0
$\begingroup$

I always get stuck when I've to show something is differentiable ,like in the following question:

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

show that $f$ is differentiable at $(0,0)$ ...

alright both partial derivatives exists and are equal,so now we have to show that :they are continuous near $(0,0)$..

$\endgroup$
  • $\begingroup$ This question looks a lot like math.stackexchange.com/questions/976998/… (except for the factor $\frac{-xy^3}{x^2+y^2}$). Is it related? $\endgroup$ – konewka Oct 16 '14 at 17:53
  • 1
    $\begingroup$ @konewka CHECK THE LINK ,it doesn't show anyother question.. $\endgroup$ – spectraa Oct 16 '14 at 17:59
3
$\begingroup$

First compute the partials derivatives, which are easily seen to be $\partial f_x(0,0)=0$ and $\partial f_y(0,0)=0$. In order to show that $f$ is differentiable we have to show that
$$\lim_{(h,k)\rightarrow (0,0)} \frac{\| f(h,k)-f(0,0)-(0,0)\cdot (h,k)\|}{\| (h,k)\|}=\lim_{(h,k)\rightarrow (0,0)}hk\frac{h^2-k^2}{(h^2+k^2)^{3/2}}=0$$

To compute the limit we can change to polar coordinates. Now, the equation is

$$\lim_{r \rightarrow 0}r^2\sin\theta\cos\theta\frac{r^2cos^2\theta-r^2\sin^2\theta}{(r^2)^{3/2}}=\lim_{r \rightarrow 0}r(\sin\theta\cos^3\theta-\sin^3\theta\cos\theta)=0 $$

where the last equality holds since $\sin \theta $ and $\cos\theta$ are bounded. Thus, $f$ is differentiable at $(0,0)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ how did you write directly $(0,0).(h,k)$ in your first step of computation of limit ..i.e. the value of partial derivatives are $0,0$ only at point $(0,0)$ ...I can't understand ... $\endgroup$ – spectraa Oct 17 '14 at 7:02
  • $\begingroup$ @spectraa The derivative is a linear function whose components are given by the partial derivatives. $\endgroup$ – azarel Oct 17 '14 at 11:11
  • $\begingroup$ but why did you compute components at a point... $\endgroup$ – spectraa Oct 17 '14 at 12:09
0
$\begingroup$

we have $\frac{\partial f}{\partial x}(0,0)=\lim_{t \to 0}\frac{f(t,0)-f(0,0)}{t}=\lim_{t->0}\frac{0}{t}=0$
$\frac{\partial f}{\partial y}(0,0)=\lim_{t \to 0}\frac{f(0,t)-f(0,0)}{t}=0$ further we have $\lim_{(x,y)\to(0,0)}\frac{xy(x^2-y^2)}{(x^2+y^2)\sqrt{x^2+y^2}}=0$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ You seem to be hinting that from the existence of both partial derivatives at the origin it follows that the functions is differentiable there. This is false. $\endgroup$ – Timbuc Oct 16 '14 at 18:04
  • $\begingroup$ @Timbuc If we show that these derivatives are continuous we can get differentiability.. $\endgroup$ – spectraa Oct 16 '14 at 18:07
  • 1
    $\begingroup$ That I know, @spectraa ...and the answer doesn't show that $\endgroup$ – Timbuc Oct 16 '14 at 18:10
  • $\begingroup$ With the editing of the answer it seems now like it is going for the straight definition of differentiability. $\endgroup$ – Timbuc Oct 16 '14 at 18:15
0
$\begingroup$

We have that

$$f(x,y)=xy\left(1-\frac{2y^2}{x^2+y^2}\right)=xy-\frac{2xy^3}{x^2+y^2}\;,\;\;\text{so:}$$

$$\begin{align*}\bullet&\;\;f_x=y-\frac{2y^3(x^2+y^2)-4x^2y^3}{(x^2+y^2)^2}=y\left(1-2\frac{y^4-x^2y^2}{(x^2+y^2)^2}\right)\\{}\\ \bullet&\;\;f_y=x-\frac{6xy^2(x^2+y^2)-4xy^4}{(x^2+y^2)^2}=x\left(1-2\frac{y^4+3x^2y^2}{(x^2+y^2)^2}\right)\end{align*}$$

The partial derivatives are clearly defined and continuous at $\;(x,y)\neq (0,0)\;$ , and as for the origin you can show, using polar coordinates say, these derivatives are continuous also at the origin and thus the function differentiable there. For example:

$$\left|1-2\frac{y^4-x^2y^2}{(x^2+y^2)^2}\right|\stackrel{\text{pol. coor.}}\longrightarrow \left|1+2\sin^2t\cos2t\right|\le3$$

and thus we get $\;f_x\xrightarrow[(x,y)\to(0,0)]{}0\;$ , and likewise with $\;f_y\;$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.