CM2 Prac Question 7.5

[Laura]

Hi all,
Would you be able to further explain how to obtain the differentiation of the RHS of the equation?

I'm unsure why the summation term for i for both terms disappears.

Thanks!

 

[Steve Hales]

In a more general case let's consider:
\[I = \sum_{i=1}^{n}\sum_{j=1}^{n}x_i x_j\]
When taking partial derivatives with respect to \(x_k\) we assume that all other \(x\)'s remain constant. So the only terms that will have a non-zero value are those which involve \(x_k\) specifically, ie:
\begin{align}
\frac{\partial I}{\partial x_k} & = \sum_{i=1}^{n}\sum_{j=1}^{n}\frac{\partial x_i x_j}{\partial x_k}\ \\
& = \underbrace{\sum_{i=1}^{n}x_i}_{\rm{when}\: j=k} + \underbrace{\sum_{j=1}^{n}x_j}_{\rm{when}\: i=k} \\
& = 2\sum_{i=1}^{n}x_i \\
\end{align}
Please let me know if that's not what you're asking about.