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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?

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(a) Right.

(b) Right.

(c) Right.

(d) Right. And since it is not continuous, it is not differentiable.

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  • $\begingroup$ @Math_QED I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Sep 20 '18 at 15:54
  • $\begingroup$ Thanks for the verification. $\endgroup$ – user370967 Sep 20 '18 at 15:56
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All points (a), (b), (c) and (d) are correct.

For (d) as an alternative we can consider any path $x=y \to 0$.

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  • $\begingroup$ @Math_QED Opsssss...yes you are right! I fix that. $\endgroup$ – user Sep 20 '18 at 16:01
  • $\begingroup$ Thanks for the verification (+1) $\endgroup$ – user370967 Sep 20 '18 at 16:04
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    $\begingroup$ @Math_QED Thanks also for your verification of mine verification! (+1) :P $\endgroup$ – user Sep 20 '18 at 16:05

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