Differentiate to get error on weighted mean (summation fraction) I need help differentiating a fraction containing two summations.
Given the weighted mean
$
\bar{x} = \frac{\sum\limits_{i=1}^{\rm N} x_{i}/\sigma_{i}^{2}}{\sum\limits_{i=1}^{\rm N} 1/\sigma_{i}^{2}}$
where $x_{i}$ and $\sigma_{i}$ are uncorrelated.
The error propagation formula
$\sigma_{f}^2 = \sigma_{f}^2 \left(\frac{\delta f }{\delta x}\right)^2 +  \sigma_{f}^2 \left(\frac{\delta f }{\delta y}\right)^2 + ...$
yields the error on $\bar{x}$ as:
$\sigma_{\bar{x}}^2 = \sum\limits_{i=1}^{\rm N} \left(\frac{\delta \bar{x} }{\delta x_{i}}\right)^2  \sigma_{i}^2$
The literature* solves the derivative as:
$\frac{\delta \bar{x} }{\delta x_{i}} = \frac{\delta }{\delta x_{i}} \frac{\sum\limits_{i=1}^{\rm N} x_{i}/\sigma_{i}^2}{\sum\limits_{i=1}^{\rm N} 1/\sigma_{i}^2}
 = \frac{1/\sigma_{i}^2}{\sum\limits_{i=1}^{\rm N} 1/\sigma_{i}^2}$
I cannot figure out how this result is achieved. By my working, if $\sigma_{i}$ can be treated as a constant, then $\frac{\delta \bar{x} }{\delta x_{i}}=1$.
Am I missing something? I have tried the Quotient rule but I don't see how it applies in this case.
Eqn. 4.19, Bevington, Data Reduction and Error Analysis for the Physical Sciences
http://astro.cornell.edu/academics/courses/astro3310/Books/Bevington_opt.pdf
 A: Note that:
$\frac{\delta }{\delta x_{i}} \frac{\sum\limits_{i=1}^{\rm N} x_{i}/\sigma_{i}^2}{\sum\limits_{i=1}^{\rm N} 1/\sigma_{i}^2}
= \frac{1}{\sum\limits_{i=1}^{\rm N} 1/\sigma_{i}^2}\frac{\delta }{\delta x_{i}} \sum\limits_{i=1}^{\rm N} x_{i}/\sigma_{i}^2
~~~\mathbf{\neq}~~~
\frac{1}{\sum\limits_{i=1}^{\rm N} 1/\sigma_{i}^2}\sum\limits_{i=1}^{\rm N}\frac{\delta }{\delta x_{i}}(x_i/\sigma_i^2)
$
It's very, very easy to mistake the inequality for an equality because the same letter $i$ in the leftmost and central terms is used to denote three different index variables, inside and outside of the two summations: there's a very high risk of confusing two different things both called $i$ when the $\frac{\delta }{\delta x_{i}}$ operator is "brought inside" the summation in the rightmost term. Using for the three different indices  three different names $h,i,j$ (a highly recommended "safety practice") yields a slightly uglier but far less misleading:
$\frac{1}{\sum\limits_{i=1}^{\rm N} 1/\sigma_{i}^2}\frac{\delta }{\delta x_{i}} \sum\limits_{i=1}^{\rm N} x_{i}/\sigma_{i}^2
=
\frac{1}{\sum\limits_{h=1}^{\rm N} 1/\sigma_{h}^2}\frac{\delta }{\delta x_{i}} \sum\limits_{j=1}^{\rm N} x_{j}/\sigma_{j}^2
=
\frac{1}{\sum\limits_{h=1}^{\rm N} 1/\sigma_{h}^2} \sum\limits_{j=1}^{\rm N} \frac{\delta }{\delta x_{i}} (x_{j}/\sigma_{j}^2)
=
\frac{1}{\sum\limits_{h=1}^{\rm N} 1/\sigma_{h}^2} (1/\sigma_i^2)$
since $\frac{\delta }{\delta x_{i}} (x_{j}/\sigma_{j}^2)=0$ if $i\neq j$.
