Before I post this I would just like to state that I know that there is a very similar question with a very similar function but I've gone through the answer and it doesn't really help me.
Consider a function:
$f(x,y) = \frac{xy}{x^4+y^4} $ if $x \not = 0$ and $0$ otherwise.
Show that the directional derivative exists in any direction at the point (0,0)
We've also been given the definition of directional derivative in direction $e = (e_1, e_2)$ as:
$ \frac{d f(a,b)}{d e}= \lim_{h \to 0} \frac {f(a + he_1, b + he_2) – f(a,b)}{h}$
Now using this definition I get to the solution as follows:
$ \frac{df(0,0)}{de} = \lim_{h \to 0} \frac{\frac{he_1he_2}{(he_1)^4 + (he_2)^4}}{h}$
$ = \lim_{h \to 0} \frac{\frac{h^2e_1e_2}{h^4(e_1^4 + e_2^4)}}{h}$
$ = \lim_{h \to 0} \frac{h^2e_1e_2}{h^5(e_1^4 + e_2^4)}$
$ = \lim_{h \to 0} \frac{e_1e_2}{h^3(e_1^4 + e_2^4)} $
but clearly this limit diverges??? So the limit does not exist? Did I make a stupid algebraic mistake somewhere? Or am I missing something entirely?
Thanks.
Best Answer
On the subset $x = y$, $$f(t,t) = \frac{t^2}{2t^4} = \frac1{2t^2}$$ is unbounded near $(0,0)$, i.e. $f$ is discontinuous on the subset $x = y$ and the corresponding directional derivative can't exist.
But the partial derivatives exists because on the axes $f$ is zero.