If $f:\mathbb{R}^{m}-\lbrace 0 \rbrace \longrightarrow \mathbb{R}$ is defined by $f(x) = |x|^{a}$, with $a \in \mathbb{R}$, then $df(x).v=a|x|^{a-2}a\langle x,v\rangle$ for all $v \in \mathbb{R}^{m}$.
$|.|$ is euclidian norm.
I tried writte:
$$f(x+h) - f(x) - a|x|^{a-2}a\langle x,h\rangle = r(h)$$
and show that $\displaystyle \lim_{h\to 0}\frac{r(h)}{|h|} = 0$. But,I could not find the correct manipulation for this.
Does this work? Any hint? Thank for the advance!
UPDATING
I still cannot use grad... this will be set later!
The partial derivatives of $f$ all exist and are continuous away from the origin, so $f$ is differentiable away from the origin and its gradient is given by $ df(x) = [\partial_1 f(x), \ldots, \partial_m f(x)]$. These are easy to compute:
$$\frac{\partial f}{\partial x_k}(x) = \frac{\partial}{\partial x_k} \left( x_1^2 + \cdots + x_m^2 \right)^{a/2} = \frac a2 \left( x_1^2 + \cdots + x_m^2 \right)^{a/2-1}2x_k = a |x|^{a-2} x_k$$
so that $$df(x) = a |x|^{a-2} [x_1,\ldots,x_m].$$