How to compute derivatives of functions with vectors inside? Suppose $\vec{w}=\frac{g}{||\vec{v}||} \vec{v}$, what is the derivative of $\vec{w}$ w.r.t. $\vec{v}$?
Don't know how to deal with the norm of $\vec{v}$ here...
Thanks in advance. :-)
Edit: 
$L$ is a function of $\vec{w}$ and $g$. Based on $\vec{w}=\frac{g}{||\vec{v}||} \vec{v}$, we have
$$\nabla{g}{L}=\frac{\nabla{\vec{w}}{L} \cdot \vec{v}}{||\vec{v}||}$$
$$\nabla{\vec{v}}{L}=\frac{g}{||\vec{v}||}\nabla{\vec{w}}{L}-\frac{g\nabla{g}{L}}{||\vec{v}||^2}\vec{v}$$
Could you show how to get exactly the second equation? It seems a bit weird to me.
 A: Hint.
Apply the chain rule, using the fact that $$f(\vec{v})=\frac{\vec{v}}{||\vec{v}||} = F(\vec{v})\vec{v}$$ where $$F(\vec{v})= \frac{1}{\sqrt{\Vert \vec{v} \Vert^2}}$$ and $$h(\vec{v})=\Vert \vec{v} \Vert^2= (\vec{v},\vec{v})$$ is a bilinear map so its Fréchet derivative is $$h^\prime(\vec{v}).\vec{r} = 2 (\vec{v},\vec{r})$$ and applying the chain rule $$F^\prime(\vec{v}).\vec{r}=-\frac{(\vec{v},\vec{r})}{\Vert \vec{v} \Vert^3}$$
Applying again the chain rule to $f$:
$$f^\prime(\vec{v}).\vec{r}=(F^\prime(\vec{v}).\vec{r})\vec{v} + F(\vec{v})\vec{r}$$ you finally get
$$f^\prime(\vec{v}).\vec{r} = -\frac{(\vec{v},\vec{r})}{\Vert \vec{v} \Vert^3}\vec{v}+\frac{\vec{r}}{\Vert \vec{v} \Vert}$$
If $g$ is also a map depending on $\vec{v}$, you need to applying the chain rule once more.
A: The ''derivative'' of a vector function of a vector is not a single number, but a matrix that contains all the partial derivative of the vector function with respect to the components of the independent vector, called the Jacobian matrix.
In your case, supposing that, $\vec W$ and $\vec v$ are vectors of   a $n-$ dimensional real vector space, we have:
$$
\frac{\partial \vec w}{\partial \vec v}=
\begin{bmatrix}
\frac{\partial w_1}{\partial  v_1}&\frac{\partial w_1}{\partial  v_2} &\cdots&\frac{\partial w_1}{\partial  v_n}\\
\frac{\partial w_2}{\partial  v_1}&\frac{\partial w_2}{\partial  v_2} &\cdots&\frac{\partial w_2}{\partial  v_n}\\
\cdots\\
\frac{\partial w_n}{\partial  v_1}&\frac{\partial w_n}{\partial  v_2} &\cdots&\frac{\partial w_n}{\partial  v_n}\\
\end{bmatrix}
$$
Using this definition you can evaluate the elements of the matrix for your function. If $n=2$,  we have:
$$
\begin{bmatrix}
w_1\\
w_2
\end{bmatrix}
=
\begin{bmatrix}
\frac{gv_1}{\sqrt{v_1^2+v_2^2}}\\
\frac{gv_2}{\sqrt{v_1^2+v_2^2}}
\end{bmatrix}
$$ 
so we can find the partial derivatives as:
$$
\frac{\partial w_1}{\partial v_1}=\frac{gv_2^2}{\sqrt{\left(v_1^2+v_2^2\right)^3}}
\qquad
\frac{\partial w_1}{\partial v_2}=\frac{-gv_1v_2}{\sqrt{\left(v_1^2+v_2^2\right)^3}}
$$
and so one...  and you can do the same if $n>2$.
