Define $f:\mathbf{R}^{2}\rightarrow\mathbf{R}$ by $f(x,y) = \frac{x^{2}y}{x^{4}+y^{2}}, (x,y)\neq (0,0)$ with $f(0,0):=0$
I am attempting to show that the directional derivative, $\partial_{(a,b)}f(0,0)$ exists for all real $(a,b)\neq (0,0)$
I have managed to show that the directional derivative is $\frac{a^{2}}{b}$, for $b\neq0$, but I am unable to figure out how to prove it for the $b=0$ case.
Here is my workings for $b\neq0$:
The directional derivative exists if the limit
$\lim_{t\rightarrow0}\frac{f(t(a,b))-f(0,0)}{t}$ exists.
$\partial_{(a,b)}f(0,0) = \lim_{t\rightarrow0}\frac{\frac{(at)^{2}(bt)}{(at)^{4}+(bt)^{2}}}{t}=\lim_{t\rightarrow0}\frac{a^{2}b}{t^{2}a^{4}+b^{2}}=\frac{a^2}{b}$
This clearly is not the case for $b=0$, but, provided $a\neq0$, the directional derivative should exist.
Best Answer
For $b=0$, it means that $y=0$. Go back to the original function, $f(x,y)\equiv0$ in this direction, so the directional derivative of $(a,0)$ with $a\neq0$ is $0$