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What will be the result of derivative below. Also if possible kindly provide the link of any site that explains how to calculate derivatives of summations. Here a,b,c and $\sigma$ are scalar constants. For $i \in N$

\begin{align} \dfrac{\partial}{\partial p_{i}}\sum_{k=1}^N\dfrac{a}{\sigma^{2}+\sum_{n=1,n\neq k}^N p_{n}b +p_{k}c} \end{align}

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At first we recall the differential operator $\frac{\partial}{\partial p_i}$ is linear. So, the following is valid \begin{align*} \frac{\partial}{\partial p_i}\sum_{j}\alpha_j f(p_j) =\sum_{j}\alpha_j \frac{\partial}{\partial p_i}f(p_j) =\alpha_i\frac{\partial}{\partial p_i}f(p_i) \end{align*} Note that all summands $i\ne j$ cancel, since $p_j$ is treated as constant for $i\ne j$ and \begin{align*} \frac{\partial}{\partial p_i}f(p_j)=0\qquad\qquad i\ne j \end{align*}

Since we focus at the variable $p_i$ the summands have the structure \begin{align*} \frac{a}{\sigma^{2}+\sum_{n=1,n\neq k}^N p_{n}b + p_{k}c}=\frac{A}{B+C p_i}\tag{1} \end{align*} with $A,B,C$ constants. Differentiation yields \begin{align*} \frac{\partial}{\partial p_i}\left(\frac{A}{B+C p_i}\right)=-\frac{AC}{\left(B+C p_i\right)^2}\tag{2} \end{align*}

In the following we use the linearity of the differential operator and conveniently split the sums according to occurrences of $p_i$. We obtain \begin{align*} \frac{\partial}{\partial p_{i}}&\sum_{k=1}^N\frac{a}{\sigma^{2}+\sum_{n=1,n\neq k}^N p_{n}b +p_{k}c}\\ &=\frac{\partial}{\partial p_i}\sum_{{k=1}\atop{k\ne i}}^N \frac{a}{\underbrace{\sigma^{2} +\displaystyle{\sum_{{1\leq n\leq N}\atop{n\ne k, n\ne i}}p_{n}b} +p_{k}c}_B+p_ib} +\frac{\partial}{\partial p_i}\left(\frac{a}{ \sigma^{2}+\displaystyle{\sum_{{1\leq n\leq N}\atop{n\ne i}} p_{n}b} +p_{i}c}\right)\tag{3}\\ &=\sum_{{k=1}\atop{k\ne i}}^N\frac{\partial}{\partial p_i}\left( \frac{a}{\sigma^{2} +\displaystyle{\sum_{{1\leq n\leq N}\atop{n\ne k, n\ne i}}p_{n}b} +p_{k}c+p_ib}\right) +\frac{\partial}{\partial p_i}\left(\frac{a}{ \sigma^{2}+\displaystyle{\sum_{{1\leq n\leq N}\atop{n\ne i}} p_{n}b} +p_{i}c}\right)\tag{4}\\ &=-\sum_{{k=1}\atop{k\ne i}}^N \frac{ab}{\left(\sigma^{2} +\displaystyle{\sum_{{1\leq n\leq N}\atop{n\ne k, n\ne i}}p_{n}b} +p_{k}c+p_ib\right)^2} -\frac{ac}{\left( \sigma^{2}+\displaystyle{\sum_{{1\leq n\leq N}\atop{n\ne i}} p_{n}b} +p_{i}c\right)^2}\\ &=-\sum_{{k=1}\atop{k\ne i}}^N \frac{ab}{\left(\sigma^{2} +\displaystyle{\sum_{{1\leq n\leq N}\atop{n\ne k}}p_{n}b} +p_{k}c\right)^2} -\frac{ac}{\left( \sigma^{2}+\displaystyle{\sum_{{1\leq n\leq N}\atop{n\ne i}} p_{n}b} +p_{i}c\right)^2}\\ \end{align*}

Comment:

  • In (3) we extract in the outer sum as well as in each sum in the denominator the term with index $n=i$. Observe the structural similarity of both summands with (2), whereby in the first term the constant part of the denominator is marked as $B$.

  • In (4) we use the linearity of the differential operator and can now apply the differentiation according to (1).

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    Is there any tool like mathematica or matlab in which I can solve such problems? – 2017-02-09
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    @ZainAli: Yes, you can type in the whole expression in tools like Mathematica and they will solve it. But the benefit in this case is small. The real challenge here is to think about which parts are constant, which parts contain the variable $p_i$ and how to conveniently treat them, to make differentiation feasible. In fact, I've used WA as verification for the expression $\frac{\partial}{\partial p_i}\frac{A}{B+Cp_i}$ and nothing else. It is a good training to think about this somewhat complicated expression. This will help you to solve similar problems more and more routinely in the future. – 2017-02-09