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Why $$\sqrt{1+x^2+y^2}$$ is not differentiable at origin?

Jeel Shah
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    Not differentiable where? With respect to what? And who said it isn't? –  Jul 03 '17 at 14:44
  • my calculus teacher said that it is continuous at the origin but it is not differentiable... but i don't understand why... As long as I remember, for a function to be differentiable, just be continuous and exist on the point. – Augusto Amaral Jul 03 '17 at 14:49
  • It looks perfectly differentiable (with gradient zero) to me. – hmakholm left over Monica Jul 03 '17 at 15:07
  • @AugustoAmaral, you could try follow the definition of multi-variable derivative – Covvar Jul 03 '17 at 15:08
  • Hello and welcome to math.stackexchange. All partial derivatives of this function exist and are continuous, so the function is differentiable infinitely often everywhere. – Hans Engler Jul 03 '17 at 15:09
  • Hello and welcome to math.stackexchange. All partial derivatives of this function exist and are continuous, so the function is differentiable infinitely often everywhere. – Hans Engler Jul 03 '17 at 15:09
  • @Covvar , I'm sorry for bad English :( – Augusto Amaral Jul 03 '17 at 15:10
  • Consider $f(x,y)=\sqrt{x^2+y^2}$ instead; that function is continuous but not differentiable at the origin. – Hans Lundmark Jul 03 '17 at 17:22
  • @AugustoAmaral, "continuous and existing" aren't enough for differentiability. The limit that defines the derivative must exist (although maybe that's what you meant). For example, $|x|$ exists and is continuous at $x=0$, but it is not differentiable there because the limit that defines the derivative does not exist, since the limit from the left is $-1$ and from the right is $1$. –  Jul 07 '17 at 17:04

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This function is differentiable at the origin. The easiest way to see this is that the function has continuous partials at $(0,0)$. You can look at this post for this theorem. Proof that continuous partial derivatives implies differentiability