derivability don't imply partial to be continuous ? example

Question

Is $$f(x,y) =\begin{cases} x^2+2x+2y & \text{ for } (x,y)\neq (0,0) \\ y^2 & \text{ for } (x,y)=(0,0) \end{cases}$$ derivable? But its partials are not continuous?

Answer 1

If a function $f(x,y)$ is continuous at a point in its domain and its partial derivatives exist at that point, then $f(x,y)$ is said to be derivable at that point.

Here, $f(x,y)$ is continuous at $(0,0)$.

And if both $f^\prime_x$ and $f^\prime_y$ exist at $(0,0)$, then $f(x,y)$ is derivable at $(0,0)$.

So, $f^\prime_x =\lim_{h \to 0}$ $\frac{f(h,0)-f(0,0)}{h}$

Hence, $f^\prime_x = 2$

Similarly, $f^\prime_y =\lim_{h \to 0}$ $\frac{f(0,h)-f(0,0)}{h}$

Hence, $f^\prime_y = 2$

So, $f(x,y)$ is derivable at $(0,0)$.

And at all points in $R^2$,except $(0,0)$, it's pretty obvious that $f(x,y)$ is derivable as $f(x,y)$ is simply a polynomial in that domain.

Hence, $f(x,y)$ is derivable on $R^2$.