Computing the directional derivative of a functional

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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.

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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.

3D

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*}