Why Does This Partial Derivative Exist Everywhere?

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.

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.

• "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. – A Slow Learner Jan 19 at 9:53
• @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. – Henno Brandsma Jan 19 at 11:00
• so the only way to compute $f_x(0,0)$ is to set $y=0$? – A Slow Learner Jan 19 at 11:20
• @ASlowLearner indeed. – Henno Brandsma Jan 19 at 11:21
• 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? – A Slow Learner Jan 19 at 11:25

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 $$f_x$$.