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

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?

• Do you know why existence of all partial derivatives does not mean that a function is differentible? – Arthur Aug 20 '16 at 11:25
• @Arthur Yes, I do – TheQuantumMan Aug 20 '16 at 14:47
• I don't follow your "intuition". It is actually possible for a tangent plane to exist even if the partials are not continuous, and I don't see why it is "only logical" for the tangent plane to exist if they are continuous. – Eric Wofsey Aug 22 '16 at 7:25
• @EricWofsey If all the partials exist and are continuous at a neighborhood around a point, then why shouldn't there be a tangent plane? – TheQuantumMan Aug 22 '16 at 8:08

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 $$(x,y)$$.

• So here you ended up with: we need that $D_2 f$ is continuous. Does the theorem weaken to this? Can we say the derivative exists if all the partials exist and all but one is continuous? – Keshav Dec 22 '18 at 20:25