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How can I show that this problem isn't differentiable at $(0,0)$? at first, I think that this function isn't continuous at $(0,0)$ that implies it's not differentiable. Am I right? Is there any other solution to prove whether this function is differentiable? Thanks. $$f(x,y) = \begin{cases} 2xy\frac{x^2-y^2}{x^2+y^2}, & \mbox{if } x^2+y^2 \neq 0 \\ 0, & \mbox{otherwise}\end{cases}$$

Botond
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zeinab
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2 Answers2

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The OP has a duplicate here Differentiability at $(0,0)$.

However, I think it could be useful make a general comment here on the topic of Differentiability and Continuity for functions of several variables. Any other contribution or observation will be appreciate.

1. Firstly you might check whether or not f is continuos in (0,0) by calculating the limit as $(x,y)\to(0,0)$. If it is discontinuos then it can't be differentiable, infact:

NOTE

continuity is a necessary condition since differentiability implies continuity

2. Then you have to check whether or not partial derivatives in (0,0) exist by calculating them as limits of the incremental ratio; if the partial derivatives do not exist then $f$ can't be differentiable, infact:

NOTE

existence of partial derivatives is a necessary condition since differentiability implies their existence

Even is $f$ is continuos in $(0,0$ and partial derivatives exist in $(0,0)$ you can't conclude anything yet about differentiability at $(0,0)$.

3. To be sure you need to check that partial derivatives are continuos in $(0,0)$ by calculating $f_x$ and $f_y$ for $(x,y)\neq(0,0)$ and the calculationg their limit at (0,0). If their are continuos you are done, infact:

NOTE "Differentiability theorem"

if all the partial derivatives exist and are continuous in a neighborhood of (0,0) then (i.e. sufficient condition) the function is differentiable at $(0,0)$

4. If partial derivatives are not continuos at $(0,0)$ you can't yet conclude anything about differentiability. You need to check directly differentiability by definition that:

$$\lim_{(h,k)\rightarrow (0,0)} \frac{\| f(h,k)-f(0,0)-(f_x(0,0),f_y(0,0))\cdot (h,k)\|}{\| (h,k)\|}=0$$

NOTE

In this case you can also skip "3" and try directly with "2" and"4" to check differentiability.

user
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  • Ok. I show that this function is continuous at (0,0) ( I set y=mx) . – zeinab Dec 12 '17 at 07:00
  • @zeinab to show that its continuos you should use polar coordinates, works on line it's not a correct manner to calulate limits for several variables – user Dec 12 '17 at 07:04
  • So, it'll need more time to show that this function is differentiable at (0,0). This's a difficult question. – zeinab Dec 12 '17 at 07:50
  • For functions as this, all the difficulties are in the evaluation of the limits which are not simple for functions of several variables. In all the others point you need only to verify that partial derivatives exist by simple derivation ( it’s a sufficient condiction). – user Dec 12 '17 at 07:55
  • yes, You are right. Thanks a lot for your helping. – zeinab Dec 12 '17 at 08:09
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From $f(x,0)=f(0,y)\equiv0$ it follows that $f_x(0,0)=f_y(0,0)=0$. This implies that $df(0,0)=0$ if $f$ is at all differentiable at $(0,0)$. We therefore have to check whether $$\lim_{(x,y)\to(0,0)}{f(x,y)-f(0,0)-0\over\sqrt{x^2+y^2}}=0\ .$$ To this end we express $f$ in polar coordinates $r$, $\phi$ and obtain $${f(x,y)-f(0,0)-0\over\sqrt{x^2+y^2}}={\hat f(r,\phi)\over r}={1\over2} r\sin(4\phi)\to0\qquad(r\to0+)\ .$$ This shows that in fact $df(0,0)=0$.