Mathematics Asked on December 13, 2021
Let $varphi$ be differentiable. Show that $$phi: mathbb R^n to mathbb R^n, x mapsto varphi(lVert xrVert_2) x$$ 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?
Quick Proof
We know that
$varphi$ is differentiable. $$|cdot |:xto |x|$$ 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|) +Dvarphi(|x|)(f(x)(h) +o(h)) $
Can you finish from here?
Answered by EDX on December 13, 2021
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