# Differentiability at $(0,0)$.

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 $(0,0)$..

• 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? – konewka Oct 16 '14 at 17:53
• @konewka CHECK THE LINK ,it doesn't show anyother question.. – spectraa Oct 16 '14 at 17:59

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 $(0,0)$.

• 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 ... – spectraa Oct 17 '14 at 7:02
• @spectraa The derivative is a linear function whose components are given by the partial derivatives. – azarel Oct 17 '14 at 11:11
• but why did you compute components at a point... – spectraa Oct 17 '14 at 12:09

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 $\lim_{(x,y)\to(0,0)}\frac{xy(x^2-y^2)}{(x^2+y^2)\sqrt{x^2+y^2}}=0$

• 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. – Timbuc Oct 16 '14 at 18:04
• @Timbuc If we show that these derivatives are continuous we can get differentiability.. – spectraa Oct 16 '14 at 18:07
• That I know, @spectraa ...and the answer doesn't show that – Timbuc Oct 16 '14 at 18:10
• With the editing of the answer it seems now like it is going for the straight definition of differentiability. – Timbuc Oct 16 '14 at 18:15

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 $\;f_y\;$