# Counting a Directional derivative.

Let f be a function that satysfies : $$f: \mathbb{R}^2 \rightarrow \mathbb{R} , (x,y) \mapsto \begin{cases} 0 & \text{for } (x,y)=(0,0) \\ \frac{x^3}{x^2+y^2} & \text{for } (x,y) \neq (0,0) \end{cases}$$ I want to show that f has directional derivative at point $$(0,0)$$ at every direction. So, let $$v=(v_1,v_2) \in R^2, (v_1,v_2) \neq (0,0)$$. f has directional derivative in $$(0,0)$$ if and only if following limit exist : $$\lim_{t->0}\frac{f(a+tv)-f(a)}{t}=\lim_{t->0}\frac{f(tv_1,tv_2)}{t}=\lim_{t->0}\frac{t^3v_1^3}{t(t^2v_1^2+t^2v_2^2)}=\frac{v_1^3}{v_1^2+v_2^2}$$.

So i show that above limit exist and it depends of choosing vector v. So the directional direvative exist at point $$(0,0)$$ in every direction. The main question is :

## Am i thinking correcly ?

Yes, you are right. The directional derivative exists at $$(0,0)$$ in every dirction $$v=(v_1,v_2)$$ and is $$=\frac{v_1^3}{v_1^2+v_2^2}.$$