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I am trying to prove the existence of directional derivatives of the function

$$f: \mathbb R^2 \rightarrow \mathbb R $$ $$ f(x,y) = \begin{cases} \dfrac {2xy^2}{x^2+y^4} & (x,y)\ne 0 \\\\ 0 & (x,y)=0 ~ \end{cases} $$

The directional derivative $D_u f(0,0)$ of $f$ at $(0,0)$ in the direction $u = (u_1,u_2) \ne (0,0).$ then $f$ is defined as

$$D_u f(0,0)=\lim_{t \to 0}\frac{f(0+tu)-f(0)}{t}$$

which is

$$D_u f(0,0)=\lim_{t \to 0} \frac{2(tu_1)(tu_2)^2}{(tu_1)^2+(tu_2)^4} \frac{1}{t}=\lim_{t \to 0} \frac{2t^3u_1u_2^2}{t^3(u_1^2+t^2u_2^4)}=\lim_{t \to 0} \frac{2u_1u_2^2}{(u_1^2+t^2u_2^4)}=\frac{2u_1u_2^2}{u_1^2} $$

which is obviously not defined for any $u\neq 0$ with $u_1=0$.

However, if we look at the values of $f$ along the $y$-axis, its obvious that its $0$, as suggested in this thread, which would mean that the directional derivatives exist. Why can't I deduce that from the definition of directional derivatives? How do I justify treating the cases separately?

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It is only your last step that fails when $u_1 = 0$. Take a hard look at $\lim_{t \to 0} \frac{2u_1u_2^2}{(u_1^2+t^2u_2^4)}$ and see what happens there when $u_1 = 0$ (and then, by assumption $u_2\neq 0$, and by definition of limit $t\neq 0$). In the other case, when $u_1\neq 0$, you can continue with the argument you had.

As for how you can justify treating the cases separately, I know there are some who think that a proof by cases is a less worthy proof, morally. But logically, there is no issue, and has never been an issue, with doing a proof by cases (at least in standard logic). If it really bothers you, you can split the proof the moment you're given a $u$, and do the two cases entirely separately.

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  • $\begingroup$ What was bothering me about the cases was one implying that the inexistence while the other proved its existence. But you cleared that up, thank you. $\endgroup$ – B.Swan Jun 13 '17 at 12:57
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Your attempt seems to be fine. Let $u = (u_1, u_2) \in \mathbb R^2\backslash\{0\}$. If $u_1 \neq 0$, then $$ \mathrm D_u f(0,0) = \frac{2 u_1 u_2^2}{u_1^2} \; , $$ like you have shown above.

If $u_1 = 0$, then $$ \frac{f(0+tu) - f(0)}{t} = 0 \quad \text{for all } t \neq 0 \; , $$ so $$ \lim_{t \to 0} \frac{f(0+tu) - f(0)}{t} = 0 \; , $$ i.e. $\mathrm D_u f(0,0) = 0$.

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