Partial derivatives of $f(x,y)=\sqrt{|xy|}$
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
Updated on August 01, 2022Comments
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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?
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Git Gud over 9 yearsHave you tried using the definition? What are you having problems with?
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user130916 over 9 yearsOh. I got it. Using definition, I got both partial derivatives are $0$. Let me ask my next question instead.
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Git Gud over 9 yearsIf you don't know how to solve a problem, work from the definitions. Edit fast before the MSE police sees this.
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Git Gud over 9 yearsAgain: have you tried using the definition?
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user130916 over 9 yearsTried now. But I get all directional derivatives exist.
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user130916 over 9 yearsI see my miscomputation. Got it. thx.
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Git Gud over 9 years@user130916 You're welcome.
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sabeelmsk over 3 yearsIs the given function continuous at (0,0)
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Git Gud over 3 years@sabeelmsk Yes.