Directional derivative exists does not imply total differentiable

Question

Let $f:\mathbb{R}^2\to\mathbb{R}$ be given by $f(x,y)=\begin{cases} \dfrac{x^3}{x^2+y^2},\text{if $(x,y)\neq (0,0)$}\\ 0,\text{if $(x,y)=(0,0)$}\end{cases}$

Show that for every $v=(x,y)\in\mathbb{R}^2$ the directional derivative $D_vf(0,0)$ of $f$ exists in the point $(0,0)$, but $f$ is not totally differentiable.

So, I calculate the directional derivative easily by:

$D_vf(x,y)=\lim_{t\to 0\\ t\neq 0}\dfrac{1}{t}\left(f(tx,ty)-f(0,0)\right)$

$=\lim_{t\to 0\\ t\neq 0}\dfrac{1}{t}\left(\dfrac{t^3x^3}{t^2x^2+t^2y^2}\right)=\dfrac{x^3}{x^2+y^2}$ as $t$ just cancels out.

But how do I show that $f$ is not totally differentiable?

I want to test for total differentiability in $(0,0)$.

When I calculate the derivative we get:

$\dfrac{3x^2(x^2+y^2)-2x^4}{(x^2+y^2)^2}-\dfrac{2yx^3}{(x^2+y^2)^2}=\dfrac{3x^2(x^2+y^2)-2x^4-2yx^3}{(x^2+y^2)^2}$

Plugging everything in the formula:

$\lim_{(x,y)\to (0,0)\\ (x,y)\neq (0,0)}\dfrac{1}{\sqrt{x^2+y^2}}\left(\dfrac{x^3}{x^2+y^2}-\dfrac{3x^2(x^2+y^2)-2x^4-2yx^3}{(x^2+y^2)^2}\right)$

Now I use polar coordinates: $x=r\sin(\varphi)$ and $y=r\cos(\varphi)$, and get:

$\lim_{r\to 0} \sin^3(\varphi)-3\dfrac{\sin^2(\varphi)}{r^3}-\dfrac{\sin^4(\varphi)-2\cos(\varphi)\sin(\varphi)}{r}$

But this limit does not exist, and in particular is not $0$.

So $f$ would not be totally differentiable in $(0,0)$.

Is this correct?

Thanks in advance.

Answer 1

If $f$ was differentiable at $(0,0)$, then, for each $v\in\Bbb R^2$, $f'(0,0)(v)=D_vf(0,0)$. But that's impossible, since the map $v\mapsto D_vf(0,0)$ is simply $f$, which is not linear.

Answer 2

For the differential, you should compute $$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-xD_1f(0,0)-yD_2f(0,0)}{\sqrt{x^2+y^2}}$$ where $D_1$ and $D_2$ are partial derivatives.

According to your calculations, $D_1f(0,0)=1$ and $D_2f(0,0)=0$, so $f$ would be differentiable at the origin iff $$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-x}{\sqrt{x^2+y^2}}=0.$$

But, in this case, $$\lim_{(x,y)\to(0,0)}\frac{f(x,y)-f(0,0)-x}{\sqrt{x^2+y^2}}=\lim_{(x,y)\to(0,0)}\frac{\frac{x^3}{x^2+y^2}-x}{\sqrt{x^2+y^2}}=\lim_{(x,y)\to(0,0)}\frac{-xy^2}{(x^2+y^2)^{3/2}}$$ so taking the directional limit when $y=x$, you'll get that the limit cannot be 0.