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We know for sure that for functions of several variables the existence of partial derivatives does not guarantee continuity. However, if partial derivatives exist and $f$(x) has continuous first partials in a neighborhood of some point x, then $\nabla f$(x) exist, which implies there is a vector satisfies the definition of differentiability;

$f$(x+h) $-$ $f$(x) $=\nabla f$(x) $\cdot$ h $+$ $o$(h)

Therefore $f$(x) is differentiable and continuous but that seems to contradict with our first statement, since the existence of partial derivatives guarantee differentiability, which leads to continuity. So where is my mistake or misconception?

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    $\begingroup$ I don't understand what the problem is. The existence of partials does not imply continuity, but the existence of continuous partials does. Stronger hypothesis, stronger conclusion - what's the problem? $\endgroup$ Jul 10 at 15:04
  • $\begingroup$ Also the second term on the right hand side is not $O(h)$ but rather $g(h)O(h)$ where $g$ goes to zero as h goes to zero $\endgroup$
    – Leonid
    Jul 10 at 15:06
  • $\begingroup$ The problem is, the existence of partials implies differentiability, which implies continuity. Actually this is my question, thanks for cooperation. @DavidC.Ullrich $\endgroup$ Jul 10 at 15:09
  • $\begingroup$ @Leonid it's the exact form of definition you'll find in Calculus Salas, where little-o (h) tends to zero as h goes to zero. mathworld.wolfram.com/Little-ONotation.html $\endgroup$ Jul 10 at 15:14
  • $\begingroup$ "The existence of partials does not imply continuity, but the existence of continuous partials does" But in the way we've followed hypothesis I found that the existence of partials implies continuity which is absolutely false, I just need to know the mistake!! @DavidC.Ullrich $\endgroup$ Jul 10 at 15:20
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The problem is, the existence of partials implies differentiability, which implies continuity

This is incorrect. A very common counterexample is here

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  • $\begingroup$ Thanks that was really helpful, but still having just one small question if you don't mind; in the counter example you've mentioned to me, $f$ has continuous first partials in a neighborhood of $(0,0)$, we can check that by studying the curve arises from the intersection of surface $f(x,y)$ and $y=0$ plane or $x=0$ plane. Anyway Calculus Salas (10th edition, p791) established that if f has continuous first partials in a neighborhood of $x$, then $f$ is differentiable. Would you please explain me the theorem with respect to the counter example. thanks a lot. $\endgroup$ Jul 10 at 15:51
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    $\begingroup$ @HumamAbuAl-Shaikh No, the function in that example does not have continuous partials. $\endgroup$ Jul 10 at 16:48

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