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I've just been taught that a two-variables function is called derivable if there exist ALL the directional derivatives.

I've also been taught that a two-variables function is named differentiable if there exist the two partial derivatives.

I wonder if a function can exist which has existing two partial derivative but not all the directional ones, making it differentiable, but not derivable.

Thanks

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Yes. There exists, you can construct it. For example you can take the function: $f(x,y) = 0 \text{ for } x\ne y$ and $f(x,y) = |x| \text{otherwise}$. You can notice that for this function the partial derivatives exists but it can't be derived in the direction $v=(\sqrt{2}/2,\sqrt{2}/2)$

Anyway those are quite strange definitions. One usually says that a function of several variables is derivable if there exists all the partial derivatives, and differentiable (in a point $x_0 \in \Bbb{R}^n$ if there exist a linear operator $T: \Bbb{R}^n \to \Bbb{R}$ such that:

$\lim_{x \to x_0} \frac{|f(x_0 + x ) -f(x)-T(x-x_0)|}{|x-x_0|} =0$

Note that differentiability implies derivability (you cam show that the linear operator matches with the so called gradient of the function) but not the convers

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  • $\begingroup$ "Note that differentiability implies derivability " isn'it just the opposite....?! derivability implies differentiability ? if not I am just more confused... $\endgroup$ – f126ck Jan 10 '17 at 21:13
  • $\begingroup$ @f126ck differentiability, according to the definition that i gave, implies derivability. $\endgroup$ – jJjjJ Jan 10 '17 at 21:17
  • $\begingroup$ For example $f(x,y) = \frac{x^2y}{x^2 + y^2} \text{for} (x,y)\ne(0,0)$ and $f(0,0)=0$ has the two partial derivatives but it's not differentiable in(0,0) $\endgroup$ – jJjjJ Jan 10 '17 at 21:25
  • $\begingroup$ I've never heard “derivable” in English. The usual phrase is “partially differentiable”. $\endgroup$ – Hans Lundmark Jan 11 '17 at 7:58

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