Showing that $\boldsymbol{\nabla}r^n=nr^{n-2}\mathbf{x}$

Question

In three dimensions, use suffix notation and the summation convention to show that $$\boldsymbol{\nabla}r^n=nr^{n-2}\mathbf{x},$$ where $\mathbf{a}$ is any constant vector and $r=|\mathbf{x}|.$

My try: $\nabla f=\frac{\partial f}{\partial x_i}\mathbf{e}_i$ and $r=\sqrt{x_i x_i}=\sqrt{x_1^2+x_2^2+x_3^2}$. Computing $i$-th component of $\nabla r^n$ we get $\left(\nabla r^n\right)_i$.

Using chain rule, $\frac{\partial}{\partial x_{i}}(r^{n})=nr^{n-1}\frac{\partial r}{\partial x_{i}}$. Since $r=|\mathbf{x}|, nr^{n-1}\frac{\partial}{\partial x_{i}}|\mathbf{x}|$. Then I get $\frac{\partial|\mathbf{x}|}{\partial x_{i}}=\frac{\partial r}{\partial x_{i}}=\frac{x_{i}}{r}=\frac{x_{i}}{|\mathbf{x}|}$. I can't really show $ \frac{\partial|\mathbf{x}|}{\partial x_i}=\mathbf{x}$.

Maybe I missed some minor yet crucial details? I am assuming this what I used $x_i=\mathbf{x}=(x_1,x_2,x_3)$ is correct.

Answer 1

You got, $$ \frac{\partial (r^n)}{\partial x_i}=nr^{n-1}\frac{\partial r}{\partial x_i}. $$ Now, $$ \frac{\partial r^2}{\partial x_i} = \frac{\partial (\vec x\cdot \vec x)}{\partial x_i} \\ \implies r\frac{\partial r}{\partial x_i}=\vec x\cdot\frac{\partial \vec x}{\partial x_i}=\vec x\cdot\vec e_i=x_i. $$ Combining above results, $$ \frac{\partial (r^n)}{\partial x_i}=nr^{n-2}\times r\frac{\partial r}{\partial x_i} \\ \implies \frac{\partial (r^n)}{\partial x_i}=nr^{n-2}x_i. $$ So, we can conclude that $\nabla (r^n)=nr^{n-2}\vec x$.

Answer 2

Observe that it is $nr^{n-2}\vec{x}$ not $nr^{n-1}\vec{x}$. ($nr^{n-1}\nabla \vec{x}=nr^{n-1}\frac{x_i \vec{e_i}}{r})$). You have solved it correctly... congrats!