# Can a two variable function be differentiable but not derivable?

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

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:
$\lim_{x \to x_0} \frac{|f(x_0 + x ) -f(x)-T(x-x_0)|}{|x-x_0|} =0$
• 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) – jJjjJ Jan 10 '17 at 21:25