I'm pretty stuck on the following problem.
Define $f: R^2 \rightarrow R$ by
$$f(x,y) = \frac{xy^2}{x^2+y^2} \quad\text{ if }\quad (x,y) \neq (0,0),$$
$$f(x,y) = 0 \quad\quad\quad\text{ if }\quad\quad (x,y) =(0,0).$$
Prove $D_uf$ exists for all $u$.
So I know that this is the directional derivative, and that $u$ can be any unit vector.
But there are infinite possibilities for $u$, so how can I show the derivative exists for all of them?
Best Answer
Let $u = (a,b)$ and $\|u\| = 1$.
According to the definition of directional derivative, it results that \begin{align*} D_{u}f(0,0) & = \lim_{t\to 0}\frac{f(0 + ta, 0 + tb) - f(0,0)}{t} =\frac{ab^{2}}{a^{2} + b^{2}} = ab^{2} \end{align*}
since $\|u\| = 1$, and we are done.
Hopefully this helps!