Prove that if an even function $f(x)$ is $C^2$, $f(|x|)$ is also $C^2$ Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a $C^2(\mathbb{R})$ function which is also even (ie, $f(-x) = f(x)$). Prove that the function $F: \mathbb{R^n} \rightarrow \mathbb{R}$ defined by $F(x) = f(|x|)$ (where $|x|$ denotes the Euclidean norm of the vector $x$) is also $C^2(\mathbb{R^n})$.

It feels so simple, but I'm just not getting it, so any help would be appreciated.

Thanks!
Hint $F(x)=f(x)$ for $x\ge 0$, and for $x&lt;0$ we have $F(x)=f(|x|)=f(-x)=f(x)$, i.e. $F(x)=f(x)$ for all $x\in \mathbb{R}$
Hint $F(x)=f(x)$ for $x\ge 0$, and for $x&lt;0$ we have $F(x)=f(|x|)=f(-x)=f(x)$, i.e. $F(x)=f(x)$ for all $x\in \mathbb{R}$