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Math Help - A question about partial derivatives and directional derivative.

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    A question about partial derivatives and directional derivative.

    Assume f is a real-valued function defined on an open set E of \mathbb R^n and all partial derivatives of f exist on E. Is it possible that there is some {\bf {a}}\in E and {\bf u}\in\mathbb R^n such that the directional derivative of f in the direction of \bf u at \bf a does not exist?
    Thanks!
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    Quote Originally Posted by zzzhhh View Post
    Assume f is a real-valued function defined on an open set E of \mathbb R^n and all partial derivatives of f exist on E. Is it possible that there is some {\bf {a}}\in E and {\bf u}\in\mathbb R^n such that the directional derivative of f in the direction of \bf u at \bf a does not exist?
    On \mathbb{R}^2, let f(x,y) = \begin{cases}\frac{xy}{x^2+y^2}&((x,y)\ne(0,0)),\\ 0&((x,y)=(0,0)).\end{cases}

    Then both partial derivatives of f exist at the origin (and everywhere else), but the directional derivatives in the directions of (1,\pm1) do not exist.
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    If f is differentiable at a point, then the directional derivative in all directions exists. But, as Opalg showed, it may happen that has partial derivatives at a point but is not "differentiable" there.

    Another example is
    <br />
f(x,y) = \begin{cases}1 &((x,y)\ne(0,0)),\\ 0&((x,y)=(0,0)).\end{cases}<br />

    At (0,0) both partial derivatives exist and are 0. But the function is not differentiable and directional derivatives do not exist except in the directions of the axes.
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    Quote Originally Posted by HallsofIvy View Post
    Another example is
    <br />
f(x,y) = \begin{cases}1 &((x,y)\ne(0,0)),\\ 0&((x,y)=(0,0)).\end{cases}<br />

    At (0,0) both partial derivatives exist and are 0. But the function is not differentiable and directional derivatives do not exist except in the directions of the axes.
    Umm, I hate to disagree, but that example does not have partial derivatives at the origin:

    \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(h,0)-f(0,0)}h = \lim_{h\to0}\frac{1-0}h = (ouch!)

    I think you probably meant <br />
f(x,y) = \begin{cases}1 &(xy\ne0),\\ 0&(x \text{ or }y =0).\end{cases}<br />
But no, on second thoughts that won't work either, because that function doesn't have partial derivatives on either of the axes (except at the origin). I have a feeling that the example I gave previously can't really be simplified.
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    Thank you! Now I have a further question: Assume the conditions in my first post still hold, that is, f is a real-valued function defined on an open set E of \mathbb R^n and all partial derivatives of f exist on E. If the directional derivative of f in the direction of {\bf u}=(u_1,...,u_n) at a point {\bf a}\in E does exist, is it always true that the value of the directional derivative equals the inner product (D_1f({\bf a}),...,D_nf({\bf a}))\cdot(u_1,...,u_n)? Thanks!
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    Quote Originally Posted by Opalg View Post
    I think you probably meant <br />
f(x,y) = \begin{cases}1 &(xy\ne0),\\ 0&(x \text{ or }y =0).\end{cases}<br />
But no, on second thoughts that won't work either, because that function doesn't have partial derivatives on either of the axes (except at the origin). I have a feeling that the example I gave previously can't really be simplified.

    Why does this not work?

    f=0 everywhere on the x and y axis, and f=1 otherwise, so:

    \frac{\partial f}{\partial x}(x,0)=\lim\limits_{h\to0}\frac{f(x+h,0)-f(x,0)}{h}=\lim\limits_{h\to0}\frac{0-0}{h}=0, for all x\in\mathbb{R}
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    Quote Originally Posted by hjortur View Post
    Why does this not work?

    f=0 everywhere on the x and y axis, and f=1 otherwise, so:

    \frac{\partial f}{\partial x}(x,0)=\lim\limits_{h\to0}\frac{f(x+h,0)-f(x,0)}{h}=\lim\limits_{h\to0}\frac{0-0}{h}=0, for all x\in\mathbb{R}
    This function has a partial x-derivative at points on the x-axis, but not a partial y-derivative. If x\ne0 then \frac{\partial f}{\partial y}(x,0)=\lim_{k\to0}\frac{f(x,k)-f(x,0)}{k}=\lim_{k\to0}\frac{1-0}{k}, which does not exist.

    On the y-axis, the function has a partial y-derivative but not a partial x-derivative.
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    Quote Originally Posted by Opalg View Post
    Umm, I hate to disagree, but that example does not have partial derivatives at the origin:

    \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(h,0)-f(0,0)}h = \lim_{h\to0}\frac{1-0}h = (ouch!)

    I think you probably meant <br />
f(x,y) = \begin{cases}1 &(xy\ne0),\\ 0&(x \text{ or }y =0).\end{cases}<br />
But no, on second thoughts that won't work either, because that function doesn't have partial derivatives on either of the axes (except at the origin). I have a feeling that the example I gave previously can't really be simplified.
    Ouch! You are right. That is the example I meant to give. It does have partial derivatives at (0,0) but is not differentiable there. \frac{\partial f}{\partial x} exists at (x, 0) but not \frac{\partial f}{\partial y} and vice-versa at (0, y).

    That is not a good example because the initial post asked for an example where the partial derivatives exist on some open set. Thank you.
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    Quote Originally Posted by HallsofIvy View Post
    That is not a good example because the initial post asked for an example where the partial derivatives exist on some open set. Thank you.
    I would still like to express my thanks for your consideration of my problem. I benefit a lot from you all.
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