# Showing $\frac{x^2y^3}{x^2+y^4}$ is differentiable everywhere, and if the first order partials are continuous

So the function is defined as above, except at (0,0) the function is defined to be 0. So I computed the first order partial derivatives, and it's obviously continuous at all points other than (0,0), and then I used the definition of the limit to compute the partial derivatives by x and y at (0,0), which are both 0.

Now I'm trying to satisfy the condition $\frac{||f(x,y)-f(0,0)-T(0,0)||}{||h||}$, which could be simplified to $\frac{x^2y^3}{(x^2+y^4)\sqrt{x^2+y^2}}$, and now I'm stuck at showing that this goes to 0 as h goes to zero (where h is (x,y)).

Also, could someone just let me know how to show that the partials are continuous? I don't really know how to show they're continuous at (0,0). Thanks in advance!

• To show something is continuous at $(0,0)$, compute the limit at $(0,0)$. Can you compute limits?
– Pedro
Dec 5, 2014 at 21:05
• The usual trick for these sorts of limits is polar coordinates. In your case you have $\lim_{r \to 0} \frac{r^5 \cos(\theta)^2 \sin(\theta)^3}{r \left ( r^2 \cos(\theta)^2 + r^4 \sin(\theta)^4 \right )}$. You need to check whether this depends on $\theta$. Does it? (Note that this actually isn't enough to check that the limit exists, but if it does depend on $\theta$ then the limit definitely doesn't exist.)
– Ian
Dec 5, 2014 at 21:06
• Using l'hopitals rule I obtained $\frac{cos^2(\theta)}{sin(\theta)}$, so it does depend on $\theta$ if I did it right, but I'm supposed to prove that it is differentiable and this contradicts this right?
– Jeff
Dec 5, 2014 at 21:19

Let $m=\min(|x|,|y|).$

Note that

$$\frac{|x||y|}{\sqrt{x^2+y^2}} \leqslant m.$$

Hence,

$$0 \leqslant \frac{|x|^2|y|^3}{(x^2+y^4)\sqrt{x^2+y^2}} =\frac{|x||y|^2}{(x^2+y^4)} \frac{|x||y|}{\sqrt{x^2+y^2}} \\ \leqslant \frac{|x||y|^2}{(m^2+m^4)} m =\frac{|x||y|^2}{(m+m^3)} \\\leqslant \frac{|x||y|^2}{m},$$

and

$$\lim_{(x,y)\rightarrow (0,0)}\frac{|x||y|^2}{\min(|x|,|y|)}= \lim_{(x,y)\rightarrow (0,0)}\left(|y|^21_{\{|x|\leqslant|y|\}} + |x||y|1_{\{|x|>|y|\}}\right) = 0.$$

By the squeeze principle,

$$\lim_{(x,y)\rightarrow(0,0)} \frac{|x|^2|y|^3}{(x^2+y^4)\sqrt{x^2+y^2}} = 0,$$

and the derivative at $(0,0)$ is the zero linear operator.