Given is the function $f: \mathbb{R^2} \to \mathbb{R}$ with
$ f(x,y) := \frac{2x^2y}{x^4 + y^2} \text{ for all } (x,y) \neq (0,0) \text{ and } 0 \text{ for all } (x,y) = (0,0)$
How can one prove that in all (bilateral) directional derivatives of $f$ exist in $(0,0)$?
I would have calculated the gradients
$$f_x = \frac{\partial}{\partial x} \frac{2x^2y}{x^4 + y^2} = \dfrac{4yx}{x^4+y^2}-\dfrac{8yx^5}{\left(x^4+y^2\right)^2}$$
and
$$f_y = \frac{\partial}{\partial y} \frac{2x^2y}{x^4 + y^2} = \dfrac{2x^2}{y^2+x^4}-\dfrac{4x^2y^2}{\left(y^2+x^4\right)^2}$$
and then I would have calculated $f_x(0,0)$ and $f_y(0,0)$, but that would lead to division by zero.
I also thought that we need a vector for directional derivatives.
What am I misunderstanding?
Best Answer
According to the definition of directional derivative at the origin, if we let $v = (a,b)$, we get that: \begin{align*} D_{v}f(0,0) & = \lim_{t\to 0}\frac{f((0,0) + t(a,b)) - f(0,0)}{t}\\\\ & = \lim_{t\to 0}\frac{2t^{3}a^{2}b}{t^{3}(t^{2}a^{4} + b^{2})}\\\\ & = \lim_{t\to 0}\frac{2a^{2}b}{t^{2}a^{4} + b^{2}}\\\\ & = \frac{2a^{2}b}{b^{2}}\\\\ & = \frac{2a^{2}}{b} \end{align*}
and we are done.
Hopefully this helps!