I have some trouble understanding how my prof. got to this conclusion.
I'm asked to find the directional derivative at point $\zeta=(0,0)$ of $$\left\{ \begin{array}{c} f(x,y)=\frac{x+xy}{\sqrt{x^2+y^2}} ,\forall(x,y)\neq(0,0) \\f(0,0)=0 \end{array} \right.$$
Applying the definiton of the directional derivative and using $v_1^2+v_2^2=1$, I get: $$\lim_{h\to 0} \frac{f(hv_1,hv_2)-f(0,0)}{h} =\lim_{h\to 0}\frac{hv_1+h^2v_1v_2}{h^2\sqrt{v_1^2+v_2^2}}=\lim_{h\to 0} \frac{v_1+hv_1v_2}{h}=undef.$$
Given this result, my professor now says that at $f(0,0)$ the directional derivative is not defined.
My question now is:
Shouldn't the dericational derivative for $v_1=0$ do exist? Cause if that's the case, I have: $$\lim_{h\to 0} \frac{f(0,hv_2)-f(0,0)}{h}=\lim_{h\to 0}0=0$$
Which would imply that that the directional derivative does exist and is not undefined.
Does this make any sense to you guys or is my logic and understanding of directional derivatives just flawed?
You are right, the directional derivative in direction $(0,1)$ exists and is zero. This directional derivative is nothing else than $\dfrac{\partial f}{\partial y} (0,0)$.
However, strictly speaking it does not make sense to say that the directional derivative exists at some point - directional derivatives always depend on a given direction. I guess your professor wants to say that the directional derivative of $f$ exists at some point $(a,b)$ if directional derivatives in all directions exist. This is not the case at $(0,0)$.