# Show that function $f(a,b)$ is differentiable

Show that function

$f: \mathbb{R}^2 \rightarrow \mathbb{R}$

$f(a,b) = \dfrac{a^3+b^3}{\sqrt{a^2+b^2}}$ if $(a,b)\in\mathbb{R}^2 -\{(0,0)\}$

and $0$ if $(a,b)=(0,0)$

is differentiable on $\mathbb{R}^2$. Is it class $C^1$ on $\mathbb{R}^2$?

As far as I understood I need to calculate partial derivatives and prove that they exist and are continuous.However for both partials I get not really pretty fractions which I'm not sure if are continuous.

Also will appreciate any help on how to start proof on continuous differentiability(class $C^1$).

• It is not sufficient to just check the partial derivatives, I suggest you look over the definition once more, probably there are examples of non-C^1 functions too. – AD. Mar 3 '15 at 8:39
• There is an example on wikipedia en.wikipedia.org/wiki/Function_of_several_real_variables – AD. Mar 3 '15 at 8:41
• It is indeed sufficient to check the partials exist and are continuous. Compare this question which references a relevant theorem of Rudin: math.stackexchange.com/questions/563680/… – Jason Mar 3 '15 at 8:47

Let's calculate the partial derivatives. Given the symmetry, it is sufficient to look at the first coordinate. For $(x,y)\ne(0,0)$, we have $$f_x(x,y)=\frac{3x^2(x^2+y^2)^{1/2}-x(x^2+y^2)^{-1/2}(x^3+y^3)}{x^2+y^2}=\frac{3x(x^2+y^2)-(x^3+y^3)}{(x^2+y^2)^{3/2}}x$$ which is clearly continuous for $(x,y)\ne(0,0)$. Moreover, $$\lim_{(x,y)\to(0,0)}f_x(x,y)=\lim_{r\to0}\frac{3r^3\cos\theta-r^3(\cos^3\theta+\sin^3\theta)}{r^3}r=0.$$ Now we just need to verify the partials are zero at the origin: $$f_x(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}{h}=\lim_{h\to0}\frac{\frac{h^3}{|h|}-0}{h}=0$$ so the partials indeed do exist and are continuous everywhere.
• So it IS class $C^1$ function? – John Lennon Mar 3 '15 at 9:06
• Yes, $f\in C^1(\mathbb R^2)$. – Jason Mar 3 '15 at 9:06
For any point $(a,b)\ne(0,0)$ we have $$f'_a(a,b)=\frac{3a^2(a^2+b^2)-a(a^3+b^3)}{(a^2+b^2)^{3/2}},$$ and $$f'_a(0,0)=0.$$ So $f'_a$ is continuous at every point different from $(0,0)$. Now one just need to show that $$\lim_{(a,b)\rightarrow(0,0)}f'_a(a,b)=0,$$ which is easy to show if we put $$a=r\cos\phi,b=r\sin \phi.$$