$f:\mathbb{R^m} \rightarrow \mathbb{R}, f(x) = |x|^a, a\in \mathbb{R}$. Calculate the derivative of f. 
Let $f:\mathbb{R^m} \rightarrow \mathbb{R}, f(x) = |x|^a, a\in \mathbb{R}$ and $|\cdot |$ is the euclidean norm ($|x| = \sqrt{<x,x>}$). Show that $df(x)\cdot v = a |x|^{a-2}<x,v>$

$df(x)\cdot v = \frac{\partial f(x)}{\partial v} = \lim_{t \to 0}\frac{f(x+tv)-f(x)}{t}$ by definition.
$\lim_{t \to 0}\frac{f(x+tv)-f(x)}{t} = \lim_{t \to 0}\frac{|x+tv|^a-|x|^a}{t} =  \lim_{t \to 0}\frac{(<x+tv,x+tv>)^{a/2}-(<x,x>)^{a/2}}{t}$
$$=\lim_{t\to 0}\frac{(<x,x>+2t<x,v>+t^2<v,v>)^{a/2}-<x,x>^{a/2}}{t}$$
From this part I can't make anything else, any hints?
 A: If $x\ne 0$ then $f(x)=(x_1^2+\ldots+x_m^2)^{a/2}$ is differentiable at $x$ since it is given by the composition of differentiable functions. 
By the chain rule you have $\frac{\partial f}{\partial x_i}(x)=\frac{a}2 (x_1^2+\ldots+x_m^2)^{a/2-1}2x_i=a|x|^{a-2}x_i$ and so 
$df(x)(v)=\nabla f(x)\cdot v=a|x|^{a-2}\cdot v$. The only problem is $x=0$, where you need to use the definition. You have 
$$\frac{f(0+t e_i)-f(0)}{t}=\frac{|t|^a-0}{t}=\frac{|t|^a}{t}\to 0$$
iff $a>1$, while it does not exist if $a\le 1$. So you now assume $a>1$. Then $\frac{\partial f}{\partial x_i}(0)=0$ and so to study the differentiability at $x=0$ you consider
$$\frac{f(x)-f(0)-\nabla f(0)\cdot (x-0)}{|x|}=\frac{|x|^a-0-0}{|x|}=|x|^{a-1}\to 0$$
when $a>1$. Hence, $f$ is differentiable at $x=0$ iff $a>1$.
A: Starting with where you left off, L'Hôpital's Rule is useful:
\begin{align*}
    \lim_{t\to 0} \frac{\left(\left<x,x\right> + 2t \left<x,v\right> + t^2 \left<v,v\right>\right)^{a/2} - \left<x,x\right>^{a/2}}{t}
  &=\lim_{t\to 0} \frac{a}{2}\left(\left<x,x\right> + 2t \left<x,v\right> + t^2 \left<v,v\right>\right)^{a/2-1}\left(2\left<x,v\right> + 2t\left<v,v\right>\right)\\
  &=\frac{a}{2}\left<x,x\right>^{a/2-1}\cdot 2 \left<x,v\right> = a\left<x,x\right>^{(a-2)/2}\left<x,v\right>\\
  &=a \left|x\right|^{a-2} \left<x,v\right>
\end{align*}
If $x=0$, use the definition, just like Gio:
\begin{align*}
    \lim_{t\to 0} \frac{\left(\left<0,0\right> + 2t \left<0,v\right> + t^2 \left<v,v\right>\right)^{a/2} - \left<0,0\right>^{a/2}}{t}
    &= \lim_{t\to 0} \frac{\left(t^2\left<v,v\right>\right)^{a/2}}{t} \\
    &= \left<v,v\right>\lim_{t\to 0} \frac{|t|^a}{t}
\end{align*}
which is zero iff $a>1$.
