Gradient 1D of the total variation regularization

Question

I try to calculate the gradient of the total variation regularization for a parameter which is a 1D scalar field, but I am doing something wrong... ???

Here is the latex code:

\documentclass{article}
\usepackage{mathtools}
\RequirePackage{amsmath}
\RequirePackage{amsthm}
\RequirePackage{amssymb}
\RequirePackage{dsfont}
\usepackage[dvipsnames]{xcolor}
\newcommand{\norm}[1]{\left\lVert#1\right\rVert}
\newcommand{\scal}[2]{\langle#1,#2\rangle}
\newcommand{\dpartial}[2]{\dfrac{\partial#1}{\partial#2}}
\newcommand{\dpartials}[3]{\dfrac{\partial^{#3} #1}{\partial #2 ^{#3}}}
\begin{document}
\begin{equation}
\begin{split}
G_{r} &= div \left( \dfrac{\vec{grad(m)}}{\norm{\vec{grad(m)}}_{1}} \right)\
&= div \left( \dfrac{\vec{grad(m)}}{\sqrt{\vec{grad(m)^{2}}}} \right) \
&= \dpartial{}{x} \left( \dfrac{\dpartial{m}{x}}{\sqrt{\left(\dpartial{m}{x}\right)^{2}}} \right) \
&= \dpartial{}{x} \left( \dfrac{\dpartial{m}{x}}{\sqrt{\left( \dpartial{m}{x}\right)^{2}}} \right) \
&= \dpartials{m}{x}{2} \dfrac{1}{\sqrt{\left( \dpartial{m}{x}\right)^{2}}}

\dpartial{m}{x} \dpartial{}{x} \left( \dfrac{1}{\sqrt{\left( \dpartial{m}{x}\right)^{2}}} \right) \

&= \dfrac{\dpartials{m}{x}{2} \left( \dpartial{m}{x}\right)^{2}}{\left(\left( \dpartial{m}{x}\right)^{2}\right)^{3/2}}

\dpartial{m}{x} \dpartial{}{x} \left( \dfrac{1}{\sqrt{\left( \dpartial{m}{x}\right)^{2}}} \right) \

&= \dfrac{\dpartials{m}{x}{2} \left( \dpartial{m}{x}\right)^{2}}{\left(\left( \dpartial{m}{x}\right)^{2} \right)^{3/2}}

\dpartial{m}{x} \left( - \dfrac{2 \dpartial{m}{x} \dpartials{m}{x}{2}}{2 \left(\left( \dpartial{m}{x}\right)^{2} \right)^{3/2}} \right) \

&= \dfrac{\dpartials{m}{x}{2} \left( \dpartial{m}{x}\right)^{2}}{\left(\left( \dpartial{m}{x}\right)^{2} \right)^{3/2}}

\dfrac{ \left( \dpartial{m}{x}\right)^{2} \dpartials{m}{x}{2}}{\left(\left( \dpartial{m}{x}\right)^{2} \right)^{3/2}} \

&= 0
\end{split}
\end{equation}
\end{document}