Steps used to calculate the partial derivative x-component used to calculate divergence?

Question

Given a spherically symmetric vector field with amplitude increasing as the square of the distance from the origin.

Thus $$\vec A=r^2\hat r$$

$$r^2=(x^2 + y^2 + z^2)$$

$$\hat r =\frac {x\hat i+y\hat j+z\hat k}{\sqrt{ x^2 + y^2 + z^2}}$$

What are the steps required to get to this?

$${\partial A_x\over \partial x}= (x^2 + y^2 + z^2)^{(1/2)} +x(1/2)(x^2 + y^2 + z^2)^{(-1/2)}(2x)$$

Answer 1

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\hat{r} \stackrel{\mathrm{def.}}{=} {\,\,\vec{r} \over r}}$.

\begin{align} \color{#f00}{\partiald{A_{x}}{x}} & = \partiald{\pars{r^{2}\hat{r}}_{x}}{x} = \partiald{\pars{r\vec{r}}_{x}}{x} = \partiald{\pars{rx}}{x} = \overbrace{\partiald{r}{x}}^{\ds{x \over r}}\ \,x\ +\ r\ \overbrace{\partiald{x}{x}}^{1} = \color{#f00}{{x^{2} \over r} + r} \end{align}

Note that $\ds{r^{2} = x^{2} + y^{2} + z^{2}\ \imp\ 2r\,\partiald{r}{x} = 2x\ \imp\ \partiald{r}{x} = {x \over r}}$