Gradient of dot product using suffix notation

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Homework Help Overview

The discussion revolves around finding the gradient of the expression \underline{\nabla}(\underline{a}\cdot\underline{r})^n, where \underline{a} is a constant vector, using suffix notation and the chain rule. The problem is situated in the context of vector calculus and involves differentiation techniques.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the differentiation of the expression, with some questioning the validity of applying the power rule in the context of the dot product. There is a focus on the correct application of the chain rule and the treatment of the inner product in the differentiation process.

Discussion Status

Participants are actively engaging with each other's reasoning, with some providing clarifications and corrections regarding the differentiation steps. There is a recognition of potential confusion in the notation and expressions used, and suggestions for more elegant formulations are being discussed.

Contextual Notes

There is an indication of uncertainty regarding the application of the chain rule and the treatment of the inner product in the context of the problem. Participants are also reflecting on the implications of their notation choices and the clarity of their expressions.

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Homework Statement


Find the gradient of \underline{\nabla}(\underline{a}\cdot\underline{r})^n where a is a constant vector, using suffix notation and chain rule.

Homework Equations


On the previous problem,s I found that grad(a.r)=a and grad(r)=\underline{\hat{r}}

The Attempt at a Solution


<br /> \underline{e_i}(\frac{\partial }{\partial x_i})(\underline{a}\cdot\underline{r})^n=\underline{e_i}(\frac{\partial }{\partial x_i}(a_jx_j)^n)=\underline{e_i}(n(a_jx_j)^{n-1}(a_j\delta_{ij}))<br />
I'm sure that the last step is wrong so could someone lead me to the right direction?
Thank you!
 
Last edited:
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The derivative of (ax)^n is NOT n(ax)^{n-1}.
 
HallsofIvy said:
The derivative of (ax)^n is NOT n(ax)^{n-1}.
That is what I thought. Could you explain that step?
 
Just after I wrote that I noticed that you also had "a_j\delta_{ij}". The derivative of (ax)^n= a^nx^n is na^nx^{n-1} but that can also be written as (ax)^{n-2}(a) where the last "a" is due to the chain rule. Was that what you meant?
 
##na^nx^{n-1}## is confusing. The inner product ##(a_j x_j)^{n-1}## is lost.

When I write it out and then put it back together again ##
\underline{e_i}(n(a_jx_j)^{n-1}(a_j\delta_{ij}))## seems OK.
##
na_i \underline{e_i}\;(a_jx_j)^{n-1}## might be somewhat more elegant,
and even that can be simplified further !
 
Last edited:

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