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The following function is taken from my textbook example. \begin{cases} f(x,y)=\frac{2xy^2}{x^2+y^4}, &(x,y) \neq (0,0)\\ f(x,y)=0, &(x,y)=(0,0) \end{cases} My textbook asserts that the partial derivatives of $f$ exists everywhere, which I do not understand. I tried to solve this myself, but it seems like I am misunderstanding something because my work shows some inconcsistencies:

To calculate $f_x$ at $(0,0)$: $$ \lim_{x \to 0}\frac{f(x,y)-f(0,0)}{x}=\lim_{x \to 0} \frac{2y^2}{x^2+y^4}=\frac{2}{y^2} \to \infty \text{ as } y \to 0 $$ However, if I fix $y=0 \to f(x,0)=0 \space \forall x$, then: $$ \lim_{x \to 0, y=0}\frac{f(x,0)-f(0,0)}{x}=0 $$

Does this mean $f_x$ does not exist at $(0,0)$? But this contradicts with my textbook's claim.

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For $f_x(0)$ you must compute $\lim_{x \to 0} \frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis); the first computation is irrelevant. It's clear that $f(x,0) = 0$ for all $x \neq 0$ from the formula and also $f(0,0)=0$, hence the limit and the partial derivative is just $0$.

Similarly the other partial derivative is $\lim_{y \to 0} \frac{f(0,y) - f(0,0)}{y}$, which is similarly seen to be $0$ too.

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  • $\begingroup$ "For $f_x(0)$ you must compute $\lim_{x \to 0} \frac{f(x,0) - f(0,0)}{x}$ instead (so $y=0$ all the time, because we approach along the $x$-axis)". Does this mean I can also fix, for instance, $y=3$, to compute $f_x$? This is essentially what the first calculation did: I wanted to approach (0,0) from an arbitrary horizontal line. $\endgroup$ – A Slow Learner Jan 19 at 9:53
  • $\begingroup$ @ASlowLearner If you take $y=3$ and the limit as $x$ tends to $0$ we have the partial derivative at $(0,3)$ in the $x$-direction. The partial derivative at points not $(0,0)$ can be computed by the formula. Only at $(0,0)$ do we need the explicit limits. $\endgroup$ – Henno Brandsma Jan 19 at 11:00
  • $\begingroup$ so the only way to compute $f_x(0,0)$ is to set $y=0$? $\endgroup$ – A Slow Learner Jan 19 at 11:20
  • $\begingroup$ @ASlowLearner indeed. $\endgroup$ – Henno Brandsma Jan 19 at 11:21
  • $\begingroup$ Ok. But looking back in my post, where $\frac{2}{y^2} \to 0$ as $y \to 0$. This limit should be equal to $0$ is $f_x(0,0) = 0$, as done in the second calculation. But it is not. Am I still missing something? $\endgroup$ – A Slow Learner Jan 19 at 11:25
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To compute the partial derivative of $f$ in $(x_0, y_0)$ with respect to $x$, you only vary $x$ around $x_0$ and keep $y=y_0$ the whole time.

Your second calculation is in fact $f_x(0,0)$: you fix $y=0$ and see how the differences behave when you send $x$ to $0$. So you've shown that $f_x$ exists in $(0,0)$.

The first calculation isn't anything standard, because you're varying $x$ but your $y$ is not fixed: you have $y=y$ in one term and $y=0$ in the other. For instance, changing $\lim_{x \to 0} \frac{f(x,y) - f(0,0)}{x}$ to $\lim_{x \to 0} \frac{f(x,y) - f(0,y)}{x}$ would give you $f_x(0,y)$ for arbitrary $y \ne 0$. Taking the limit of that for $y \to 0$ gives you $\lim_{y \to 0} f_x(0,y)$.

Note: this change does not affect the result, though. So you still end up getting $\lim_{y \to 0} f_x(0,y) = \infty$. That gives you insight into the (dis-)continuity of $f_x$.

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