Define $f(x,y) = \frac{x^2y^2}{x^2y^2 + (y-x)^2}$ if $(x,y) \neq (0,0)$, $f(0,0) = 0$ on $\mathbb{R}^2$.
(a) For which vectors $u \neq 0$ does $f'(0,u)$ exist? Evaluate it when it exists
(b) Do $D_1f, D_2f$ exist at $0$?
(c) Is $f$ differentiable at $0$?
(d) Is it continuous at $0$?
My attempt:
(a)$$f'((0,0),(u_1,u_2)) = \lim_{t \to 0} \frac{u_1^2 u_2^2t}{t^2 u_1^2 u_2^2 + (u_1-u_2)^2}$$
exists only if $u_1 \neq u_2$, and then equals $0$.
(b) Since $D_1f(0,0) = f'(0,(1,0))$, it follows that $D_1f(0,0) = 0$. Similalrly for $D_2f(0,0)$
(c) No, not all directional derivatives exist, and also no because of (d)
(d) $(1/n,1/n) \to 0$ but $f(1/n,1/n) = 1 \not \to 0 = f(0,0)$.
Hence, $f$ isn't continuous at $0$.
Is this correct?
Best Answer
(a) Right.
(b) Right.
(c) Right.
(d) Right. And since it is not continuous, it is not differentiable.