Proof: All directional derivatives $\frac{\partial f}{\partial e}$ of $\frac{sin(x^3+y^3)}{x^2+y^2}$ are in the origin

Question

Let $M := (0,\infty) \subset \mathbb{R^2}$ and $f:\mathbb{R}^2 \to \mathbb{R}$.

How can one prove that all directional derivatives $\frac{\partial f}{\partial e}$ of $f(x,y)$ are existing in the origin and calculate them?

Is the following correct?

Let $(a_n)_{n\in\mathbb{R}}$ be a sequence with $a_n = (\frac{1}{n},\frac{1}{n})$.

If I use $a_n$ in the function $f(x,y)$, I get

$$\frac{sin(\frac{1}{n}+\frac{1}{n})^3}{(\frac{1}{n})^2+(\frac{1}{n})^2} $$

Then $\lim n \to \infty = 0$

Therefore all directional derivatives exist in the origin.

I know that

$$\frac{\partial f}{\partial x} = \dfrac{3x^2\cos\left(x^3+y^3\right)}{x^2+y^2}-\dfrac{2x\sin\left(x^3+y^3\right)}{\left(x^2+y^2\right)^2}$$

and

$$\frac{\partial f}{\partial y} = \dfrac{3y^2\cos\left(y^3+x^3\right)}{y^2+x^2}-\dfrac{2y\sin\left(y^3+x^3\right)}{\left(y^2+x^2\right)^2}$$

By using the sequence $a_n$ we get $\lim n \to \infty = 0$

So $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exist in the point $(0,0)$

Is that correct? I didn't use $\frac{\partial f}{\partial e}$

Answer 1

No, the reasoning is not correct. Your discussion merely partially shows that when $\boldsymbol e = (1/\sqrt 2, 1/\sqrt 2)$, $\partial f/\partial \boldsymbol e$ might exist.

Just apply the definition: the directional derivative of $f$ along the direction $\boldsymbol e$ at $\boldsymbol x \in \mathbb R^2$ is $$ \lim_{t\to 0} \frac {f(\boldsymbol x + t\boldsymbol e) - f(\boldsymbol x)}t. $$ Given any unit vector $\boldsymbol e =( \cos \theta, \sin \theta)$, the limit at $\boldsymbol x = (0,0)$ is [WRONG EXPONENT HERE] $$ \lim_{t\to 0}\frac {\sin(t^3(\cos^3\theta + \sin^3\theta))}{t^2 (\sin^2\theta + \cos^2\theta)} = \lim_{t\to 0}\frac {\sin(t^3(\cos^3\theta + \sin^3\theta))}{t^2} $$ use $|\sin x | \leqslant |x|$ when $|x| < 1$, we have $$ \left\vert \frac {\sin(t^3(\cos^3\theta + \sin^3\theta))}{t^2} \right\vert \leqslant |t| |\cos^3\theta +\sin^3\theta| \leqslant 2|t| \xrightarrow{t \to 0} 0, $$ so the limit exists for all direction $\boldsymbol e$ by the squeeze theorem, and $(\partial f/\partial \boldsymbol e) (0,0) = 0$.

EDIT

The limit should be $$ \lim_{t\to 0}\frac {\sin(t^3(\cos^3\theta + \sin^3\theta))}{t^{\color{red}3} (\sin^2\theta + \cos^2\theta)} = \lim_{t\to 0}\frac {\sin(t^3(\cos^3\theta + \sin^3\theta))}{t^{\color{red}{3}}}, $$ Now use $\sin (x) \sim x[x \to 0]$, we conclude that $$ \frac {\partial f}{\partial \boldsymbol e} = \sin^3\theta + \cos^3\theta, $$ not $0$ as above.

Answer 2

It is not correct. We have $f(a_n) \ne \frac{sin(\frac{1}{n}+\frac{1}{n})^3}{(\frac{1}{n})^2+(\frac{1}{n})^2}$ !

With the definition: $\frac{\partial f}{\partial e}(0,0)= \lim_{t \to 0}\frac{f(te)-f(0,0)}{t}$, if the limit exists.

Let $e=(u,v)$ be a direction. WLOG we can assume that $u^2+v^2=1$. Then we have

$\frac{f(te)-f(0,0)}{t}=\frac{\sin(t^3(u^3+v^3)}{t^3}$.

If $u^3+v^3=0$, then $\frac{\partial f}{\partial e}(0,0)=0$.

Now assume that $u^3+v^3 \ne 0$. It follows that

$\frac{f(te)-f(0,0)}{t}=\frac{\sin(t^3(u^3+v^3)}{t^3}=(u^3+v^3)\frac{\sin(t^3(u^3+v^3)}{t^3(u^3+v^3)} \to u^3+v^3$ as $t \to 0$, since $\frac{\sin x}{x} \to 1$ as $x \to 0$.

Conclusion:

$\frac{\partial f}{\partial e}(0,0)=u^3+v^3$.