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Math Help - 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.

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

    And this is what I did:

    \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  \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
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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.

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

    And this is what I did:

    \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  \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 :

    \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, \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 \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  \partial f / \partial y. Thus is is not differentiable at the indicated point. Is this correct?
    Not quite. What you proved is that \frac{\partial f}{\partial x} cannot be continuously extended at 0 (the limit depends on the sign of x when it approaches 0).
    The following works: if f were differentiable at (0,0), then the function x\mapsto f(x,0)=|x| would be differentiable at 0 (it is just saying that if f is differentiable at (0,0), then \frac{\partial f}{\partial x} exists at (0,0), since this is a "directional derivative" at (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
    \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 f:x\mapsto x^2\sin\frac{1}{x} is continuous at 0, f' is not continuous at 0 (as you can check by computation). However, f is differentiable at 0 ( \frac{f(x)}{x}\to 0 as x tends to 0)...

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

    Your argument could be made rigorous using additional properties. You can find two directions ( x\mapsto(\pm x,0) for x>0) where different finite limits exist for \frac{\partial f}{\partial x}. And then use the following fact: if a continuous function is differentiable on (a,b) and such that its derivative admits a finite limit \ell at b, then the function is differentiable at b and its derivative at b is \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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