I am an undergrad physics student working through 4ed of Griffith's Electrodynamic book. This question is purely mathematical, however. Do not fret.
In Griffith's (pg. 198), these equivalences are stated:
$$\frac{1}{2}\Delta(\textbf{D}\cdot\textbf{E}) = \frac{1}{2}\Delta(\epsilon{E}^2) = \epsilon(\Delta\textbf{E})\cdot\textbf{E} = (\Delta\textbf{D})\cdot\textbf{E}$$
Where $\textbf{D} = \epsilon\textbf{E}$, bold indicates a vector, and $\Delta$ indicates the change in a quantity. If you are curious, going from the right most quantity to the left most quantity allows one to write nicely the expression for energy associated with bringing free charges into a linear dielectric material.
I do not understand how to get from the 2nd (from the left) quantity to the next quantity. Or, how to get from one to the other--the order doesn't really matter. I remember showing that these two quantities are equal by integrating them w.r.t. something and showing that the integrands (the quantities) then must be equal. But, I seem to have lost these notes and cannot retrace my steps. I also am hoping that there is a much simpler way to see this equality. Lastly, are there fundemental things to know when manipulating quantities with $\Delta$ in them? I have tried explicitly writing out what $\Delta{quantity}$ represents , to no avail.
Thanks!
If $\Delta$ is just change, then there is a simple heuristic argument for this. Allow $\Delta E=E’-E$. Notice that we can factor out the epsilon as it is a constant. $$\begin{align}\frac{\varepsilon}{2}\Delta(E\cdot E)&=\frac{\varepsilon}{2}(E’\cdot E’-E\cdot E)\\&=\frac{\varepsilon}{2}(E’\cdot(E’-E)+(E’-E)\cdot E)\\&=\frac{\varepsilon}{2}(E\cdot(E’-E)+(E’-E)\cdot E)+\frac{\varepsilon}{2}(E’-E)^2\\&=\varepsilon(\Delta E)\cdot E+\mathcal{O}((\Delta E)^2)\end{align}$$
Given the context, exact equality is not what is being offered here. Rather, for the purpose of first order approximation, infinitesimal analysis or limit-taking, we have shown the first and second order terms of this change. If you were for example taking an integral with respect to this (considered as infinitesimal), the second order term would vanish.