Computing the directional derivative of a functional I'm studying the numerical applications of the total variation using Vogel's "Computational methods forinverse problems", but I'm stuck with some (presumably easy) calculus issues.
At a certain point I have to compute the gradient of a functional $J$ defined by:
\begin{equation}
J(\mathbf f) = \frac{1}{2}\sum_{i=0}^n \psi \bigl((D_i \mathbf f)^2 \bigr) \Delta x \end{equation}
where $\psi$ is a smooth function such that $\psi'(t) > 0$ for $t>0$,
$\mathbf f$ is a vector
$D_i$ is the row of a matrix (so, a vector)
$\Delta x$ the length of an interval.
In order to compute the gradient, the author fistly computes the directional derivative: for any $\mathbf v \in \mathbb R^n$:
\begin{equation*}
\frac{d}{d \tau} J(\mathbf f + \tau \mathbf v) = \sum^n_{i=1} \psi'\left([D_i \mathbf f]^2\right)(D_i \mathbf f)(D_i \mathbf v)\Delta x
\end{equation*}
 and here he's lost me.
First (silly) question: why have the parentheses in the first formula become a square bracket in the last formula? (I've copied the formulas from the book, and I don't understand if it has a particular meaning)
And, more important, how do I get from formula 1 to formula 2?
Thanks in advance.
 A: In the probably more familiar case of 2 dimensions, there's only one direction to take the derivative in.
But if you imagine a three dimensional surface, differentiating at a point on that surface can be done along any direction.
 
So if you had a plane surface $G(x)$ (green in the diagram), that you wanted to differentiate at a point called $A$, then you need to also choose a direction  $h$ (or $v$ in your case) to differentiate along. 
The red line is $A + \tau h$
So you differentiate along the straight line through $h$.
$\frac{d}{d\tau} G(A + \tau h)$
EDIT
See here for the directional derivative
$D G(A)[v] = [\frac{d}{d\tau} G(A + \tau v)]_{\tau=0}$
or in our case
\begin{equation*} \frac{d}{d \tau} J(\mathbf f + \tau \mathbf v) = [ \frac{1}{2}\frac{d}{d \tau} \sum^n_{i=1} \psi\left([D_i (\mathbf f + \tau \mathbf v) ]^2\right)\Delta x ]_{\tau=0} \end{equation*}
Which becomes
\begin{equation*} \frac{d}{d \tau} J(\mathbf f + \tau \mathbf v) = [\sum^n_{i=1} \psi'\left([D_i (\mathbf f + \tau \mathbf v) ]^2\right)(D_i (\mathbf f + \tau \mathbf{v}))(D_i \mathbf v)\Delta x ]_{\tau=0} \end{equation*}
\begin{equation*} \frac{d}{d \tau} J(\mathbf f + \tau \mathbf v) = \sum^n_{i=1} \psi'\left([D_i \mathbf f  ]^2\right)(D_i \mathbf f )(D_i \mathbf v)\Delta x  \end{equation*}
