Find partial derivatives of $f(x,y)= \sqrt {xy}$ at the point $(0,0)$

Question

I'm trying to find partial derivatives $f_x$ and $f_y$ at $(0,0)$, assuming they exist (although I believe they don't), of the function $f(x,y)= \sqrt {xy}$

when I use the chain rule, I get partial derivatives that would result in ${0\over 0}$

e.g. $y\over 2\sqrt{\left(xy\right)}$ and $x\over 2\sqrt{\left(xy\right)}$

I assume I need to use the first principle approach to find a valid answer.

So far I have, for $f_x$ using the point $(0,0)$

$f_x =\lim_{h\to 0}{\sqrt {\left((x+h)y\right)} \ - \sqrt{\left(xy\right)}\over{h}}$

$=\lim_{h\to 0}{\sqrt{(\left(0+h)0\right)} \ - \sqrt{\left(0\right)}\over{h}}$

which is the same as

$=\lim_{h\to 0}{\sqrt0-\sqrt0\over{h}}$

which equals zero

Is this a valid way to find that the partial derivative for $f_x = 0$? Is it even zero?

I have followed the same method for $f_y$

Am I on the correct track here or am I barking up the wrong tree?

Answer 1

The partial derivatives of $f$ at the origin are both $0$, and you calculated $f_x(0,0)$ correctly. $f_y$ follows in more or less the same manner.