Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

$$\frac{df}{dx}(0,0) = \frac{d}{dx}f(x,0)\vert_{y=0} = 0$$ 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.

share|cite|improve this question
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
up vote 5 down vote accepted

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 $x\mapsto f(x,0)$.

share|cite|improve this answer
+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
Thanks for the great answer. I wonder what is method 1 telling me exactly, anyway, i think i should do fine if i just apply method 2/3 whenever i need to determine P.D. – adsisco Sep 14 '11 at 14:12
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
what about f(x,y) = [x^2 - y^2]/[x^2 + y^2] when x^2 + y^2 >0 and = 0, when x=y=0 Seems like method 2 and 3 are giving me different solution. Its undefined when using limit definition but 0 when using the restricted method. – adsisco Sep 14 '11 at 15:39
Well, what about it? Since f(x,0)=1 for every x, methods 2 and 3 do concur. – Did Sep 14 '11 at 15:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.