1
$\begingroup$

Prove or disprove: This function has directional derivative in every direction at $(0,0)$.
$f(x,y)=\frac{\arctan(xy)}{\sqrt{x^2+y^2}}$ whenever $(x,y)\ne (0,0)$, and whenever $(x,y)=(0,0)$ it's $0$.

How would you approach a question like this?

My approach:
I know that if a function is differentiable in a point then all of it's directional derivatives exist at any direction at that point. But if it's not differentiable, the directional derivatives can still exist, so this method is good only if the function is differentiable.

Another method I thought of is this (Check by definition):
$\vec n=(cos(t),sin(t))$.
$$\frac{\partial f}{\partial n}(0,0)=\lim_{h \to 0}\frac{f(\cos(t)h,\sin(t)h)-0}{h}=\lim_{h \to 0}\frac{\arctan(\cos(t)\sin(t)h^2)}{h}$$ And currently stuck dealing with this limit, I'm having some troubles dealing with $\arctan$ here (side question: Can I say that $\arctan(x)=\frac{\cos(x)}{\sin(x)}$?).

I would aprreciate any help and feedback to my attempts, thanks in advance!

$\endgroup$
4
  • 1
    $\begingroup$ Regarding the side question: No, $\arctan(x)$ is not at all the same thing as $\cos(x)/\sin(x)$. $\endgroup$ Commented Jul 5, 2021 at 8:48
  • 1
    $\begingroup$ For limits, math.stackexchange.com/questions/1653248/… $\endgroup$
    – Math Lover
    Commented Jul 8, 2021 at 9:53
  • 1
    $\begingroup$ Yes by definition, directional derivative exists in every direction at $(0, 0)$. $\endgroup$
    – Math Lover
    Commented Jul 8, 2021 at 9:58
  • $\begingroup$ it should be $h^2$ in the denominator $\endgroup$
    – AlvinL
    Commented May 28, 2022 at 11:37

2 Answers 2

3
$\begingroup$

By the mean value theorem we see that

$$| \arctan t| \le |t|$$

for all $t$. Hence

$$ |\frac{\arctan(\cos(t)\sin(t)h^2)}{h}| \le |h|.$$

Conclusion ?

$\endgroup$
1
  • $\begingroup$ The conclusion should be that the directional derivative exists at any direction in $(0,0)$ since this limit exists, but what's weird is that the final answer is false..? is there a problem in the question or am i missing something? $\endgroup$
    – Pwaol
    Commented Jul 5, 2021 at 8:26
2
$\begingroup$

First, $f$ is $\mathcal C^\infty$ away from $(0,0)$, so it has all its directional derivatives at those points.

For $(x,y)\in\mathbb R^2$, we can compute the directional derivative : $$D_{(x,y)}f(0,0) = \lim_{h\to 0} \frac{f(hx,hy)}h = \lim_{h\to 0}\frac{\arctan(h^2xy)}{h^2\sqrt{x^2+y^2}} = \frac{xy}{\sqrt{x^2+y^2}} $$

since $\arctan(h^2xy) = h^2 xy + o(h^2)$.

$\endgroup$
2
  • $\begingroup$ Thanks for the answer, could you please emphasize on the notation you used $f$ is $C^{\infty}$, and the last line $\arctan(h^2xy)=h^2xy + o(h^2)$. (I guess that $o(h^2)$ means a component that is much less than $h^2$). But how did u find it out? $\endgroup$
    – Pwaol
    Commented Jul 5, 2021 at 8:20
  • $\begingroup$ $\mathcal C^\infty$ means that $f$ is differentiable, its differential is also differentiable and so on. $o(h^2)$ means some function that can be written $\epsilon(h)h^2$ with $ \epsilon(h)\to 0$. See Taylor's theorem for the statement about $\arctan$ $\endgroup$ Commented Jul 5, 2021 at 8:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .