Is $f(x, y)$ differentiable at $(0, 0)$? Moreover, is $f(x, y)$ a continuously differentiable function

derivativesreal-analysis

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.

Best Answer

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?

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