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Thread: Differentiability

  1. #1
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    Differentiability

    Hey everyone, I just want to make sure that I did this problem correctly. I have to show that this is not differentiable at the indicated point.

    $\displaystyle f (x,y) = \sqrt{x^2+y^2} ; (0,0)$

    And this is what I did:

    $\displaystyle \partial f / \partial x = \frac{1}{2} (x^2 + y^2) ^\frac{-1}{2} = \frac{x}{\sqrt{x^2+y^2}}$

    So at (0,0) this is undefined and it is the same for $\displaystyle \partial f / \partial y$. Thus is is not differentiable at the indicated point. Is this correct? Thanks for your help!
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Layla View Post
    Hey everyone, I just want to make sure that I did this problem correctly. I have to show that this is not differentiable at the indicated point.

    $\displaystyle f (x,y) = \sqrt{x^2+y^2} ; (0,0)$

    And this is what I did:

    $\displaystyle \partial f / \partial x = \frac{1}{2} (x^2 + y^2) ^\frac{-1}{2} = \frac{x}{\sqrt{x^2+y^2}}$

    So at (0,0) this is undefined and it is the same for $\displaystyle \partial f / \partial y$. Thus is is not differentiable at the indicated point. Is this correct? Thanks for your help!
    No.

    For example, we have :

    $\displaystyle \frac{\partial f}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}$

    If it is differentiable, then the limit as (x,y) approaches (0,0) must be the same, whatever direction you take.

    In particular, $\displaystyle \lim_{x \to 0^+} \lim_{y \to 0} \frac{x}{\sqrt{x^2+y^2}}=\lim_{x \to 0^-} \lim_{y \to 0} \frac{x}{\sqrt{x^2+y^2}}$

    But this is not true, as you can see if you calculate the limit $\displaystyle \frac{x}{\sqrt{x^2}}$
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  3. #3
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    Quote Originally Posted by Layla View Post
    So at (0,0) this is undefined and it is the same for $\displaystyle \partial f / \partial y$. Thus is is not differentiable at the indicated point. Is this correct?
    Not quite. What you proved is that $\displaystyle \frac{\partial f}{\partial x}$ cannot be continuously extended at 0 (the limit depends on the sign of $\displaystyle x$ when it approaches 0).
    The following works: if $\displaystyle f$ were differentiable at $\displaystyle (0,0)$, then the function $\displaystyle x\mapsto f(x,0)=|x|$ would be differentiable at $\displaystyle 0$ (it is just saying that if $\displaystyle f$ is differentiable at $\displaystyle (0,0)$, then $\displaystyle \frac{\partial f}{\partial x}$ exists at (0,0), since this is a "directional derivative" at $\displaystyle (0,0)$). However, as you know, this is false.
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  4. #4
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    Quote Originally Posted by Moo View Post
    $\displaystyle \frac{\partial f}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}$

    If it is differentiable, then the limit as (x,y) approaches (0,0) must be the same, whatever direction you take.
    The function $\displaystyle f:x\mapsto x^2\sin\frac{1}{x}$ is continuous at 0, $\displaystyle f'$ is not continuous at 0 (as you can check by computation). However, $\displaystyle f$ is differentiable at 0 ($\displaystyle \frac{f(x)}{x}\to 0$ as $\displaystyle x$ tends to 0)...

    Your argument could be made rigorous using additional properties. You can find two directions ($\displaystyle x\mapsto(\pm x,0)$ for $\displaystyle x>0$) where different finite limits exist for $\displaystyle \frac{\partial f}{\partial x}$. And then use the following fact: if a continuous function is differentiable on $\displaystyle (a,b)$ and such that its derivative admits a finite limit $\displaystyle \ell$ at $\displaystyle b$, then the function is differentiable at $\displaystyle b$ and its derivative at $\displaystyle b$ is $\displaystyle \ell$. But this is a bit overcomplicated.
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  5. #5
    Moo
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    Quote Originally Posted by Laurent View Post
    The function $\displaystyle f:x\mapsto x^2\sin\frac{1}{x}$ is continuous at 0, $\displaystyle f'$ is not continuous at 0 (as you can check by computation). However, $\displaystyle f$ is differentiable at 0 ($\displaystyle \frac{f(x)}{x}\to 0$ as $\displaystyle x$ tends to 0)...

    Your argument could be made rigorous using additional properties. You can find two directions ($\displaystyle x\mapsto(\pm x,0)$ for $\displaystyle x>0$) where different finite limits exist for $\displaystyle \frac{\partial f}{\partial x}$. And then use the following fact: if a continuous function is differentiable on $\displaystyle (a,b)$ and such that its derivative admits a finite limit $\displaystyle \ell$ at $\displaystyle b$, then the function is differentiable at $\displaystyle b$ and its derivative at $\displaystyle b$ is $\displaystyle \ell$. But this is a bit overcomplicated.
    Awww... My remnants of multivariable calculus look far behind lol! Thanks for your input

    (I think I was trying to do the thing you described in the latter part ^^)
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