# On Partial Derivative

I am having issues with this mathematical concept, but i couldn't point out where, I've tried rereading thomas and stewart for more than 3 times already but still had no clue. I'll try to explain my thought process and will anyone of you please point out my mistakes?

given a function $f(x,y)=|xy|^{0.5}$, how do we determine if the partial derivative is defined at a point, lets say (0,0)? I've figured out there are 3 methods.

method 1:
$$\frac{d}{dx} |xy|^{0.5} =\frac{(x^2*y^2)^{1/4}}{2x}$$
then at $(0,0)$, it is undefined due to division-by-zero, therefore the P.D is undefined.

method2:
$f(x,0)=0$
$$\frac{df}{dx}(0,0) = \frac{d}{dx}f(x,0)\vert_{y=0} = 0$$ then at $(0,0)$, $f_x(0,0) = 0$. (P.D is defined)

method 3:
$$\lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h} = 0$$
therefore, $f_x(0,0) = 0$ (P.D is defined)

So what the differences between these method?? Did i apply those methods correctly? If so, why do i have contradicting solutions?

Thanks a lot for the help.

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The reason that Method 1 fails is that the equality is only valid for $xy \neq 0$, so it doesn't make sense to try to evaluate it at $(0,0)$. – Hans Lundmark Sep 14 '11 at 12:54

Methods 2 and 3 are correct. The partial derivative of $f$ with respect to $x$ at a given point $(x_0,y_0)$ involves the function $g:x\mapsto f(x,y_0)$ and only this function (and moreover, only the values of $g$ in a neighborhood of $x_0$). The values of $f$ at $(x,y)$ with $y\ne y_0$ are simply not relevant (nor its values at $(x,y)$ for $y=y_0$ and $|x-x_0|\ge\varepsilon$ for any given positive $\varepsilon$).
As pointed out by @Srivatsan Narayanan in a comment, method 3 is essentially the same as method 2. To see why, unroll the definition of the derivative of the single-variable function $x\mapsto f(x,0)$.
+1. @adsisco Actually, it's also worth pointing out that method 3 is essentially the same as method 2. To see why, just unroll the definition of a derivative of the single-variable function $x \mapsto f(x,0)$. – Srivatsan Sep 14 '11 at 11:30
The trouble you meet trying to use method 1 is a hint that $f$ might not be differentiable at $(0,0)$ (as is the case). – Did Sep 14 '11 at 14:18