8
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Do you know why existence of all partial derivatives does not mean that a function is differentible? $\endgroup$ – Arthur Aug 20 '16 at 11:25
  • $\begingroup$ @Arthur Yes, I do $\endgroup$ – TheQuantumMan Aug 20 '16 at 14:47
  • $\begingroup$ 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. $\endgroup$ – Eric Wofsey Aug 22 '16 at 7:25
  • $\begingroup$ @EricWofsey If all the partials exist and are continuous at a neighborhood around a point, then why shouldn't there be a tangent plane? $\endgroup$ – TheQuantumMan Aug 22 '16 at 8:08
5
+50
$\begingroup$

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 $(x,y)$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Keshav Dec 22 '18 at 20:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.