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I have the following problem of two variables differentiability:

$f(x,y)=xy\frac{x^2-y^2}{x^2+y^2}$

The value of function at $(0,0)$ is $0$.I have proved that the function is continuous at $(0,0)$. Also, both of the partial derivatives exists and equal to zero. But, I am not able to prove the differentiability of the above Function. I know that the continuous partial derivatives means that the function is differentiable but isn't the existence of partial derivatives proves continuity of partial derivatives. Because, I have obtained the partial derivatives using the limit definition. I think the equality of both partial derivatives means the continuity of them. I am really confused in this problem. Please help and give me set of advice while checking the differentiability of two variables function. Thanks in advance.

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By definition $f(x,y)$ is differentiable in $(a,b)$ if there are $\varepsilon_1$ and $\varepsilon_2$ such that $$f(x,y)-f(a,b)=f_x(a,b)(x-a)+f_y(a,b)(y-b)+\varepsilon_1(x,y)(x-a)+\varepsilon_2(x,y)(y-b)$$ where $\displaystyle\lim_{(x,y)\to(a,b)}\varepsilon_1(x,y)=0$ and $\displaystyle\lim_{(x,y)\to(a,b)}\varepsilon_2(x,y)=0$. For $(a,b)=(0,0)$ the let $\varepsilon_1(x,y)=y\dfrac{x^2}{x^2+y^2}$ and $\varepsilon_2(x,y)=x\dfrac{-y^2}{x^2+y^2}$.

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  • $\begingroup$ Thanks for your answer. my teacher told us to check the continuity of partial derivatives to check the differentiability. What is $\varepsilon(x,y)$ here?. Is it partial derivatives? $\endgroup$ – RAHUl JHa Jan 13 '18 at 10:39
  • $\begingroup$ Have you another formula for differentiability? $\endgroup$ – Nosrati Jan 13 '18 at 10:46
  • $\begingroup$ Is this way of checking differentiability is universal? Can we apply it in most of the situation? $\endgroup$ – RAHUl JHa Jan 13 '18 at 10:47
  • $\begingroup$ Yes. It's universal. Another equivalent definition is $$f(x_0+h,y_0+k)-f(x_0,y_0) = \dfrac{\partial f}{\partial x}(x_0,y_0)h+\dfrac{\partial f}{\partial y}(x_0,y_0)+\epsilon(h,k) \sqrt{h^2+k^2}$$ I think the equality of both partial derivatives means the continuity of them, do you mean $f_x=f_y$ implies continuity of them? $\endgroup$ – Nosrati Jan 13 '18 at 13:40
  • $\begingroup$ there is a factor $k$ missing at $f_y$ in your formula in the comment. $\endgroup$ – Thomas Jan 13 '18 at 13:45
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The easiest way to show differentiability is to use the definition. Denote $v:=(x,y)$ and note that $|x^2 - y^2|\le \min\{x^2, y^2\} \le ||v||^2$ and $||v||=(x^2 + y^2)^\frac{1}{2} \ge \max\{|x|, |y|\}$

Then

$$\left| \frac{f(v)-f(0)}{||v-0||} \right| = \left|\frac{xy(x^2-y^2)}{||v||^{3}}\right| \le \left|\frac{xy||v||^{2}}{||v||^{3}}\right|=\frac{|xy|}{||v||}\le \frac{|xy|}{\max\{|x|,|y|\}} $$

which tends to $0$ if $v$ does .

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