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Where is $f(x,y)=\frac{xy}{x^2-y^2}$ (0 at singularities) differentiable? How about $g(x,y)=\frac{x|y|}{\sqrt{x^2+y_2}}$ (0 at singularity)?

I am not just looking for answers here, but an algorithm that always works. This is kind of a remedial question for me. I can compute the partials, but I don't recall if it suffices to check continuity along the axes, or whether a more general argument is needed. And beyond that, whether there is a better approach altogether, such as the abstract definition of derivative or looking at the directional derivatives.

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$C^1$ implies differentiability as far as I remember. That is, continuity of both partial derivatives suffices to prove that a function is differentiable (existence isn't enough). However, with some functions (like your second one) it can be easier to just check the definition of differentiability than to actually differentiate and check continuity. –  Javier Badia Nov 6 '12 at 2:14

2 Answers 2

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Most of the cooked up examples for such issues as continuity and differentiability in two variables can be understood easily by moving into polar coordinates. I'll demonstrate it on your two examples.

  1. Writing $f$ in polar coordinates, we have $$ f(r, \theta) = \frac{r\cos(\theta)r\sin(\theta)}{r^2\cos^2(\theta) - r^2\sin^2(\theta)} = \frac{1}{2} \tan(2\theta). $$ From here, you immediately see that this function does not depend on $r$, depends on $\theta$, and you expect problems at $\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$. And indeed, your function is not defined at $y = \pm x$, and as you approach on a ray with angle $\theta$, you are constant on that ray, but with a result that depends on $\theta$, so you have multiple limits, and the function has no limit at $(x,y) = (0,0)$.
  2. Writing $g$ in polar coordinates, we have $$ g(r, \theta) = r\cos(\theta)|\sin(\theta)|. $$ This one looks better. It is defined everywhere, as $(x,y) \rightarrow (0,0)$, you must have that $r \rightarrow 0$, and so the function approaches $0$ and hence continuous. Considering the derivative, we see that the absolute value might screw things up when $\sin(\theta) = 0$. That is, at $\theta = 0$, you have two different tangents when you approach from the left or from the right on $|\sin(\theta)|$. This also implies that the problem will be only with $\theta$, not with $r$. And indeed, the partial derivatives $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ exist at $(0,0)$ and are both equal to $0$ but still, the function is not differentiable. To prove that, you know that if the function would be differentiable, the differential would be the zero map. Thus, you would have $$ \lim_{(x,y) \rightarrow 0} \frac{g(x,y) - g(0,0) - 0x - 0y}{\sqrt{x^2 + y^2}} = \lim_{(x,y) \rightarrow 0} \frac{x|y|}{x^2 + y^2} = 0. $$ As this is not the case (again, write it in polar coordinates if you don't see why), $g$ is not differentiable at $(0,0)$.
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Per my calculus book (University Calculus-Early Transcendentals by Hass, Weir, and Thomas):

If a function $f(x,y)$ is differentiable at $(x_0, y_0)$, then $f$ is continuous at $(x_0,y_0)$.

And:

If the partial derivatives $f_x$ and $f_y$ of a function $f(x,y)$ are continuous throughout an open region $R$, then $f$ is differentiable at every point of $R$.

If anyone has the book and wants a page number, see pg 711.

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