I’m reading through A Physical Introduction to Suspension Dynamics and I am having trouble understanding the jump between equations $(2.5)$ and $(2.6)$ in the photo. How do the partial derivatives in $(2.5)$ lead to the functions in $(2.6)$? There is really no explanation for the series, and Batchelor does not expand it like this.
Basically, how can we go from $\frac{\partial}{\partial x}\left(\frac 1r \right)$ to $\frac{x}{r^3}$?
Note that $r^2 = x_j x_j$ (using summation convention), so \begin{align*} 2r \frac{\partial r}{\partial x_i}&= 2x_j \frac{\partial x_j}{\partial x_i} \\ r\frac{\partial r}{\partial x_i} &= x_j \delta_{ij} \\ \frac{\partial r}{\partial x_i} &= \frac{x_i}{r}. \end{align*} Thus, \begin{align*} \frac{\partial}{\partial x_i}\left(\frac 1r \right) &= -\frac{1}{r^2} \frac{\partial r}{\partial x_i} \\ &= -\frac{1}{r^2} \cdot\frac{x_i}{r} \\ &= -\frac{x_i}{r^3}, \end{align*} as required. Similarly, \begin{align*} \frac{\partial^2}{\partial x_i \partial x_j}\left(\frac 1r \right) &= \frac{\partial }{\partial x_i} \left(-\frac{x_j}{r^3}\right) \\ &= -\left(\frac{\partial x_j}{\partial x_i} \frac{1}{r^3} + x_j \frac{\partial}{\partial x_i} \left(\frac{1}{r^3}\right)\right) \\ &= -\left(\frac{\delta_{ij}}{r^3} - \frac{3 x_j}{r^4} \frac{\partial r}{\partial x_i}\right) \\ &= -\left(\frac{\delta_{ij}}{r^3} - \frac{3 x_j}{r^4} \cdot \frac{x_i}{r} \right) \\ &= -\left(\frac{\delta_{ij}}{r^3} - \frac{3 x_i x_j}{r^5}\right). \\ \end{align*}