What is the dirctional derivative of a norm?

Question

Usually norm is not differentiable function. But norm has directional derivative. What is the form dirctional derivative of a norm in a direction d?

Answer 1

The norm$$f(x):=\|x\|_2=\sqrt{x_1^2+x_2^2+\ldots+x_n^2}$$ is differentiable at all points $x\ne0$. One has $${\partial f\over\partial x_i}={2x_i\over 2\sqrt{x_1^2+x_2^2+\ldots+x_n^2}}={x_i\over\|x\|_2}\qquad(1\leq i\leq n)\ ,$$ and therefore $$\nabla f(x)={x\over\|x\|_2}\qquad(x\ne0)\ .$$ If $u$ is a given unit vector representing a certain direction then $$D_uf(x):=\lim_{t\to0+}{f(x+t u)-f(x)\over t}=\nabla f(x)\cdot u={x\cdot u\over\|x\|_2}\ .$$ The directional derivatives as defined here exist even at $0$: One has $$D_uf(0)=1$$ for all unit vectors $u$.