Differentiability of $\sqrt{\frac{2x^4+2y^4+2z^4+3x^2y^2+3y^2z^2}{(x^2+y^2+z^2)^2}}$

Question

Let $$f(x,y,z)=\sqrt{\frac{2x^4+2y^4+2z^4+3x^2y^2+3y^2z^2}{(x^2+y^2+z^2)^2}}.$$

Is $f\in C^2(\mathbb R^3\setminus\{0\})$?

My thoughts: According to wolfram alpha we have $f(x,y,z)\in[1,\sqrt{2}]$. The only points where $f$ wouldn't be differentiable are given by the roots of the fraction. Everywhere else is the function $f$ obviously differentiable. Since $f\geq1$ everyhwere, we don't have any roots and the function $f$ is everywhere differentiable.

Then I've tried to consider the range of the partial derivatives to see if they are continuous and differentiable, but since I can't find the range and it's pretty difficult to find all the roots, I am asking myself if there is a easier way to verifiy $f\in C^2(\mathbb R^3\setminus\{0\})$.

My question: How can you verify $f\in C^2(\mathbb R^3\setminus\{0\})$ easily?

Answer 1

Hint.

Changing to spherical coordinates

$$ x = r\sin\theta\cos\phi\\ y = r \sin\theta\sin\phi\\ z = r\cos\theta $$

we obtain after simplifications

$$ f(x,y,z) \equiv g(\theta,\phi) $$

so $f(x,y,z)$ is not continuous at $(0,0,0)$ but $g(\theta,\phi)\in \mathbb{C}^{\infty}$ then in $\mathbb{C}^3\setminus \{0\}$ we have using the chain rule

$$ \frac{\partial f}{\partial x_i} = \frac{\partial g}{\partial\theta}\frac{\partial\theta}{\partial x_i}+\frac{\partial g}{\partial\phi}\frac{\partial\phi}{\partial x_i} $$

and those operations are well defined and continuous for $\sqrt{x^2+y^2+z^2}> 0$

etc.

A glimpse at $g(\theta,\phi)$

Answer 2

Hint: In fact $f$ is $C^\infty$ on this domain. To see this, note the numerator is a positive polymomial raised to the $1/2$ power. The denominator is a positive polynomial on this domain.