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Find partial derivatives of $f(x,y)=\sqrt{|xy|}$ at $f_y(-\frac{1}{4}, -9)$ and $f_{xy}(-4,9)$

I'm doing $f_y(-\frac{1}{4}, -9)$ by the definition of partial derivative: $$ f_y(-\frac{1}{4}, -9)=\lim_{h \to 0} \frac{f(x_0, y_0+h)-f(x_0,y_0)}{h}=\\ =\lim_{h \to 0}\frac{\sqrt{|-\frac{1}{4}(-9+h)|}-\sqrt{{9 \over 4}}}{h}=\\ =\lim_{h \to 0}\frac{\sqrt{|{9 \over 4}-{h \over 4}|}-{3 \over 2}}{h} $$

Is it correct at this point to plug in $h=0$ and get that $\lim=\infty$? If it's correct then our function doesn't have a derivative at $(-\frac{1}{4}, -9)$.


Regarding $f_{xy}(-4,9)$ I split the original function into piecewise representation: $$ f(x,y)= \begin{cases} \sqrt{xy} \quad xy \ge0 \\ \sqrt{-xy} \quad xy<0 \end{cases} $$

Then: $$ f_{xy}(x,y)= \begin{cases} \frac{1}{2}(-{1 \over 2}\cdot \frac{xy}{\sqrt{(xy)^3}}+\frac{1}{\sqrt{xy}}) \quad xy > 0 \\ \frac{1}{2}({1 \over 2}\cdot \frac{xy}{\sqrt{(-xy)^3}}+\frac{1}{\sqrt{-xy}}) \quad xy<0 \end{cases} $$ Here's Wolfram Alpha calculation: http://www.wolframalpha.com/input/?i=d%2Fdx+d%2Fdy+sqrt(xy)

From this we can see that $f_{xy >0}$ does not exist while $f_{xy<0}$ does exist.

I'm not sure but even though we have $f_{xy<0}$ I think we don't have a partial derivative at $f_{xy}(-4,9)$ because in my understanding $f_{xy>0}$ and $f_{xy<0}$ need to be the same.

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To your first question: plugging in $h=0$ at that point does not result in $\lim f=\infty$, because the fraction would simplify to $\frac00$, which is an indeterminate form, and does not imply divergence to infinity. What you can do is drop the absolute value signs, because when $h$ is small, the quantity inside them is clearly positive. Then you can multiply top and bottom by the conjugate of the top, and simplify from there.

For your second question, we're only interested in the function's behavior in the neighborhood of the point $(-4,9)$ here, so you can ignore the possibility that $xy>0$. As long as you're close to $(-4,9)$, you know that $xy<0$, and you can just work with $\sqrt{-xy}$ accordingly.

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  • $\begingroup$ I completely missed that the fractions in the nominator will cancel each other. Could I apply L'hopital rule amd get that the limit is 0? $\endgroup$
    – Yos
    May 16, 2017 at 13:50
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    $\begingroup$ L'Hopital's rule should work, although the conjugate trick is easier, because derivatives of square roots are a pain $\endgroup$ May 16, 2017 at 15:26

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