What does it mean for partial derivative to exist but not continuous?

Continuous partial derivatives imply differentiability and differentiability imply continuity of a function and the existence of partial derivatives. I also understand that any other implication of this statement is false.

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    $\begingroup$ And your question is… ? $\endgroup$ Mar 12 '19 at 10:19
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    $\begingroup$ @JoséCarlosSantos The title, I think. $\endgroup$
    – user284001
    Mar 12 '19 at 10:24
  • $\begingroup$ A function can exist but not be continuous. Take such a function $f(x)$ on an interval and let $F(x)$ be an antiderivative of $f(x)$. Then $F'(x)$ exists and is not continuous. $\endgroup$
    – KCd
    Mar 12 '19 at 10:29
  • $\begingroup$ The function $f(x) = x^2 \sin(1/x)$, $f(0) = 0$ is differentiable, but its derivative is not continuous. See Wikipedia $\endgroup$ Mar 12 '19 at 15:32
  • $\begingroup$ math.stackexchange.com/questions/1822023/… $\endgroup$ Mar 12 '19 at 15:32

It is possible to have a function of several variables that has partial derivatives everywhere, but which is not fully differentiable everywhere. In such a case the partial derivatives will be discontinuous.

A typical example of the kind of functions we're talking about would be

$$ \begin{align} f(r\cos\theta,r\sin\theta) &= r\sin(2\theta) & \quad \text{for }r>0 \\ f(0,0) &= 0 \end{align} $$

This has both partial derivatives everywhere, but they are not continuous. We have $\frac{\partial f}{\partial y}=0$ at $(0,0)$ but $\frac{\partial f}{\partial y}=2$ at $(x,0)$ for every $x>0$. And the function is not differentiable at $(0,0)$, because the linear mapping induced by the partial derivatives doesn't approximate it well enough in directions other than the one the partial derivatives were taken in.

Even in one dimension, it is possible to be differentiable everywhere, but not continuously differentiable; the classical example is $$ f(x) = \begin{cases} 0 & \text{when }x=0 \\ x^2\sin(1/x) & \text{otherwise} \end{cases} $$


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