Show total differentiability of $\phi: \mathbb R^n \to \mathbb R^n, x \mapsto \varphi(\lVert x\rVert_2) x$ where $\varphi$ is differentiable

Question

Let $\varphi$ be differentiable. Show that $$\phi: \mathbb R^n \to \mathbb R^n, x \mapsto \varphi(\lVert x\rVert_2) x$$ is (total) differentiable where $x \neq 0$.

How can I show this? I know that $\varphi(\lVert x \rVert_2)$ is differentiable by the chain rule but I don't know any "multidimensional product rule". How can I show differentiability instead?

Answer 1

Quick Proof

We know that

$\varphi$ is differentiable. $$\|\cdot \|:x\to \|x\|$$ is differentiable everywhere except at zero.

By composition :

$$ \phi=(\varphi \circ \| \cdot \|) ×Id$$

is differentiable.

Definition proof

$f(x) \triangleq D(\| \cdot \|) (x)$ exits for all non null $x$

$\phi(x+h) =\varphi(\|x+h\|)(x+h) $

Yet

$\varphi(\|x+h\|)=\varphi(\|x\|+f(x)(h) +o(\|h\|)) =\varphi(\|x\|) +D\varphi(\|x\|)(f(x)(h) +o(h)) $

Can you finish from here?