Showing a function is differentiable at $x$. Let $v: \mathbb{R^n} \rightarrow \mathbb{R^m}$ be a function such that $v(y) \neq 0,\forall y \in \mathbb{R^n}$ that is differentiable at $ x \in \mathbb{R^n}$. 
Show that $v/|v|:\mathbb{R^n}\rightarrow \mathbb{R^m}$ is differentiable at $x$ and that $$\triangledown (\frac{v}{|v|})(x)=\frac{1}{|(x)|^3}(|v(x)|^2\triangledown v(x)-v(x) \otimes (\triangledown v(x))^Tv(x))$$
I am unsure as how to go about this problem. I was thinking the chain rule, but I am not sure as how to apply it.
 A: You indeed need to use chain rule. I'll show part of the calculation because it sounds like you've never seen something like this before. It's also poorly explained in a lot of beginner textbooks.

Derivative: If $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$, then $\nabla f(x)$ is a linear map $\mathbb{R}^n \rightarrow \mathbb{R}^m$ such that
  $$
\nabla f(x)u = \lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon} (f(x + \epsilon u) - f(x))
$$

It's very important, in my opinion, to view $\nabla f(x)$ as a linear map. It makes the chain rule make sense as a composition of maps:

Chain rule: If $h \circ f$ is defined and both $h,f$ are differentiable, then
  $$
\nabla(h \circ f)(x) = \nabla h(f(x)) \circ \nabla f(x).
$$

For your problem, you need to compute the derivative of $n(v(x))$, where $n$ is the normalization map $n(v) = v/|v|$. You can probably find a formula for $\nabla n$ in textbooks or online, but I will compute it here so you can see how the process works for vector functions.
\begin{align}
\frac{1}{\epsilon} (n(v + \epsilon u) - n(v)) & = \frac{1}{\epsilon} \left( \frac{v + \epsilon u}{|v + \epsilon u|} - \frac{v}{|v|} \right) \\
& = \frac{1}{\epsilon} \frac{(v + \epsilon u) |v| - v|v + \epsilon u|}{|v + \epsilon u||v|}
\end{align}
Now, write $u = u_{\parallel} + u_{\perp}$, the parallel and perpendicular parts to $v$, respectively. Then
$$
|v + \epsilon u| = \sqrt{ (|v| + \epsilon |u_{\parallel}|)^2 + (\epsilon |u_{\perp}|)^2},
$$
so by expanding this in Taylor series we obtain
$$
|v + \epsilon u| = |v| + \epsilon |u_{\parallel}| + O(\epsilon^2)
$$
Returning to our previous calculation, we obtain
\begin{align}
\frac{1}{\epsilon} \frac{(v + \epsilon u) |v| - v|v + \epsilon u|}{|v + \epsilon u||v|} & = \frac{1}{\epsilon} \frac{ v|v| + \epsilon u |v| - v|v| - \epsilon v |u_{\parallel}| + O(\epsilon^2)}{ |v + \epsilon u||v|} \\
& = \frac{u|v| - v|u_{\parallel}| + O(\epsilon)}{ |v + \epsilon u||v|}.
\end{align}
Now, notice that $|u_{\parallel}| = v^T u / |v|$. Then, taking the limit as $\epsilon \rightarrow 0$ yields
$$
\nabla n(v)u = \frac{ |v|^2 u - v v^T u}{|v|^3} = \frac{ |v|^2 u - v \otimes (u^T v)}{|v|^3}
$$
Try and see if you can follow this calculation. It can also be much more easily calculated if you first prove the appropriate generalization of the Quotient Rule for Derivatives. Then, after this, apply the chain rule as indicated above to obtain the result you need.
