# How to prove that a function has directional derivative?

I want to prove that the function the given function has directional derivative for all $x\in\mathbb{E}^2$ and for each $v\in \mathbb{E}^2 \setminus \{0\}$.

$$f(x)= \begin{cases} \frac{x_1^2x_2}{x_1^4+x_2^2}, & \text{if } x \in \mathbb{E}^2 \setminus \{\mathbf{0}\}, \\ 0, & \text{if } x = \mathbf{0}. \end{cases}$$

I was thinking about trying to let $x\in\mathbb{E}^2$ and $v\in\mathbb{E}^2 \setminus \{0\}$ and try to show that $$D_vf(x) = \lim_{t\to 0}\dfrac{f(x+tv)-f(x)}{t}$$ exists. But I ended up getting really messy equation. Also, if the function has directional derivatives for each $x\in\mathbb{E}^2$, does this mean it is also differentiable? Also, does existence of all partial derivatives respect to each $x\in\mathbb{E}^2$ imply the function is differentiable?

• – Nosrati May 3 '17 at 3:03
• @MyGlasses Doesn't this only try to show that the directional derivative exist at $(0,0)$ and not for all $x\in\mathbb{E}^2$? – user3000482 May 3 '17 at 3:06
• let $x_2 = x_1^2$ and check out the limit as $x_1$ goes to $0.$ Is the function continous? – Doug M May 3 '17 at 3:10
• For $p\neq(0,0)$ directional derivative is $\nabla f(p).\vec{v}$ with unit $\vec{v}$. – Nosrati May 3 '17 at 3:10
• @DougM If I let $x_2=x_1^2$, then the function would not be continuous at $(0,0)$ but this shows that function is not differentiable at $(0,0)$. I want to show that the directional derivative D_vf(x) exists at each $x\in\mathbb{E}^2$ for each $v\in\mathbb{E}^2\setminus \{0\}$ – user3000482 May 3 '17 at 3:24