Equivalent conditions for differentiability of radially symmetric functions Let $f : B(0, 1) \subset \mathbf R^n \to \mathbf R$ be a $C^k$-function, namely all partial derivatives up to order $k$ are continuous. Suppose that $f$ is also radially symmetric with respect to the origin $0 \in \mathbf R^n$, namely
$$f(x) = f(y) \quad \text{for all } \, x,y \in B(0,1) \, \text{ satisfying 
 }\,|x|=|y|.$$
Having this property, $f$ can also be regarded as a single-variable function, say of the distance $r$, in $[0,1)$ because $f(x) = f(|x|)$; then we can talk about $f^{(i)}$ with $1 \leq i \leq k$ without difficulty. My question is can we have an equivalent set of conditions for the $C^k$-differentiability of $f$ in terms of $f$ and its all derivative $f^{(i)}$ with $1 \leq i \leq k$ in $[0,1)$? For example, if $f$ is continuous, then $f$ must be continuous in $[0,1)$. My expectation is that if $f$ is of class $C^k$, then for each $1 \leq i \leq k$, the derivative $f^{(i)}$ exists in $(0,1)$ and the limit $$\lim_{r \searrow 0} f^{(i)}(r)$$ exists.

First a comment: you should not have the same notation for $f(x)$ and $f(\lvert x \rvert)$. Below, I denote by $F$ the map $F:\mathbb R^n \to \mathbb R$ and by $f$ the map from $\mathbb R$ to $\mathbb R$.
Coming back to the main point of your question, you would also need that $f^{(i)}(0^+)=0$ for $1 \le i \le k$.
For example for $n=k=1$ with $f(x)=x$, you have $F(x) = f(\lvert x \rvert) = \lvert x \rvert$ which is not differentiable at $0$.
However if $f^{\prime}(0^+)=0$, then $F : \mathbb R^n \to \mathbb R$ is differentiable everywhere. This is trivial for $0 \neq x \in \mathbb R^n$. And around $0$, you can write
$$f(t) = f(0) + t \epsilon(t) \text{ with } \lim\limits_{h \to 0^+} = 0.$$ Hence
$$\frac{F(x) - F(0)}{\lvert x \rvert} = \epsilon(\lvert x \rvert)$$ converges to $0$ as $x \to 0$, proving that $F$ is differentiable at $0$ and that its derivative vanishes there.
You can proceed by induction for $k \gt 1$.
