What is the explanation for the confusion in Griffiths' energy derivation?

[albega]

In Griffiths section 4.4.3, he derives the energy in a dielectric system as
W=0.5∫D.Edτ.
Part of the derivation involves the relation
0.5Δ(D.E)=0.5Δ(εE²)=ε(ΔE).E=(ΔD).E
for infinitesimal increments, using D=εE. Now the part 0.5Δ(εE²)=ε(ΔE).E loses me so I was wondering if anybody could explain it. Thanks.

 

[AlephZero]

If you can't "see" it, write out the components. I'm not going to do it in full but this shows what's happening:

$\frac 1 2 \nabla(\mathbf{D}.\mathbf{E}) = \frac 1 2 ( \frac{\partial}{\partial x}(D_x.E_x) \cdots) = \frac 1 2 \epsilon ( \frac{\partial}{\partial x}(E_x.E_x) \cdots) = \frac 1 2 \epsilon(2\frac{\partial E_x}{\partial_x}E_x \cdots)$ etc.

 

[albega]

If you can't "see" it, write out the components. I'm not going to do it in full but this shows what's happening:

$\frac 1 2 \nabla(\mathbf{D}.\mathbf{E}) = \frac 1 2 ( \frac{\partial}{\partial x}(D_x.E_x) \cdots) = \frac 1 2 \epsilon ( \frac{\partial}{\partial x}(E_x.E_x) \cdots) = \frac 1 2 \epsilon(2\frac{\partial E_x}{\partial_x}E_x \cdots)$ etc.

Hmm I was using deltas not nablas...

Anyway I think I've worked it out. Effectively we have d(E²) and because d(E²)/dE=2E, d(E²)=2EdE. The fact the book used deltas blinded me from this...

Only issue now is understanding why εEΔE=εE.ΔE - how do I know they have the same direction? Note E is just the field due to some free charge distribution ρ_f and ΔE is just the change in E due to the addition of an amount Δρ_f of the free charge.