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I must study the derivability in different directions at point $\overline A$ of $$f(x,y)=\begin{cases} \dfrac{xy-x}{x^2+{(y-1)}^2}&\wedge&(x,y)\neq (0,1)\\0&\wedge&(x,y)=(0,1) \end{cases},\qquad\overline A=(0,1).$$

I did the incremental quotient by definition for a generic versor $\hat r=(u,v)$:

$$\lim_{h\to0}{\dfrac{f((0,1)+h(u,v))-f(0,1)}{h}}= \lim_{h\to0}{\dfrac{f(hu,1+hv)-0}{h}}= \lim_{h\to0}{\dfrac{\frac{hu(1+hv)-hu}{{(hu)}^2+{(hv)}^2}}{h}}= \lim_{h\to0}{\dfrac{hu(1+hv-1)}{h^2\underbrace{\left(u^2+v^2\right)}_{=\; 1}}\cdot\dfrac 1h}= \lim_{h\to0}{\dfrac{huv}{h^2}}= \lim_{h\to0}{\dfrac{uv}h}$$

... but I don't know how to proceed. Is my statement correct? If so, how do I proceed?

Thanks!

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  • $\begingroup$ Do you know the formula for the directional derivative in terms of the partial derivatives? $\endgroup$ – saulspatz Apr 15 '18 at 18:13
  • $\begingroup$ Yes, it is $f'\left(\vec{x_0};(1,0)\right)=f'_x\left(\vec{x_0}\right)$ and $f'\left(\vec{x_0};(0,1)\right)=f'_y\left(\vec{x_0}\right)$. But in this exercise they ask for any address, not for the canonical, right? $\endgroup$ – manooooh Apr 15 '18 at 18:16
  • $\begingroup$ That's why I decided to do it by the most "general" definition (i.e. the formal one). $\endgroup$ – manooooh Apr 15 '18 at 18:18
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    $\begingroup$ I dont see any mistake in your calculations. Now observe that $\lim_{h\to 0}\frac{uv}h$ doesnt exists when $uv\neq 0$. $\endgroup$ – Masacroso Apr 15 '18 at 18:23
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    $\begingroup$ observe that $h\in\Bbb R$ and $h\neq 0$ in the limit (check the definition of functional limit), then you have $\lim_{h\to 0}\frac0h=0$. And $\lim_{h\to 0^+}\frac1h=+\infty\neq\lim_{h\to 0^-}\frac1h=-\infty$ $\endgroup$ – Masacroso Apr 15 '18 at 18:38
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I will answer my own question. We have $$\lim_{h\to0}{\frac{uv}{h}}.$$ If we pick $a=0\;\vee\;b=0$ then the quotient is $0$ before calculating the limit, so in those four cases, $ (0,1), (0,-1), (1,0)$ and $(-1,0)$ exists the limit and it's $0$.

Hence for $a$ and $b$ not null does not exist, for $a$ or $b$ null yes and it is $0$.

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