In my book, I found a problem, asked to check the differentiability of the function $f(x,y)$ at $(0,0)$ where the function f$(x,y)$ is defined as follows

$$ f(x,y) = \begin{cases} \large\frac{xy}{\sqrt {x^2+y^2}}, & \text{if $x^2+y^2 \neq (0,0)$} \\ 0, & \text{if $x^2+y^2 = (0,0)$} \end{cases} $$

Now in the book, they give a solution using alternative definition of differentiability at a point $(0,0)$ and there they have approached to a contradiction that $\frac{1}{2} = 0$

So, they proved that the function is not differentiable at all.

But when I tried to solve the problem by proving that the $f_x$ and $f_y$ exist at $(0,0)$ and the function is continuous at that point too. So basically, I prove that the given function is differentiable at that very point.

But the given solution in my book makes me confuse. I don't understand what actually happen, Did I do something wrong if I did please make me clear? Because at the same time these two solution is not possible for this function.

And another thing can I use "sufficient condition for differentiability" for this function?

Pardon me if I did any wrong.

Thank you.

  • $\begingroup$ If $x^2+y^2 \neq 0$ at a point $(x,y)$ then I think $f$ is differentiable at that point by using usual methods of finding derivatives. Is your question specifically about whether it's differentiable at $(0,0)$? $\endgroup$ – coffeemath Apr 21 '17 at 10:53
  • $\begingroup$ In order for a function to be continuous at a point $a$ it's limit must exist and be EQUAL to it at $x=a$. Thus it is not enough for it's limit to exist, as that explicitly does not require it be defined at the point in question. So $f(x,y)$ is NOT continuous at $(0,0)$. $\endgroup$ – marshal craft Apr 21 '17 at 10:54
  • $\begingroup$ Oh sorry, didn't notice your $=0$ just sitting there. Thought it was a typo but you are defining it or something? $\endgroup$ – marshal craft Apr 21 '17 at 11:07
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    $\begingroup$ yes @coffeemath $\endgroup$ – Austin20 Apr 21 '17 at 11:16
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    $\begingroup$ It doesn't apply, since the first partials aren't continuous at the origin. (They can't be, since that would imply that $f$ is differentiable at the origin, which it isn't.) $\endgroup$ – Hans Lundmark Apr 21 '17 at 11:33

Let me try to show this using the very definition of differentiability. Suppose the function $f$ is differentiable at $(0,0)$. Then there is a matrix $D = (a,b)$ (the derivative of $f$ at zero) such that $$ 0 = \lim_{(x,y)\to (0,0)} \frac{|f(x,y) - f(0,0) - D \cdot (x,y)^T|}{\|(x,y) - (0,0)\|} = \lim_{(x,y)\to(0,0)} \frac{1}{\sqrt{x^2 + y^2}} \left| \frac{xy}{\sqrt{x^2+y^2}} -ax -by \right|. $$ Let us take two particular ways how to approach zero, say $(x,y) = (h,h)$ as $h \to 0_\pm$ (left/right). Then we have $$ 0 = \lim_{h \to 0} \frac{1}{\sqrt{2h^2}} \left| \frac{|h|}{\sqrt{2}} - (a+b)h \right| = \frac{1}{\sqrt{2}} \lim_{h\to 0} \left| \frac{1}{\sqrt{2}} - (a+b)\mathrm{sign}(h) \right| = \frac{1}{\sqrt{2}} \left| \frac{1}{\sqrt{2}} \mp (a+b) \right|. $$ So, equalities $a+b = \frac{1}{\sqrt{2}}$ and $a+b = - \frac{1}{\sqrt{2}}$ would have to hold at the same time. This is clearly a contradiction.

  • $\begingroup$ +1, better than my previous (erroneous) "answer". $\endgroup$ – coffeemath Apr 21 '17 at 13:54

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