A: It is not necessary to apply the quotient rule, since the function is linear in each $x_i, 1\leq i\leq N$. Let's denote the constant
\begin{align*}
\sum_{i=1}^N\frac{1}{\sigma_i^2}=C
\end{align*}
The function can then be written for $1\leq i\leq N$
\begin{align*}
\overline{x}&=\frac{1}{C}\sum_{j=1}^N\frac{x_j}{\sigma_j^2}\\
&=\frac{1}{C\sigma_i^2}x_i+\frac{1}{C}\sum_{{j=1}\atop{j\ne i}}^N\frac{x_j}{\sigma_j^2}
\end{align*}
which is of the form 
\begin{align*}
\overline{x}(x_i)=a x_i + b\qquad\qquad a,b \quad\text{const.}
\end{align*}

The derivative $\frac{\partial}{\partial x_i}\overline{x}$ is therefore
\begin{align*}
\frac{\partial}{\partial x_i}\overline{x}&=\frac{\partial }{\partial x_i}\left(
\frac{1}{C\sigma_i^2}x_i+\frac{1}{C}\sum_{{j=1}\atop{j\ne i}}^N\frac{x_j}{\sigma_j^2}\right)\\
&=\frac{1}{C\sigma_i^2}\qquad\qquad\qquad\qquad\qquad\qquad 1\leq i\leq N
\end{align*}

Note: We have to use a different index variable $j$ for summation to avoid conflicts with the variable $x_i$.


Add-on: [2017-01-22] Some remarks to the index notation with respect to OPs comments.
It seems that Eqn 4.18 in the referred book causes some troubles due to the index notation.
  \begin{align*}
\frac{\partial}{\partial x_i}\frac{\sum\left(x_i/\sigma_i^2\right)}{\sum\left(1/\sigma_i^2\right)}
=\frac{1/\sigma_i^2}{\sum\left(1/\sigma_i^2\right)}\tag{1}
\end{align*}

At first we take a look at the left hand side of (1). The expression can be equivalently written as
\begin{align*}
\frac{\partial}{\partial x_i}\frac{\sum\left(x_i/\sigma_i^2\right)}{\sum\left(1/\sigma_i^2\right)}
=\frac{\partial}{\partial x_i}\frac{\sum\left(x_j/\sigma_j^2\right)}{\sum\left(1/\sigma_k^2\right)}
=\frac{\partial}{\partial x_i}\frac{\sum_{j=1}^{N}\left(x_j/\sigma_j^2\right)}{\sum_{k=1}^{N}\left(1/\sigma_k^2\right)}\tag{2}
\end{align*}

When looking at $x_i$ in 
  \begin{align*}
\sum\left(x_i/\sigma_i^2\right)\tag{3}
\end{align*}
  the index $i$ is a so-called bound variable. This means
  
  
*
  
*the scope of $i$ in (3)  belongs to the $\Sigma$ symbol and is limited by the parenthesis
  
  
  $$\sum\color{blue}{\left(\right.}x_i/\sigma_i^2\color{blue}{\left.\right)}$$
  
  
*
  
*any usage of $i$ somewhere else means another, different symbol
  
*the symbol $i$ in the derivative operator $\frac{\partial}{\partial x_i}$ has nothing to do with the symbol $i$ in (3)
  
*the symbol $i$ in the sum $\sum\left(1/\sigma_i^2\right)$ has nothing to do with $i$ in (3).
Hint: Since the indices $i$ in the two sums and in the derivative operator have nothing in common, it is usually more convenient to use distinct variable names, e.g. $i,j,k$ as we can see in (2).

The expression $\frac{\partial}{\partial x_i}$ stands for the derivative of any $x_i$ arbitrarily, fixed chosen from $1\leq i \leq N$.
Albeit the notation the authors use is mathematically correct, it should be avoided as overloading of the same symbol makes the text harder to read for the less experienced.
