Why is $\Delta (R_i^2)$ equal to $2R_i \Delta (R_i)$ So when deriving the equation for charge, I reached a point where
$\frac{z\eta}{4\epsilon} \sum_{i=1}^N \frac{\Delta (R_i^2)}{(z^2+R_i^2)^\frac{3}{2}}$
But my teacher explains that 
$\Delta (R_i^2)$
can be rearranged to 
$2R_i \Delta (R_i)$ 
but he cannot explain why.
He talks about taking a differential but then it becomes $dr$ and the equation turns into an integral in terms of R, so doesn't that just counter his claim?
 A: Unpacking the meaning of $\Delta$ as ...
$$ \Delta f(t)\equiv   f(t+\Delta t)-f(t) $$
We can write ...
$$ \begin{array} \\ \Delta x^2(t)
\\=x^2(t+\Delta t) - x^2(t) \\=(x(t+\Delta t)+x(t))(x(t+\Delta t)-x(t))
\\=\left [\Delta x(t) +2x(t) \right ] \Delta x(t)    \end{array} $$
So provided that $\Delta x(t) << 2x(t)$ we have ...
$$\Delta x^2(t)  \approx 2x(t) \Delta x(t)
$$
A: When taking differentials, $$x = f(t) \implies \mathrm dx = 
\frac{\mathrm df}{\mathrm dt} \, \mathrm dt$$
Let's apply the above statement. Let $u_i = R_i^2$. As $\Delta u_i$ approaches $0$, it becomes a differential $\mathrm du_i$. So we have
$$u_i = f(R_i) = R_i^2 \implies \mathrm du_i = 
\frac{\mathrm df}{\mathrm dR_i} \, \mathrm dR_i$$
Note that $\frac{\mathrm df}{\mathrm dR_i} = \frac{\mathrm d}{\mathrm dR_i}[R_i^2] = 2R_i$. Hence 
$$\mathrm du_i = 
(2R_i) \, \mathrm dR_i$$
And, remembering that $u_i = R_i^2$:
$$\mathrm d(R_i^2) = 2 R_i \, \mathrm dR_i$$
Which means that for small changes in $R_i^2$ (i.e. as we take the limit approaching $0$), $$\Delta(R_i^2) \approx 2 R_i \Delta R_i^2$$
