Given
$f(x,y)=\left\{\begin{matrix}
\frac {x^2y}{x^4+y^2} & (x,y)\neq(0,0)\\
0& (x,y)=(0,0)
\end{matrix}\right.$
I need to calculate the directional derivative at the point (0,0) in the direction making an angle of $30^{\circ}$ with the positive $x$ axis.
Does this derivative even exist? How show I treat this problem?
Best Answer
The directional derivative in the direction $(h_x,h_y)$ (with $h_y \neq 0$) at $(0,0)$ is given by $df((0,0),(h_x,h_y)) = \lim_{t \to 0} { f(th_x,t h_y) -f(0,0) \over t} = \lim_{t \to 0} { t^3 h_x^2h_y \over t^4 h_x^4 + t^2 h_y^2}$.
It is straightforward to see that the limit is zero in all cases (look at $h_y=0$ and $h_y \neq 0$ separately).
Old incorrect answer: (Thanks to @user61681 for catching that.)
The directional derivative in the direction $(h_x,h_y)$ (with $h_y \neq 0$) at $(0,0)$ is given by $df((0,0),(h_x,h_y)) = \lim_{t \to 0} { f(th_x,t h_y) -f(0,0) \over t} = \lim_{t \to 0} { t^2 h_x^2h_y \over t^4 h_x + t^2 h_y^2} = {h_x^2 \over h_y}$.
If $(h_x,h_y)$ is in the direction given, we have $(h_x,h_y) = \lambda ({\sqrt{3} \over 2},{1 \over 2})$ for some $\lambda>0$, and so $df((0,0),\lambda ({\sqrt{3} \over 2},{1 \over 2})) = {3 \over 2} \lambda$.