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Question: Let $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be defined by $f(0,0)=0$ and for $(x,y)\neq(0,0)$:

$f(x,y)=\frac{xy}{\sqrt{x^2+y^2}}$.

Show that the two-sided directional derivative of $f$ evaluated at $(x,y)=(0,0)$ exists in all directions $h\in\mathbb{R}^{2}$, but that $f$ is not differentiable at $(0,0)$.

I am unsure how to solve this problem. I believe I need to set up a limit as $h\to0$, but behind that I am stumped.

Paul Frost
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  • Do you know the definition of directional derivatives? Do you know the definition of differentiability of a multivariable function? See this related question for an example: https://math.stackexchange.com/questions/372070/f-not-differentiable-at-0-0-but-all-directional-derivatives-exist?rq=1 – Daniel Kawai May 29 '20 at 05:03
  • If you learn the theorem stated in https://math.stackexchange.com/a/3339076/568204 then this question becomes almost trivial. – peek-a-boo May 29 '20 at 05:32
  • $f$ is discontinuous at $(0,0)$ no matter what value you assign to $f(0,0) . $ – DanielWainfleet May 29 '20 at 06:44
  • So, I checked the other question out, but it is still confusing. I kinda get that i need to use a limit, but overall I have no idea what vector coordinates are needed, if they are even needed at all. More over, the frechet derivative is not something I learned. –  May 29 '20 at 08:11
  • You asked the same question three hours earlier at Question on two sided directional derivatives. Please don't do that; it wastes the site's resources. Also, please don't use generic terms like "question" in the title. All posts on this site are questions – imagine how the main page would look and how inefficient it would be if everyone did that. – joriki May 29 '20 at 08:15
  • I don’t know exactly what the definition of the directional derivative is, let alone a two sided. Do I maybe need to develop a hessian matrix? –  May 29 '20 at 08:16
  • Pardon me @joriki, I took down the other post in order to avoid further issues. I am still new to this site, so I will take note for the future. –  May 29 '20 at 08:18
  • I shall do so posthaste –  May 29 '20 at 08:19
  • @joriki are you able to assist me further with this problem? –  May 29 '20 at 08:20
  • @Tony456, the directional derivative of $f$ in $x\in\mathbb{R}^2$ with respect to a direction $h\in\mathbb{R}^2$ is the limit $\lim\limits_{t\to 0}\frac{f(x+th)-f(x)}{t}$. In other words, it is the derivative of the function $f(x+th)$ at $t=0$. – Daniel Kawai May 30 '20 at 03:31
  • @Tony456, in the other hand we say $f$ is differentiable at $x\in\mathbb{R}^2$ iff there is a linear transformation $T:\mathbb{R}^2\rightarrow\mathbb{R}$ such that $\lim\limits_{v\to 0}\frac{f(x+v)-f(x)-L(v)}{|v|}=0$. – Daniel Kawai May 30 '20 at 03:36
  • With these definitions in mind, can you understand now the solution of the related question? – Daniel Kawai May 30 '20 at 03:37
  • I think I do understand it. –  May 30 '20 at 20:39
  • Thank you very much. –  May 30 '20 at 20:39

1 Answers1

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Use polar coordinates to write the function as

$$ f(r,\phi)=r\cos\phi\sin\phi=\frac12r\sin2\phi\;. $$

Since $\sin(2(\phi+\pi))=\sin2\phi$, the angular factor is the same for opposite directions, so along any line through the origin the function behaves like an absolute value; e.g., on $(x,y)=(\lambda,\lambda)$ it takes the form $\frac1{\sqrt2}|\lambda|$. The absolute value function is not differentiable at $0$, so the two-sided directional derivatives don’t exist.

joriki
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  • Is it still possible to solve this via a limit? I kinda wanna see how one would go about doing that. –  May 29 '20 at 23:32
  • @Tony456: Perhaps you could first say something about the discrepancy with the problem statement? It asks to prove that the two-sided directional derivatives exist, but it seems that they don't (unless I've made a mistake). What's the source of the problem? – joriki May 30 '20 at 00:30
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    The source of the problem is from a math textbook published by a Dr. Sundraman. Let me check the problem from the textbook and I will get back to you. –  May 30 '20 at 02:55