$$f(x,y)=\left\{\begin{array}{cl}\dfrac{x^{2} y}{x^{2}+y^{2}}, & (x, y) \neq(0,0) \\0, & (x, y)=(0,0)\end{array}\right.$$
This function is derivable and non-differentiable at $(0,0)$ and I can prove it in mathematical language.
However , something go wrong when I try to understand it
- In a unary function , differentiable means that left derivative equals right derivative , such as $f(x)=x,f(x)=x^2$ and so on
- In a binary function , derivable means that partial derivatives exist which means that:
left derivative equals right derivative on the X axis and Y axis separately
differentiable means that : in any direction , their "left derivative" equals "right derivative"
And I find this function's "left derivative" equals "right derivative" in any direction
Obviously , this function is symmetrical about the origin
So , I don't know why my understanding of differentiable is wrong
Could you tell me the relationship between differentiable and the directional derivative
For binary functions, differentiable means that there is a tangent plane at this point.
Even if the directional derivative of a function in any direction at a point exists and the "left derivative" is equal to the "right derivative", the function may not be differentiable at that point. In particular, imagine in $\mathbb{R}^3$ that there are countless lines passing through the origin, such that they form the graph of a binary function. The directional derivative of the function described above in any direction at the origin exists and the "left derivative" is equal to the "right derivative" (because what pass through the origin are lines), but the function does not necessarily have a tangent plane at the origin, that is, it is not necessarily differentiable. (These lines are pieced together un-smoothly to form the graph of this function.) For example, $$f(x,y)=\left\{ \begin{array}{ll} x, & y=0, \\ 0,& {\rm otherwise}. \end{array} \right.$$
Generally, the directional derivative of a function along a direction at a point exists and the "left derivative" is equal to the "right derivative" means that there is a plane passing through this point and it parallels to this direction and parallels to the z axis, moreover it intersects with the graph of this function which makes us obtain a differentiable curve at the point. However, the variety between curves obtained in different directions may not be smooth. Therefore, the function may not be differentiable at the point.
To summarize, differentiability not only requires the existence of derivatives in all directions and that the left derivative is equal to the right derivative, but also requires a "smooth" relationship between directional derivatives in different directions. Therefore, it is not enough that you mention only the differentiability in each direction, and we also need a "smooth" relationship between differentiability in different directions.