Partial derivatives of $f(x,y)=\sqrt{|xy|}$

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You miscomputed something.

Note that, given $\mathbf v=(u,v)$, $$\begin{align} \nabla_\mathbf{v}f(0,0) &= \lim \limits_{h\rightarrow 0}\left(\dfrac{f(hu,hv)-f(0,0)}{h}\right)\\ &= \lim \limits_{h\rightarrow 0}\left[\dfrac {\left(\left|h^2uv\right|\right)^{1/2}}{h}\right]\\ &= \lim \limits_{h\to 0}\left(\dfrac {\color{red}| h\color{red}|\sqrt{|uv|}}{h}\right). \end{align}$$

Can you conclude?

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user130916
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user130916

Updated on August 01, 2022

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  • user130916
    user130916 over 1 year

    $f(x,y)=\sqrt{|xy|}$

    First question: How to find $f_x(0,0)$ and $f_y(0,0)$? I have figured out this using definition - Both are $0$.

    My next question is: How to show that $f_x(0,0)$ and $f_y(0,0)$ are the only directional derivative that exist at $(0,0)$?

    Again, using definition, let the direction be $\mathbf{v}=(u,v)$. $\nabla_\mathbf{v}f(0,0) = \lim_{h\rightarrow 0}(f(hu,hv)-f(0,0))/h = \lim_{h\rightarrow 0}h\sqrt{|uv|} = 0$.

    What's wrong with my argument?

    • Git Gud
      Git Gud over 9 years
      Have you tried using the definition? What are you having problems with?
    • user130916
      user130916 over 9 years
      Oh. I got it. Using definition, I got both partial derivatives are $0$. Let me ask my next question instead.
    • Git Gud
      Git Gud over 9 years
      If you don't know how to solve a problem, work from the definitions. Edit fast before the MSE police sees this.
    • Git Gud
      Git Gud over 9 years
      Again: have you tried using the definition?
    • user130916
      user130916 over 9 years
      Tried now. But I get all directional derivatives exist.
  • user130916
    user130916 over 9 years
    I see my miscomputation. Got it. thx.
  • Git Gud
    Git Gud over 9 years
    @user130916 You're welcome.
  • sabeelmsk
    sabeelmsk over 3 years
    Is the given function continuous at (0,0)
  • Git Gud
    Git Gud over 3 years
    @sabeelmsk Yes.