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Consider the following function: \begin{equation*} f(x, y) = \begin{cases} \frac{x^{2}\sin(y^2)}{x^{2} + y^{4}} & (x, y) \neq (0, 0) \\ 0 & (x, y) = (0, 0) \end{cases} \end{equation*}

  1. Regarding differentiability at $(0, 0)$: I have shown that the directional derivatives $(D_{u}f)(0, 0)$ exist and equal zero for all $u = (u_{1}, u_{2}) \in \mathbb{R}^{2} \setminus \{(0,0)\}$ where $u_{1} \neq 0$. From this I argued that the function is not differentiable. My reasoning for this is that differentiability implies the existence of all directional derivatives, and so the contrapositive would hold, but I am not sure of this argument. Is this correct?

  2. Regarding continuous differentiability: I have tried showing continuity of the partial derivatives of $f$ but to no avail. I was wondering if there is another way to determine whether $f$ is continuously differentiable. Many thanks in advance.

Math Attack
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JLGL
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  • The directional derivatives existing and equaling zero does not imply that the function is not differentiable. If it were not differentiable, it cannot be continuously differentiable (as continuously differentiable means differentiable with continuous derivative). – ViktorStein Apr 25 '23 at 19:03

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For assessing differentiability you only need the parcial derivatives... $$ f'_x(0,0) = \lim_{h \to 0} \dfrac{f(h,0)-f(0,0)}{h} = \lim_{h\to 0}\dfrac{0-0}{h}=0 $$ $$ f'_y(0,0) = \lim_{h \to 0} \dfrac{f(0,h)-f(0,0)}{h} = \lim_{h\to 0}\dfrac{0-0}{h}=0 $$

The function is differentiable at $(0,0)$ if $$ \lim_{(h_1,h_2)\to (0,0)} \dfrac{\dfrac{h_1^2 \sin(h_2^2)}{h_1^2+h_2^4}-0\cdot h_1-0\cdot h_2}{\sqrt{h_1^2+h_2^2}} = 0 $$

now, since

$$ \left|\dfrac{h_1^2 \sin(h_2^2)}{(h_1^2+h_2^4)\sqrt{h_1^2+h_2^2}} -0\right|\leq\dfrac{|\sin(h_2^2)|}{\sqrt{h_1^2+h_2^2}}\leq \dfrac{h_2^2}{\sqrt{h_1^2+h_2^2}}\leq \sqrt{h_1^2+h_2^2} \to 0 $$

we do conclude that $f$ is differentiable at $(0,0)$.

Regarding continuous differentiability, did you try to write down the partial derivatives?

Robert Z
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PierreCarre
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  • Yes I have had a look at the partial derivatives, for example I found one to be \begin{align} & \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(\frac{x^{2}\sin{y^{2}}}{x^{2} + y^{4}}\right) \ &= \frac{2xy^{4}\sin{y^{2}}}{(x^{2} + y^{4})^{2}} \end{align}. In order to show continuity of this partial derivative I found it was bounded by $sin(y^{2})$, and concluded that if $(x, y) \rightarrow 0$ then the $f_{x}(x,y) \rightarrow 0$ as well. Is this correct? – JLGL Apr 25 '23 at 19:18
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    @JLGL Your conclusion is correct, you just need to see if you got this (correct) bound in a correct way. – PierreCarre Apr 25 '23 at 20:59
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    @PierreCarre The bound is incorrect. For $x=y^2 $ and $y\to 0$ the expression tends to ${1\over 2}.$ By the way if the partial derivatives were continuous, there would be no need for proving differentiability. So I prefer to inspect continuity of the partial derivatives before undertaking differentiability. – Ryszard Szwarc Apr 25 '23 at 23:26
  • @RyszardSzwarc You are right about the bound. But I think that if you are not required to check for continuous differentiabilitity it is not very efficient to inspect the continuity of partial derivatives since if they are not continuous you still need to check differentiability in the standard way. – PierreCarre Apr 26 '23 at 13:01
  • I had in mind, if we are required to do both tasks, I would start with the second one. If we are after differentiability only, it may be simpler to check that directly, especially if we doubt that partial derivatives are continuous. – Ryszard Szwarc Apr 26 '23 at 13:24