# Derivative of the $l_p$ norm [closed]

The $l_p$-norm of the vector $\mathbf{x}$ is defined as $$\Vert \mathbf{x} \Vert_p = \left(\sum_i |x_i|^p\right)^{1/p}$$ I want to calculate the following derivative. Any hint is appreciated. $$\frac{\partial}{\partial \mathbf{x}}\Vert \mathbf{x} \Vert_p$$

Thanks.

## closed as off-topic by user223391, Empty, 6005, Claude Leibovici, user91500Oct 16 '15 at 8:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Empty, 6005, Claude Leibovici, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.

• Can we define this derivative as: $\frac{\partial}{\partial V}\|X\|_{p}=\lim_{t \to 0} \frac{\|X+tV\|_{p}-\|X\|_{p}}{t}$ – muratguner Oct 16 '15 at 2:41
• Well, despite being closed, this question helped me. +1. – Shraddheya Shendre Feb 18 '17 at 21:03
• Why is this question closed and how is it missing the context or is off-topic? – Tolga Birdal Apr 22 at 8:52

For $j = 1, 2, \ldots, N$, by chain rule, we have $$\partial_j \|\mathbf{x}\|_p = \frac{1}{p} \left(\sum_i \vert x_i \vert^p\right)^{\frac{1}{p}-1} \cdot p \vert x_j \vert^{p-1} sgn(x_j) = \left(\frac{\vert x_j \vert}{\|\mathbf{x}\|_p}\right)^{p-1} sgn(x_j)$$
For all $j\in\lbrace1,\,\dots,\,n\rbrace$, \begin{align*} \frac{\partial}{\partial x_j}{||\mathbf{x}||}_{p} &= \frac{\partial}{\partial x_j} \left( \sum_{i=1}^{n} |x_i|^p \right)^{1/p}\\ &= \frac{1}{p} \left( \sum_{i=1}^{n} |x_i|^p \right)^{\left(1/p\right)-1} \frac{\partial}{\partial x_j} \left(\sum_{i=1}^{n} |x_i|^p\right)\\ &= \frac{1}{p} \left( \sum_{i=1}^{n} |x_i|^p \right)^{\frac{1-p}{p}} \sum_{i=1}^{n} p|x_i|^{p-1} \frac{\partial}{\partial x_j} |x_i|\\ &= {\left[\left( \sum_{i=1}^{n} |x_i|^p \right)^{\frac{1}{p}}\right]}^{1-p} \sum_{i=1}^{n} |x_i|^{p-1} \delta_{ij}\frac{x_i}{|x_i|}\\ &= {||\mathbf{x}||}_{p}^{1-p} \cdot |x_j|^{p-1} \frac{x_j}{|x_j|}\\ &= \frac{x_j |x_j|^{p-2}}{{||\mathbf{x}||}_{p}^{p-1}} \end{align*}