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August 5th, 2017, 06:29 AM   #1
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A question about a sufficient condition of differentiability of a multivariable func

I see many times a sufficient condition of differentiability of a multivariable function:

Quote:
 A function $\displaystyle f(x,y)$ is differentiable at $\displaystyle (x_0,y_0)$ if both of its partial derivatives are continuous at that point.
But take the following function for example
$\displaystyle f(x,y) = \left\{ \begin{array}{*{20}{c,l}} \frac{x^2y}{x^2 + y^2}&{\rm{, if}} (x,y) \ne (0,0)\\ 0 & {\rm{, if}} (x,y) = (0,0) \end{array} \right.$.

It has constant value 0 along x- and y-axis, thus constant and therefore continuous partial derivative 0 along x- and y-axis, but it is not differentiable at (0,0) because the limit
$\displaystyle \lim \limits_{\Delta x \to 0,\Delta y \to 0} \frac{[f(\Delta x,\Delta y) - f(0,0)]-[f_x'(0,0)\Delta x + f_y'(0,0)\Delta y]}{\rho}$
where $\displaystyle \rho=\sqrt{\Delta x^2+\Delta y^2}$ does not exist (approaching along lines of different slopes).

So where is the problem? Is the sufficient condition mentioned above wrong or did I miss anything when applying it? Thank you.

August 5th, 2017, 10:11 AM   #2
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I don't know where you saw your quote:
Quote:
 A function f(x,y) is differentiable at $\displaystyle (x_0, y_0)$ if both of its partial derivatives are continuous at that point.
"many times", but it is incorrect.

A function f(x,y) is differentiable at $\displaystyle (x_0, y_0)$ if both its partial derivatives are continuous in some neighborhood of that point.

Last edited by skipjack; August 5th, 2017 at 05:00 PM.

 August 5th, 2017, 11:34 AM #3 Newbie   Joined: Jun 2017 From: Earth Posts: 16 Thanks: 0 What?
August 6th, 2017, 03:19 AM   #4
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Quote:
 Originally Posted by Country Boy I don't know where you saw your quote: "many times", but it is incorrect. A function f(x,y) is differentiable at $\displaystyle (x_0, y_0)$ if both its partial derivatives are continuous in some neighborhood of that point.
@Country Boy, I have no idea what you want to say, but I cannot help but think your math knowledge is deficient according to your answers to my questions. I'm further surprised at it after I noticed that your title is "Math Team". As for the correctness of the quoted proposition, a simple search found me the following paper (link). I excerpt it below in case you can't access it (note that the condition therein is even weaker):

August 6th, 2017, 03:35 AM   #5
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Quote:
 Originally Posted by zzzhhh I see many times a sufficient condition of differentiability of a multivariable function: But take the following function for example $\displaystyle f(x,y) = \left\{ \begin{array}{*{20}{c,l}} \frac{x^2y}{x^2 + y^2}&{\rm{, if}} (x,y) \ne (0,0)\\ 0 & {\rm{, if}} (x,y) = (0,0) \end{array} \right.$. It has constant value 0 along x- and y-axis, thus constant and therefore continuous partial derivative 0 along x- and y-axis, but it is not differentiable at (0,0) because the limit $\displaystyle \lim \limits_{\Delta x \to 0,\Delta y \to 0} \frac{[f(\Delta x,\Delta y) - f(0,0)]-[f_x'(0,0)\Delta x + f_y'(0,0)\Delta y]}{\rho}$ where $\displaystyle \rho=\sqrt{\Delta x^2+\Delta y^2}$ does not exist (approaching along lines of different slopes). So where is the problem? Is the sufficient condition mentioned above wrong or did I miss anything when applying it? Thank you.
Yes, the partial derivative is 0 at (0,0). But is it continuous there? In order to check continuity at (0,0), you need to be able to approach it from all directions, not just the coordinate axes. Yes, in order to compute the partial derivatives, you only need to coordinate axes. But in order to check continuity, you'll need to check it from all directions.

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