For single-variable functions, we have that differentiability implies continuity. However, this is not the case with the following task, which examplifies this very well:
$$f(x,y) = \begin{cases}\frac{2xy}{x^2+y^2},\:\: (x,y) \ne (0,0)\\
0,\:\: (x,y) = (0,0)\end{cases}$$ is not continuous at $(0,0)$, as the partial limits differ from the case $x = y$, where the limit is $1$. However, it is still differentiable here for both variables:$f_x(0,0) = \lim_{h \to 0} \frac{0}{h^2} = 0$
By a similar argument, $f_y(0,0) = 0$.
We have established that a function may be partially differentiable although it is not continuous. While I do understand that this follows from the definition of partial derivatives, I am asking for an intuitive explanation based on properties of derivatives (slopes, normal lines etc.)
Best Answer
Not much surprising. In the definition of $f_x(0,0)$ only values of $f(x,0)$ are considered. Similarly to define $f_y(0,0)$ only values of $f(0,y)$ are considered. This explains why the limit of $f(t,t)$ can be anything. In fact also this function has the property of being derivable in $(0,0)$ but not continuous: $$ f(x,y) = \begin{cases} 0 & \text{if $x=0$ or $y=0$}\\ 1 & \text{otherwise} \end{cases} $$