TITLE: Why is $\Delta (R_i^2)$ equal to $2R_i \Delta (R_i)$
QUESTION [1 upvotes]: 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?

REPLY [3 votes]: 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 ...
$$\Delta x^2(t)  \approx 2x(t) \Delta x(t)
$$
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