Consider the function :
$$f: \mathbb{R}^2 \rightarrow \mathbb{R} , (x,y) \mapsto
\begin{cases}
0 & \text{for } (x,y)=(0,0) \\
\frac{x^3}{x^2+y^2} & \text{for } (x,y) \neq (0,0)
\end{cases} $$
Show that $f$ not differentiable at $(0,0)$ but all directional derivatives exist.
I don't know how to tackle this problem. Can someone give some hints or solve the problem? Thanks 🙂
Best Answer
Take some (unitary) $u$. Then show that $$f'(\vec 0;u)=\lim_{h\to 0 }\frac{f(h\cdot u)-f(\vec 0)}h$$ always exists for any choice of $u$. You'll be dealing with $$\mathop {\lim }\limits_{h \to 0} \frac{\frac{{{h^3}u_1^3}}{{{h^2}(u_1^2 + u_2^2)}}}{h}$$
where $u=(u_1,u_2)$.
If your function were differentiable at the origin, then we would have $$f'(\vec 0)(u)=f'(\vec 0;u)$$ where the right hand side is the directional derivative at $\vec 0$ with direction $u$, and the left hand side is the total derivative at $(0,0)$ at evaluated at $u$. Now, $f'(\vec 0)$ would be linear, so $$f'(\vec 0)(1,1)=f'(\vec 0)(0,1)+f'(\vec 0)(1,0)$$
The right hand side is just the partial derivatives at the origin, which gives $0+1=1$. What does the left hand side give? Remember it is just the directional derivative at $\vec 0$ with direction $(1,1)$.