Existence of the directional derivative

Question

Let $f:\Bbb{R^2} \to \Bbb{R}$ $$f(x)=\begin{cases} \frac{x_1x_2^4}{x_1^2+x_2^4} &x\neq0 \\0 &x=0 \end{cases}$$

How can I determine the directional derivative $D_uf(x_0)$? Here $u,x_0 \in \Bbb{R^2}$. The question is odd, but here is my idea. Looking at $f(x)$, I notice that the function is differentiable everywhere except possible at $x=0$. Therefore $D_u f(x_0),\enspace x_0 \neq 0$ exists in all directions. This means the only case where $D_u f(x_0)$ might not be defined is when $x_0 = 0$. Indeed it can be shown the directional derivative exists in this case, provided $u \neq 0$. I think we have that $D_0 f(0 ) = 0$ always.

But this does not answer the question. I applied the limit definition of the directional derivative, expanded the polynomials. It does not seem like it is possible to refine it into some neat form for the general directional derivative of $f(x)$.

Answer 1

You are right that $D_0 f(0) = 0$ but you "dropped the ball" too early; you do get a simple form once you apply the limit definition of the directional derivative.

Indeed, for the case when $u := (u_1, u_2) \neq 0$, let $t > 0$. Then certainly $tu \neq 0$ either so that $$ f(tu) = \frac{(tu_1)(tu_2)^4}{(tu_1)^2 + (tu_2)^4} = \frac{t^5 u_1u_2^4}{t^2(u_1^2 + t^2u_2^4)} = \begin{cases} 0 &\text{ if } u_1 = 0 \\ \frac{t^3 u_1u_2^4}{u_1^2 + t^2u_2^4} &\text{ if } u_1 \neq 0\end{cases} $$ So applying the definition $$ D_u f(0) = \lim_{t \to 0; t > 0} \frac{f(0 +tu) - f(0)}{t} = \lim_{t \to 0; t > 0} \frac{f(tu)}{t} = \begin{cases} \lim_{t \to 0; t > 0} 0 & \text{ if } u_1 = 0 \\ \lim_{t \to 0; t > 0} \frac{t^2 u_1u_2^4}{u_1^2 + t^2u_2^4} & \text{ if } u_1 \neq 0\end{cases} \quad = 0 $$