 Solution

Differentiability at $p$ means there is a linear map $A=Df{|}_p$ such thatThis is the Fréchet derivative definition; the same linear approximation must work in every direction, not merely along coordinate axes.

The chain rule says that if $f$ is differentiable at $p$ and $g$ is differentiable at $f(p)$, then $g\circ f$ is differentiable at $p$ andFor its proof, put $A=Df{|}_p$, $B=Dg{|}_{f(p)}$, and $k=f(p+h)-f(p)=Ah+r(h)$. Since a linear map between these finite-dimensional normed spaces is bounded, $\Vert k\Vert =O(\Vert h\Vert )$. Differentiability of $g$ gives $g(f(p)+k)=g(f(p))+Bk+s(k)$ with $s(k)=o(\Vert k\Vert )$. ThereforeIf $k=0$, take $s(0)=0$; otherwise $\Vert s(k)\Vert /\Vert h\Vert =(\Vert s(k)\Vert /\Vert k\Vert )(\Vert k\Vert /\Vert h\Vert )\to 0$. This proves the required full linear approximation.

For the Euclidean norm, at every $x\ne 0$ the ordinary one-coordinate differentiation givesAt the origin its difference quotient in direction $e_i$ is $|t|/t$, whose right and left limits are $1$ and $-1$. Thus none of its coordinate partial derivatives exists at the origin.