Why is $\Delta (R_i^2)$ equal to $2R_i \Delta (R_i)$

Question

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?

Answer 1

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 $0$), $$\Delta(R_i^2) \approx 2 R_i \Delta R_i^2$$

Answer 2

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 ...

$$\Delta x^2(t) \approx 2x(t) \Delta x(t) $$