I have the following function: $$f(x,y)=\begin{cases}x^2+y&\wedge&x\geq 1\\3x-y&\wedge&x<1\end{cases}$$ and they ask me the following:
- Continuity domain
- Existence of directional derivatives in the point $(1,1)$ in all directions $\vec v=(a,b)$
- What can you say about the differentiability of the function at that point?
I made the map of continuity of $f$ and the differents directions of the unit versor $\vec v$ in $(1,1)$:
- $D_f=\mathbb R^2$.
- We need to calculate $f'((1,1);(a,b))$ (if exists):
$\color{red}{(1)}$ If $a=0\;\;(b=\pm 1):\displaystyle\lim_{h\to 0}{\dfrac{1^2+(1+hb)-2}h}=\boxed b$
$(2)$ If $-1\leq a<0\;\;(\forall b):\displaystyle\lim_{h\to 0}{\dfrac{3+3ha-1-hb-2}h}=\displaystyle\lim_{h\to 0}{\dfrac{h(3a-b)}h}=\boxed{3a-b}$.
$\color{blue}{(3)}$ If $0<a\leq 1\;\;(\forall b):\displaystyle\lim_{h\to 0}{\dfrac{{(1+ha)}^2+(1+hb)-2}h}=\displaystyle\lim_{h\to 0}{\dfrac{h(2a+ha^2+b)}h}=\boxed{2a+b}$.
With this I can affirm that $f$ has directional derivatives in the point $(1,1)$, which are $$\boxed{\begin{array}{llll} f'((1,1);(0,\pm 1))&=&b&\\ f'((1,1);(a,b))&=&3a-b&\text{ if }a\in[-1,0)\\ f'((1,1);(a,b))&=&2a+b&\text{ if }a\in(0,1]. \end{array}}$$
EDIT:
- I'm going to skip some steps and I'll go directly to calculate the formal limit. For that I need the partials, which are $${f'}_x(1,1)=2\qquad{f'}_y(1,1)=1.$$
So by definition: $$\displaystyle\lim_{(x,y)\to (1,1)}{\dfrac{x^2+y-[2+2(x-1)+(y-1)]}{\sqrt{{(x-1)}^2+{(y-1)}^2}}}=\displaystyle\lim_{(x,y)\to (1,1)}{\dfrac{x^2+1}{\sqrt{{(x-1)}^2+{(y-1)}^2}}}=\ldots$$ and now I don't know how to proceed :(.
Is it okay or did I lose my head?
Thanks!