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I am trying to compute the partial derivative of $|Du|^{p-2}$ with respect to the variables $x_{i}$ but the index notation is kinda wrecking my thought process. If one converts $|Du|^{p-2}$ to summation notation, then we should have

$$|Du|^{p-2}=\left(\sum_{i=1}^{n}u_{x_i}u_{x_i}\right)^{(p-2)/2}$$

Then the partial derivative of this with respect to $x_{i}$ would be

$$\frac{\partial}{\partial x_{i}}|Du|^{p-2}=\frac{p-2}{2}\left(\sum_{i=1}^{n}u_{x_i}u_{x_i}\right)^{(p-4)/2}\left(\frac{\partial}{\partial x_{i}}\sum_{i}^{n}u_{x_i}u_{x_i}\right)$$

Recasting the sum in terms of $D$ and taking the derivative of the sum on the right we get we get

$$\frac{\partial}{\partial x_{i}}|Du|^{p-2}=\frac{p-2}{2}|Du|^{p-4}\left(\sum_{i}^{n}2u_{x_ix_i}u_{x_i}\right) = (p-2)|Du|^{p-4}\left(\sum_{i}^{n}u_{x_ix_i}u_{x_i}\right)$$

If I've gotten this correct, recasting the sum on the right in terms of $D$ has me lost. Any ideas from here would be greatly appreciated!

MrStormy83
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    I think the first equality in the computation is incorrect, since $(f^k)'=kf^{k-1}f'$. – Mr. Brown Feb 15 '23 at 20:21
  • Oh dang, I totally forgot to utilize the chain rule . . . I'll edit that. – MrStormy83 Feb 15 '23 at 20:22
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    Your problem seems to be that you take the partial derivative with respect to the variable $x_i$ while also using $i$ as the index in the sum. Try to use $j$ in the sum instead. – junjios Feb 15 '23 at 21:03
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    @junjios that would give me $\sum_{j} u_{x_jx_i}u_{x_j}+u_{x_j}u_{x_jx_i} = \sum_{j} 2u_{x_jx_i}u_{x_j}$ as the derivative? – MrStormy83 Feb 15 '23 at 21:09
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    Yes, that's right. Note this is also consistent with the multidimensional chain rule: $D\left(\vert Du\vert^{p-2}\right)=(p-2)\vert Du\vert^{p-3}\frac{Du}{\vert Du\vert}D^2u=(p-2)\vert Du\vert^{p-4}DuD^2u$ where $DuD^2u$ has to be understood as the matrix product of the row vector $Du$ with the square Hessian matrix $D^2u$. – junjios Feb 16 '23 at 00:04
  • @junjios that makes sense, but suppose I have $\sum_{i} \sum_{j} u_{x_jx_i}u_{x_j}u_{x_i}$ what would that be? I am trying to make sense out of this as well. – MrStormy83 Feb 16 '23 at 00:28
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    That is the scalar that you obtain by computing $DuD^2u(Du)^T$. – junjios Feb 16 '23 at 09:50

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