# For a two variable function to be differentiable, does its partial derivatives have to be continuous?

I am currently doing an Analysis-Differential Forms module and one of the questions states:

"Give an informal interpretation of what it means for a function $$f(x,y)$$ to be differentiable at point $$c=(c_1,c_2)$$"

Taking a neighbourhood about $$f(c_1,c_2)$$. The function is differentiable if $$f_x$$ and $$f_y$$ (partial derivatives) are continuous and exist for all $$x$$ and $$y$$ values in that neighbourhood.

My question is, does its partial derivatives have to be continuous so that $$f(x,y)$$ is differentiable?

Your tutor gave a sufficient condition for differentiability, not a necessary one. The answer to your question is "no."

$$f(x,y)=\begin{cases}(x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right) & (x,y)\neq(0,0) \\ 0 & (x,y)=(0,0) \end{cases}$$
As for "give an informal interpretation," the answer should be something like: $$f(x,y)$$ is differentiable at $$c$$ if $$f$$ is well-approximated by a linear function near $$c$$. Or in other words, if the graph of $$f$$ has a well-defined tangent plane at $$c$$.
By definition, $$f$$ is differentiable at $$c$$ if there exist a (unique) linear map $$L\colon \mathbb{R}^2 \longrightarrow \mathbb{R}$$ such that $$f(c_1+h,c_2+k) = f(c_1,c_2) + L(h,k) + \rho(h,k)$$ with $$\lim\limits_{(h,k)\to (0,0)} \frac{\rho(h,k)}{\|(h,k)\|} =0$$. An informal interpretation of this is that a function $$f$$ is differentiable at $$c$$ if it has a well defined linear approximation in a neighborhood of $$c$$.
One can prove that $$L(h,k) = \frac{\partial f}{\partial x}(c) h + \frac{\partial f}{\partial y}(c) k$$