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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:

  1. Continuity domain
  2. Existence of directional derivatives in the point $(1,1)$ in all directions $\vec v=(a,b)$
  3. 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 $(1,1)$:

Domain of f

  1. $D_f=\mathbb R^2$.
  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:

  1. 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!

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    $\begingroup$ It looks right to me! $\endgroup$ – Matthew Leingang Apr 24 '18 at 0:16
  • $\begingroup$ @MatthewLeingang Thank you! It remains to study the differentiability in the point. I see how I do it and I edit my answer again. $\endgroup$ – manooooh Apr 24 '18 at 0:18
  • $\begingroup$ @MatthewLeingang I added my solution for excercise 3. Could you help me finish it, please? $\endgroup$ – manooooh Apr 24 '18 at 0:41
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    $\begingroup$ Well, again, you're going to be looking at the cases $x<1$, $x>1$, and $x=1$. $\endgroup$ – Matthew Leingang Apr 24 '18 at 0:52
  • $\begingroup$ How are you @MatthewLeingang ? I don't know how to continue after your comment :(. Could you help me to finish it, please? $\endgroup$ – manooooh Apr 24 '18 at 15:07

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