Given a function $f_i (x_1,\dots, x_N)$ that is summed over, how do I find the correct partial derivative? \begin{equation} f_i = \sum _{j=1}^Nc_{ij}x_ix_j \end{equation} Then, \begin{equation} \frac{\partial}{\partial x_k}f_i = \frac{\partial}{\partial x_k}\sum _{j=1}^Nc_{ij}x_ix_j \end{equation} I assume the derivative is distributive. \begin{equation} \frac{\partial}{\partial x_k}f_i =\sum _{j=1}^N \frac{\partial}{\partial x_k} c_{ij}x_ix_j \end{equation} I then have, \begin{equation} \frac{\partial}{\partial x_k}f_i =\sum _{j=1}^N c_{ik}x_i \delta_{jk} \qquad \qquad \qquad * \end{equation} But I can also write, \begin{equation} \frac{\partial}{\partial x_k}f_i= 2c_{kk}x_k + \sum _{j\neq k}^N c_{ij}x_j\qquad \qquad \qquad $ \end{equation} So is * or $ the correct result?
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- is right if $k\not=i$, but you had better write it as $c_{ij}x_i$. Now there's the special case $k=i$ to do.. In the second one (don't use the dollar sign), by the way, you've changed an $i$ into a $k$ for reasons I can't follow.
– ancient mathematician Apr 01 '17 at 12:34
2 Answers
The correct answer depends on the fact whether $k=i$ or $k\ne i$. It is convenient to separate these two cases.
$(k\ne i)$: We obtain \begin{align*} \frac{\partial }{\partial x_k}f_i&=\frac{\partial }{\partial x_k}\sum_{j=1}^Nc_{ij}x_ix_j\\ &=\frac{\partial }{\partial x_k}\sum_{{j=1}\atop{j\ne k}}^{N}c_{ij}x_ix_j+\frac{\partial }{\partial x_k}c_{ik}x_ix_k\\ &=0+c_{ik}x_i\\ &=c_{ik}x_i \end{align*}
$(k=i)$: We obtain \begin{align*} \frac{\partial }{\partial x_k}f_k&=\frac{\partial }{\partial x_k}\sum_{j=1}^Nc_{kj}x_kx_j\\ &=\frac{\partial }{\partial x_k}\sum_{{j=1}\atop{j\ne k}}^{N}c_{kj}x_kx_j+\frac{\partial }{\partial x_k}c_{kk}x_k^2\\ &=\sum_{{j=1}\atop{j\ne k}}^{N}c_{kj}x_j+2c_{kk}x_k \end{align*}
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \partiald{f_{i}}{x_{k}} & = \sum _{j = 1}^{N}c_{ij} \pars{\partiald{x_{i}}{x_{k}}\,x_{j} + x_{i}\,\partiald{x_{j}}{x_{k}}} = \sum _{j = 1}^{N}c_{ij} \pars{\delta_{ik}\,x_{j} + x_{i}\,\delta_{jk}} = \bbx{\ds{\delta_{ik}\sum_{j = 1}^{N}c_{ij}x_{j} + c_{ik}x_{i}}} \end{align}
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