Find the differential of an n-variable function The problem goes like this:
If $f:\mathbb{R}^n\to\mathbb{R}, f(x)=\arctan||x||^4$, prove that $Df(x)(x)=\displaystyle\frac{4||x||^4}{1+||x||^8}$
Now, I've calculated each of the partial differentials (if that's the right word) and applied that $1\times n$ matrix to a vector $(x_1, ... ,x_n)$ and I get this:
$4(\displaystyle\frac{x_1^4}{1+x_1^8}+...+\frac{x_n^4}{1+x_n^8})$
Now, the similarity between those terms and the final solution is obvious but I just can't seem to get the sum to become the above.
Am I going about this the wrong way or am I just missing something?
($||\cdot||$ is the Euclidian norm)

$\mathrm{Arctan}$ is differentiable on $\mathbb{R}$ with :
$$ \big( \mathrm{Arctan}' \big)(x) = \frac{1}{1+x^{2}} $$
And $\varphi \, : \, x \in \mathbb{R}^{n} \, \longmapsto \, \Vert x \Vert^{4}$ is differentiable on $\mathbb{R}^{n}$ with :
$$ \mathrm{D}_{x}\varphi \cdot h = 4 \Vert x \Vert^{2} \left\langle x,h \right\rangle $$
where $\left\langle \cdot,\cdot \right\rangle$ denotes the usual inner product on $\mathbb{R}^{n}$. Since $f = \mathrm{Arctan} \circ \varphi$, the chain rule gives you :
$$ 
\begin{align*}
\mathrm{D}_{x}f \cdot h & = {} \frac{1}{1+\big( \Vert x \Vert^{2} \big)^{2}} \times \mathrm{D}_{x}\varphi \cdot h \\[2mm]
 &= \frac{4 \Vert x \Vert^{2} \left\langle x,h \right\rangle}{1+\Vert x \Vert^{8}} \\
\end{align*}
$$
Which gives you the expected result for $\mathrm{D}_{x}f \cdot x$. 
