How to prove, that the partial derivatives of the expression $f(x,y) = \frac{|x|^3 + |y|^2}{|x|+|y|}$ at $(0,0)$ and $f(0,0)=0$ exist and how to determine their value in $f(0,0)$?
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1Did you mean partial derivatives at $(0,0)$? – ireallydonknow Dec 18 '13 at 18:48
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Yes in $(0,0)$, thanks for the remark! – Milan Tenk Dec 18 '13 at 18:51
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Are you also sure that all the partial derivatives exist? – ireallydonknow Dec 18 '13 at 18:58
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I'm not sure, thats why I want to know if they exist, or not. – Milan Tenk Dec 18 '13 at 19:00
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If $\phi(y) = f(0,y)$, then we see that $\phi(y) = |y|$, which is not differentiable at $y = 0$. Hence the partial of $f$ with respect to $y$ does not exist at $y=0$.
copper.hat
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$$f_{x}(0,0)=\frac{d}{dx}f(x,0)|_{x=0}=\frac{d}{dx}\frac{|x|^3}{|x|}|_{x=0}=2x|_{x=0}=0$$
$$f_{y}(0,0)=\frac{d}{dy}f(0,y)|_{y=0}=\frac{d}{dy}\frac{|y|^2}{|y|}|_{y=0}=\frac{d}{dy}|y||_{y=0}$$
Obviously the $y$-partial derivative for $f$ at $(0,0)$ does not exist.
ireallydonknow
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Here is how you advance. Using the definition, we have
$$f_x(0,0) =\lim_{h\to 0} \frac{ f(0+h,0) -f(0,0)}{h} =\dots\,.$$
You need the definition of the absolute value to finish the problem, since you need to find $\lim_{h\to 0^-}$ and $\lim_{h\to 0^+}$. Do the same for the other partial derivative with respect to $y$.
Mhenni Benghorbal
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