What is the intuition behind a function being differentiable if its partial derivatives are continuous?

Question

I have seen the proof of it, but I can't quite intuitively understand why continuous partial derivatives imply that a function is differentiable.
Thanks!

EDIT: The only reason I can think of is that if the partials exist at a point and are continuous at a neighborhood around that point, then it is only logical that the tangent plane to that point exists because the equation for the tangent plane at that point would be a good approximation of the function around that point. If the partials exist at the point but are not continuous at a neighborhood around that point, then I can't see how a tangent plane at the point can exist because would not be a good approximation of the function near the point since the function would have discontinuous slope along certain direction(s)(by the definition of a partial derivative). Is my intuition right?

Answer 1

I'll use the notation $D_i f$ to denote the $i$th partial derivative of a function $f$.

To say that a function $f:\mathbb R^2 \to \mathbb R$ is differentiable at a point $(x,y)$ means that there exists a linear transformation $L:\mathbb R^2 \to \mathbb R$ such that the approximation $$ f(x+\Delta x, y + \Delta y) \approx f(x,y) + L(\Delta x, \Delta y) $$ is good when $\Delta x$ and $\Delta y$ are small. (Exactly what "good" means can be made precise.)

So how can we approximate $f(x + \Delta x, y + \Delta y)$? Here's a way that you can visualize if you draw the points $(x,y), (x + \Delta x, y)$, and $(x + \Delta x, y + \Delta y)$:

\begin{align} f(x+\Delta x, y + \Delta y) &= f(x,y) + f(x+\Delta x, y) - f(x,y) + f(x+\Delta x, y+\Delta y) - f(x+\Delta x, y) \\ & \approx f(x,y) + D_1 f(x,y) \Delta x + D_2 f(x + \Delta x, y) \Delta y \\ \tag{1}&\approx f(x,y) + D_1 f(x,y) \Delta x + D_2 f(x , y) \Delta y. \end{align} This suggests that we can take $L$ to be the linear transformation defined by $$ L(\Delta x, \Delta y) = D_1 f(x,y) \Delta x + D_2 f(x , y) \Delta y. $$

But here's the crucial point: in equation (1), we assumed that $D_2 f(x,y) \approx D_2 f(x + \Delta x, y)$. In order to guarantee that this approximation is sufficiently accurate (for small values of $\Delta x$), we need to assume that $D_2 f$ is continuous at $(x,y)$.