# 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.

• Can we define this derivative as: $\frac{\partial}{\partial V}\|X\|_{p}=\lim_{t \to 0} \frac{\|X+tV\|_{p}-\|X\|_{p}}{t}$ Oct 16, 2015 at 2:41
• Well, despite being closed, this question helped me. +1. Feb 18, 2017 at 21:03
• Why is this question closed and how is it missing the context or is off-topic? Apr 22, 2019 at 8:52

For $$j = 1, 2, \ldots, N$$, by chain rule, we have $$\partial_j \|\mathbf{x}\|_p = \frac{1}{p} \left(\sum_i \vert x_i \vert^p\right)^{\frac{1}{p}-1} \cdot p \vert x_j \vert^{p-1} \operatorname{sgn}(x_j) = \left(\frac{\vert x_j \vert}{\|\mathbf{x}\|_p}\right)^{p-1} \operatorname{sgn}(x_j)$$
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*}
• What is $\delta_{ij}$? Mar 11, 2020 at 13:41